Create the page "1000" on this wiki! See also the search results found.

• 2016 AMC 10A Problems/Problem 2
  For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>? We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>:
  1 KB (190 words) - 10:58, 16 June 2023
• ASIA TEAM Problems/Problem 1
  ...one more than the order of <math>2n</math>, and the answer is <math>\frac{1000}{2}=\boxed{500}</math>.
  227 bytes (40 words) - 05:42, 16 February 2024
• Kansas mathematics competitions
  ...ubjects, and a team competition, Mathletics. The 2006 competition had over 1000 participants. [http://contest.kcatm.org/ website]
  3 KB (473 words) - 16:11, 16 June 2020
• Chemistry competitions
  ...xam administered in mid or late April by ACS Local Sections. Approximately 1000 students qualify from local exams to take the USNCO.
  2 KB (258 words) - 19:31, 8 March 2023
• Princeton University Mathematics Competition
  Princeton NJ 08544-1000 USA
  2 KB (295 words) - 23:19, 5 January 2019
• 2006 Alabama ARML TST Problems/Problem 8
  A bored [[mathematician]] has his computer calculate 1000 consecutive terms in the [[Fibonacci sequence]]. He notes that the smallest ...ce repeats every 16, and for every 8 numbers, there is one 0. <math>\dfrac{1000}{8}=125</math>, but we don't count the first one. <math>\boxed{124}</math>
  605 bytes (78 words) - 16:43, 17 April 2008
• Factorial
  * <math>1000! = 402387260077093773543702433923003985719374864210714632543799910429938512
  10 KB (809 words) - 16:40, 17 March 2024
• Divisibility rules
  Rule 3: Works for <math>1 \leq N \leq 1000</math>. Let <math>K = 3N</math>. If <math>K</math> is odd add 39 to <math
  8 KB (1,315 words) - 18:18, 2 March 2024
• Constructive counting
  ...i </math> is [[even integer | even]]. How many snakelike integers between 1000 and 9999 have four distinct digits?'' ...wever, this breaks our requirement that our integers must be between <math>1000</math> and <math>9999</math>, so there are no four-digit snakelike integers
  12 KB (1,896 words) - 23:55, 27 December 2023
• Michigan Mathematics Prize Competition
  ...parts. Part I is a 40 question, 100 minute multiple choice test. The top 1000 finishers of this round are selected to take Part II, which is a harder 5-q
  981 bytes (143 words) - 12:06, 16 January 2012
• Function
  n-3 & \mbox{if }n\ge 1000 \\ f(f(n+5)) & \mbox{if }n<1000
  10 KB (1,761 words) - 03:16, 12 May 2023
• Composite number
  Extra: A list of composite numbers from 1 to 1000: ...78 979 980 981 982 984 985 986 987 988 989 990 992 993 994 995 996 998 999 1000
  6 KB (350 words) - 12:58, 26 September 2023
• Base numbers
  ...rs to represent certain values (e.g. I=1, V=5, X=10, L=50, C=100, D=500, M=1000). Imagine how difficult it would be to multiply LXV by MDII! That's why t
  4 KB (547 words) - 17:23, 30 December 2020
• Fibonacci sequence
  # Except for the first two terms, each term of the sequence <math>1000, x, 1000 - x,\ldots</math> is obtained by subtracting the preceding term from the on
  6 KB (957 words) - 23:49, 7 March 2024
• Writing scholarships
  ...rm.com/scholarship/ Rizio Lipinsky Lawyer Scholarship] USD 10,000. Essay (<1000 words) about why you’re inspired to become an attorney ...ww.seniorcare.com/scholarship/ SeniorCare.com Aging Matters Scholarship] (<1000 word essay)
  7 KB (1,039 words) - 18:45, 18 January 2024
• National Scholarships
  * [[Gen and Kelly Tanabe Scholarship]] of <dollar/>1000 [http://www.gkscholarship.com/ website]
  4 KB (538 words) - 00:48, 28 January 2024
• Logarithm
  ...<math>b</math> satisfy the condition <math>\log_2(\log_{2^a}(\log_{2^b}(2^{1000})))=0.</math> Find the sum of all possible values of <math>a+b</math>.
  4 KB (680 words) - 12:54, 16 October 2023
• Rational number
  ...a finite [[decimal expansion]] is rational (say, <math>12.345=\frac{12345}{1000}</math>)
  1 KB (207 words) - 15:51, 25 August 2022
• Floor function
  * How many of the first 1000 [[positive integer]]s can be expressed in the form
  3 KB (508 words) - 21:05, 26 February 2024
• Pascal's triangle
  ...>{(0000)}_{2} = 0</math>th, <math>{(0001)}_{2} = 2^0 = 1</math>st, <math>{(1000)}_{2} = 2^3 = 8</math>th, and <math>{(1001)}_{2} = 2^3+2^0 = 9</math>th col
  5 KB (838 words) - 17:20, 3 January 2023
• Modular arithmetic/Introduction
  What is the last digit of <math>(...((7)^7)^7)...)^7</math> if there are 1000 7s as exponents and only one 7 in the middle? ...se 7 has a pattern of repetitive period 4 for the units digit. <math>(1)^{1000}</math> is simply 1, so therefore <math>7^1=7</math>, which really is the l
  15 KB (2,396 words) - 20:24, 21 February 2024
• 2006 AIME I Problems
  ...ots99!100!. </math> Find the remainder when <math> N </math> is divided by 1000. ...be constructed. What is the remainder when <math> T </math> is divided by 1000?
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  ...ill apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form a
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 13
  ...^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]]. ...rfect square. Hence <math>k</math> must be even. In particular, as <math>n<1000</math>, we have five choices for <math>k</math>, namely <math>k=0,2,4,6,8</
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 11
  ...constructed. What is the [[remainder]] when <math> T </math> is divided by 1000?
  3 KB (436 words) - 05:40, 4 November 2022
• 2006 AIME I Problems/Problem 4
  ...100!. </math> Find the remainder when <math> N </math> is divided by <math>1000</math>.
  2 KB (278 words) - 08:33, 4 November 2022
• 2006 AIME I Problems/Problem 3
  ...be <math>N</math> with the first digit deleted. Now, we know that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1, ...six numbers was possible thanks to AIME problems having answers less than 1000).
  4 KB (622 words) - 03:53, 10 December 2022
• 2005 AMC 12A Problems
  A line passes through <math>A\ (1,1)</math> and <math>B\ (100,1000)</math>. How many other points with integer coordinates are on the line and ...numbers less than 1000. How many prime-looking numbers are there less than 1000?
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  A scout troop buys <math>1000</math> candy bars at a price of five for <math>2</math> dollars. They sell
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 25
  so <math>999 = \max(a_1, a_2) \geq 1000</math>, a contradiction. Hence <math>(a_n)</math> completes at <math>i</mat
  5 KB (924 words) - 12:02, 15 June 2022
• 2005 AMC 12B Problems/Problem 1
  A scout troop buys <math>1000</math> candy bars at a price of five for <math>2</math> dollars. They sell \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\
  1 KB (179 words) - 13:53, 14 December 2021
• 2005 AIME II Problems
  For how many positive integers <math> n </math> less than or equal to 1000 is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 9
  For how many positive integers <math> n </math> less than or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true f ...on is equivalent to asking for how many [[positive integer]]s <math>n \leq 1000</math> we have that <math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\lef
  6 KB (1,154 words) - 03:30, 11 January 2024
• 2005 AIME I Problems
  ...<math> S. </math> Find the remainder when <math> 10K </math> is divided by 1000.
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME I Problems/Problem 4
  ...ou have a LOT of time (and you've memorized all your perfect squares up to 1000). ...80</math>, and since AIME answers are nonnegative integers less than <math>1000</math>, we don't have to check any higher <math>n</math>. Also, we know tha
  8 KB (1,248 words) - 11:43, 16 August 2022
• 2005 AIME I Problems/Problem 14
  ...S. </math> Find the remainder when <math> 10K </math> is divided by <math>1000</math>. ...{44^2}{10}</math>. The remainder when <math>10K</math> is divided by <math>1000</math> is <math>936</math>.
  3 KB (561 words) - 14:11, 18 February 2018
• 2005 AIME I Problems/Problem 15
  <cmath>2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000</cmath> <cmath>1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}</cmath>
  5 KB (906 words) - 23:15, 6 January 2024
• 2004 AIME I Problems/Problem 1
  ...underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}} </math><math>= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in ...be greater than 9), <math>n</math> is equal to <math>(d)+10(d+1)+100(d+2)+1000(d+3)</math> or <math>1111d +3210</math>. Now we try this number for <math>d
  2 KB (374 words) - 14:53, 27 December 2019
• 2004 AIME I Problems/Problem 6
  ...i </math> is [[even integer | even]]. How many snakelike integers between 1000 and 9999 have four distinct digits? ...the leading digit will be a zero, which is bad because all numbers between 1000 and 9999 have nonzero leading digits. So, we need to select our 4 digits on
  3 KB (562 words) - 18:12, 4 March 2022
• 2004 AIME I Problems/Problem 8
  ...are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there? Note that <math>n = 1000 = 2^{3}5^{3}.</math>
  4 KB (620 words) - 21:26, 5 June 2021
• 2004 AIME I Problems
  ...} </math> if <math> i </math> is even. How many snakelike integers between 1000 and 9999 have four distinct digits? ...are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 14
  ...777c = 7000</math>, or dividing by <math>7</math>, <math>a + 11b + 111c = 1000</math>. Then the question is asking for the number of values of <math>n = a ...is the number of multiples of <math>9</math> from <math>0</math> to <math>1000</math>, or <math>112</math>.
  11 KB (1,857 words) - 21:55, 19 June 2023
• 2004 AIME II Problems/Problem 9
  ...a_n </math> be the greatest term in this sequence that is less than <math>1000</math>. Find <math> n+a_n. </math> ...<math>n</math> such that either <math>f(n)^2\, \mathrm{or}\, f(n)f(n-1) < 1000</math>. This happens with <math>f(7)f(8) = 29 \cdot 33 = 957</math>, and th
  3 KB (538 words) - 21:33, 30 December 2023
• 2004 AIME II Problems/Problem 6
  ...l to <math>17z/9</math>. This means there are two possible solutions under 1000: 408 and 816. Trial and error can be done quickly to find the smallest poss
  6 KB (950 words) - 14:18, 15 January 2024
• 2004 AIME II Problems/Problem 5
  In order to complete a large job, <math>1000</math> workers were hired, just enough to complete the job on schedule. All ...1000</math> miles per hour and has one hour to reach its destination <math>1000</math> miles away. After <math>15</math> minutes and <math>250</math> miles
  4 KB (592 words) - 19:02, 26 September 2020
• 2004 AIME II Problems/Problem 1
  ...ind the [[remainder]] when the product <math> abcdef </math> is divided by 1000.
  2 KB (329 words) - 23:20, 4 July 2013
• 2004 AIME II Problems
  ...e. Find the remainder when the product <math> abcdef </math> is divided by 1000. In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the wo
  9 KB (1,410 words) - 05:05, 20 February 2019
• 1984 AIME Problems
  n-3 & \mbox{if }n\ge 1000 \\ f(f(n+5)) & \mbox{if }n<1000
  6 KB (933 words) - 01:15, 19 June 2022
• 1987 AIME Problems
  ...triples <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>. ...e integer]] and <math>r</math> is a positive real number less than <math>1/1000</math>. Find <math>n</math>.
  6 KB (869 words) - 15:34, 22 August 2023
• 1989 AIME Problems
  A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most fre
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1991 AIME Problems
  Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives ...\choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}</math></center>
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1992 AIME Problems
  For how many pairs of consecutive integers in <math>\{1000,1001,1002^{}_{},\ldots,2000\}</math> is no carrying required when the two i
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1993 AIME Problems
  ...rawn randomly and without replacement from the set <math>\{1, 2, 3,\ldots, 1000\}</math>. Three other numbers, <math>b_1, b_2, b_3</math>, are then drawn r
  8 KB (1,275 words) - 06:55, 2 September 2021
• 1994 AIME Problems
  ...What is the remainder when the 1994th term of the sequence is divided by 1000? ...94,\,</math> what is the remainder when <math>f(94)\,</math> is divided by 1000?
  7 KB (1,141 words) - 07:37, 7 September 2018
• 1996 AIME Problems
  ...th>. For how many positive integers <math>n</math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even
  6 KB (931 words) - 17:49, 21 December 2018
• 1997 AIME Problems
  How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonneg
  7 KB (1,098 words) - 17:08, 25 June 2020
• 1998 AIME Problems
  Except for the first two terms, each term of the sequence <math>1000, x, 1000 - x,\ldots</math> is obtained by subtracting the preceding term from the on
  7 KB (1,084 words) - 02:01, 28 November 2023
• 1999 AIME Problems
  ...whose labels divide the label on the <math>i</math>-th switch. After step 1000 has been completed, how many switches will be in position <math>A</math>?
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  ...</math> and <math>b</math> are relatively prime positive divisors of <math>1000.</math> What is the greatest integer that does not exceed <math>\frac{S}{10
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2003 AIME I Problems
  An integer between <math>1000</math> and <math>9999,</math> inclusive, is called balanced if the sum of i ...> m </math> and <math> n </math> are positive integers with <math> m + n < 1000, </math> find <math> m + n. </math>
  6 KB (965 words) - 16:36, 8 September 2019
• 2002 AIME II Problems
  ...sitive integer. Find the remainder when <math>m</math> is divided by <math>1000</math>. Find the integer that is closest to <math>1000\sum_{n=3}^{10000}\frac1{n^2-4}</math>.
  7 KB (1,177 words) - 15:42, 11 August 2023
• 2003 AIME II Problems
  ...are all different. What is the remainder when <math>N</math> is divided by 1000?
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 10
  ...0</math> to <math>1999</math>), so by complementary counting you get <math>1000-(504+27+36+1)=\boxed{432}</math> numbers.
  5 KB (855 words) - 20:26, 14 January 2023
• 1984 AIME Problems/Problem 7
  n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}</math>
  4 KB (617 words) - 18:01, 9 March 2022
• 1984 AIME Problems/Problem 12
  ...roots <math>f(x)=0</math> must have in the interval <math>-1000\leq x \leq 1000</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page In the interval <math>-1000\leq x\leq 1000</math>, there are <math>201</math> multiples of <math>10</math> and <math>2
  3 KB (588 words) - 14:37, 22 July 2020
• 1985 AIME Problems/Problem 10
  How many of the first 1000 [[positive integer]]s can be expressed in the form ...rs and so we hit <math>50 \cdot 12 = \boxed{600}</math> of the first <math>1000</math>.
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 7
  ...- t^2 = 1</math>. Then <math>s = 10, t = 3</math> and so <math>d = s^3 = 1000</math>, <math>b = t^5 = 243</math> and <math>d-b=\boxed{757}</math>.
  1 KB (222 words) - 11:04, 4 November 2022
• 1986 AIME Problems/Problem 10
  Since <math>0<100a+10b+c<1000</math>, we get the inequality <cmath>N<222(a+b+c)<N+1000</cmath>
  3 KB (565 words) - 16:51, 1 October 2023
• 1986 AIME Problems/Problem 5
  ...such that <math>n+10 \mid n^3 +100</math>, we have: <math>n+10 \mid ((n^3 +1000) - (n^3 +100) \longrightarrow n+10 \mid 900</math>. This is because of the
  2 KB (338 words) - 19:56, 15 October 2023
• 1987 AIME Problems/Problem 12
  ...r]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. ...math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>. However, we still have to make sure a sufficiently small <math>r<
  4 KB (673 words) - 19:48, 28 December 2023
• 1987 AIME Problems/Problem 7
  ...riples]] <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>. <math>1000 = 2^35^3</math> and <math>2000 = 2^45^3</math>. By [[LCM#Using prime factor
  3 KB (547 words) - 22:54, 4 April 2016
• 1988 AIME Problems/Problem 9
  ...of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4< ...is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>
  6 KB (893 words) - 08:15, 2 February 2023
• 1989 AIME Problems/Problem 11
  A sample of 121 [[integer]]s is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique [[mode]] (most ...ish to minimize or maximize <math>x</math> (in other words, <math>x \in [1,1000]</math>). Indeed, <math>D(x)</math> is symmetric about <math>x = 500.5</mat
  5 KB (851 words) - 18:01, 28 December 2022
• 1989 AIME Problems/Problem 9
  &\equiv893\pmod{1000}. &\equiv224\pmod{1000}.
  6 KB (874 words) - 15:50, 20 January 2024
• 1989 AIME Problems/Problem 3
  ...ometric series]], <math>0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}</math>. Thus <math>\frac{n}{810} = \frac{100d + To get rid of repeating decimals, we multiply the equation by 1000. We get <math>\frac{1000n}{810} = d25.d25d25...</math> We subtract the orig
  3 KB (499 words) - 22:17, 29 March 2024
• 1990 AIME Problems/Problem 3
  ...y conceivable reasoning behind this is that <math>r</math> is greater than 1000. This prompts us to look into the second case, where <math>s</math> divides
  3 KB (516 words) - 19:18, 16 April 2024
• 1991 AIME Problems/Problem 3
  Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives ...\choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}</math>
  5 KB (865 words) - 12:13, 21 May 2020
• 1992 AIME Problems/Problem 6
  For how many pairs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integer
  3 KB (455 words) - 02:03, 10 July 2021
• 1992 AIME Problems/Problem 3
  ...ath>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>. ...3}{2n+4} > .503 = \frac{503}{1000}.</cmath>Cross-multiplying, we get <math>1000(n+3) > 503(2n+4),</math> which is equivalent to <math>n < \frac{988}{6} = 1
  2 KB (251 words) - 08:05, 2 January 2024
• 1993 AIME Problems/Problem 11
  ...math>a_6=\frac{364}{729}</math>, <math>m+n = 1093 \equiv \boxed{093} \pmod{1000}</math>.
  7 KB (1,058 words) - 20:57, 22 December 2020
• 1993 AIME Problems/Problem 7
  ...rawn randomly and without replacement from the set <math>\{1, 2, 3,\ldots, 1000\}</math>. Three other numbers, <math>b_1, b_2, b_3</math>, are then drawn r There is a total of <math>P(1000,6)</math> possible ordered <math>6</math>-tuples <math>(a_1,a_2,a_3,b_1,b_2
  5 KB (772 words) - 09:04, 7 January 2022
• 1994 AIME Problems/Problem 3
  .../math> what is the remainder when <math>f(94)\,</math> is divided by <math>1000</math>?
  2 KB (252 words) - 11:12, 3 July 2023
• 1994 AIME Problems/Problem 1
  ...at is the [[remainder]] when the 1994th term of the sequence is divided by 1000? ...97-1) \cdot 3 = 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.
  946 bytes (139 words) - 21:05, 1 September 2023
• 1995 AIME Problems/Problem 2
  ...95^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>. ...^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.
  2 KB (362 words) - 00:40, 29 January 2021
• 1996 AIME Problems/Problem 15
  ...5^{\circ}</math>. The answer is <math>\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \fr
  5 KB (710 words) - 21:04, 14 September 2020
• 1996 AIME Problems/Problem 2
  ...d x. For how many positive integers <math>n</math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even ...h>n</math> must satisfy these [[inequality|inequalities]] (since <math>n < 1000</math>):
  1 KB (163 words) - 19:31, 4 July 2013
• 1997 AIME Problems/Problem 3
  ...y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}</math>, and <math>(9x-1)(9y-1000)=1000</math>. Since <math>89 < 9x-1 < 890</math>, we can use trial and error on factors of 1000. If <math>9x - 1 = 100</math>, we get a non-integer. If <math>9x - 1 = 125<
  2 KB (375 words) - 19:34, 4 August 2021
• 1997 AIME Problems/Problem 1
  How many of the integers between 1 and 1000, inclusive, can be expressed as the [[difference of squares|difference of t
  801 bytes (115 words) - 15:52, 2 March 2020
• 1998 AIME Problems/Problem 8
  Except for the first two terms, each term of the sequence <math>1000, x, 1000 - x,\ldots</math> is obtained by subtracting the preceding term from the on ...math><font color="white">aaa</font> || <math>1000 - x</math> || <math>2x - 1000</math><font color="white">a</font> || <math>2000 - 3x</math> || <math>5x -
  2 KB (354 words) - 19:37, 24 September 2023
• 1999 AIME Problems/Problem 7
  ...whose labels divide the label on the <math>i</math>-th switch. After step 1000 has been completed, how many switches will be in position <math>A</math>? The number of switches in position A is <math>1000-125-225 = \boxed{650}</math>.
  3 KB (475 words) - 13:33, 4 July 2016
• 2000 AIME I Problems/Problem 15
  ...aining, some cards have not even made one trip through yet, <math>2(1024 - 1000) = 48</math>, to be exact. Once these cards go through, 1999 will be the <m ...s initially in the deck once, in round two, you would go through all <math>1000</math> cards initially in the deck once, so on and so forth. For each round
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 14
  ...<math>\angle ACB = 80^{\circ}</math>, so the answer is <math>\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfl <math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>.
  8 KB (1,275 words) - 03:04, 27 February 2022
• 2000 AIME I Problems/Problem 11
  ...and <math>b</math> are [[relatively prime]] positive [[divisor]]s of <math>1000.</math> What is the [[floor function|greatest integer]] that does not excee Since all divisors of <math>1000 = 2^35^3</math> can be written in the form of <math>2^{m}5^{n}</math>, it f
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 9
  ...ifferent author: <math>-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.</math>
  4 KB (623 words) - 15:56, 8 May 2021
• 2000 AIME I Problems/Problem 6
  ...\sqrt{10^6} = 10^3</math>, then we can use all positive integers less than 1000 for <math>a</math> and <math>b</math>. ...square, <math>y</math> must also be a perfect square. Since <math>0 < y < (1000)^2</math>, <math>y</math> must be from <math>1^2</math> to <math>999^2</mat
  6 KB (966 words) - 21:48, 29 January 2024
• 2001 AIME I Problems/Problem 8
  Given this is an AIME problem, <math>A<1000</math>. If we look at <math>B</math> in base <math>10</math>, it must be eq
  3 KB (502 words) - 11:28, 9 December 2023
• 2001 AIME I Problems/Problem 3
  ...x</math> is a root, <math>x</math> is also a root. Thus, we pair up <math>1000</math> pairs of roots that sum to <math>\frac{1}{2}</math> to get a sum of
  2 KB (335 words) - 18:38, 9 February 2023
• 2002 AIME I Problems/Problem 7
  An equivalent statement is to note that we are looking for <math>1000 \left\{\frac{10^{859}}{7}\right\}</math>, where <math>\{x\} = x - \lfloor x
  2 KB (316 words) - 19:54, 4 July 2013
• 2003 AIME I Problems/Problem 14
  <center><math>\frac{251}{1000} \le \frac{m'}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m' < 252n \Longleftrightarrow n \le 250(4 ...ger <math>m</math> such that <math>0<\frac{m}{n}-\frac{251}{1000}<\frac{1}{1000}</math>.
  3 KB (477 words) - 14:23, 4 January 2024
• 2003 AIME I Problems/Problem 13
  .../math>'s. Find the [[remainder]] when <math> N </math> is divided by <math>1000</math>.
  4 KB (651 words) - 19:42, 7 October 2023
• 2003 AIME I Problems/Problem 12
  ...imeter]] of <math> ABCD </math> is <math> 640 </math>. Find <math> \lfloor 1000 \cos A \rfloor. </math> (The notation <math> \lfloor x \rfloor </math> mean <math>\lfloor 1000 \cos A \rfloor = \boxed{777}</math>.
  3 KB (487 words) - 22:14, 24 November 2019
• 2003 AIME I Problems/Problem 11
  ...</math> and <math> n </math> are [[positive integer]]s with <math> m + n < 1000, </math> find <math> m + n. </math>
  2 KB (284 words) - 13:42, 10 October 2020
• 2003 AIME I Problems/Problem 9
  An [[integer]] between <math>1000</math> and <math>9999</math>, inclusive, is called ''balanced'' if the sum
  4 KB (696 words) - 11:55, 10 September 2023
• 2003 AIME II Problems/Problem 13
  ...be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up, <math>342/1024</
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2003 AIME II Problems/Problem 2
  ...gits are the same. What is the remainder when <math>N</math> is divided by 1000? Therefore, the remainder when the number is divided by <math>1000</math> is <math>\boxed{120}</math>.
  1,013 bytes (162 words) - 09:00, 11 July 2023
• 2002 AIME II Problems/Problem 9
  ...ents.) Find the remainder obtained when <math>n</math> is divided by <math>1000</math>. ...th>n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}</math>.
  3 KB (404 words) - 23:07, 4 May 2024
• 2002 AIME II Problems/Problem 6
  Find the integer that is closest to <math>1000\sum_{n=3}^{10000}\frac1{n^2-4}</math>. And so, <math>1000\sum_{n=3}^{10,000} \frac{1}{n^2-4} = 1000\sum_{n=3}^{10,000} \frac{1}{4} \left( \frac{1}{n-2} - \frac{1}{n+2} \right)
  2 KB (330 words) - 05:56, 23 August 2022
• 2002 AIME II Problems/Problem 4
  ...sitive integer. Find the remainder when <math>m</math> is divided by <math>1000</math>. <math>m=361803</math>, <math>\dfrac{m}{1000}=361</math> Remainder <math>\boxed{803}</math>.
  2 KB (268 words) - 07:28, 13 September 2020
• 2005 AMC 10B Problems
  A scout troop buys <math>1000</math> candy bars at a price of five for &#36;<math>2</math>. They sell all
  12 KB (1,874 words) - 21:20, 23 December 2020
• Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions
  ...of numbers divisible by their LCM which is 36 which is <math> \left[\frac{1000}{36}\right]=27. </math> The answer is 138-27=111.
  9 KB (1,364 words) - 15:59, 21 July 2006
• Mock AIME 1 2006-2007 Problems/Problem 3
  ...integer, find the remainder when <math>n^{2007}</math> is divided by <math>1000</math>. ...1000}</math> and so <math>3^{2007} \equiv 3^7 \equiv 2187 \equiv 187 \pmod{1000}</math>
  963 bytes (135 words) - 15:53, 3 April 2012
• Mock AIME 1 2006-2007/Problems
  ...integer, find the remainder when <math>n^{2007}</math> is divided by <math>1000</math>.
  8 KB (1,355 words) - 14:54, 21 August 2020
• Mock AIME 2 2006-2007 Problems/Problem 3
  ...t square]]. Find the [[remainder]] when <math>S</math> is divided by <math>1000.</math>
  1 KB (198 words) - 10:50, 4 April 2012
• Mock AIME 2 2006-2007 Problems
  ...perfect square. Find the remainder when <math>S</math> is divided by <math>1000.</math>
  5 KB (848 words) - 23:49, 25 February 2017
• Base numbers/Conversion
  Let's try converting 1000 base 10 into base 7. Basically, we are trying to find the solution to the <center><math> 1000 = a_0 + 7a_1 + 49a_2 + 343a_3+2401a_4+\cdots</math></center>
  7 KB (1,177 words) - 15:56, 18 April 2020
• Soli Deo Gloria Home Educators
  ...ech Math Meet]], with one of the students placing first and winning a &#36;1000 scholarship. The club had the 10th place winner at the UGA meet, and place
  4 KB (637 words) - 15:26, 25 September 2006
• Mock AIME 5 2005-2006 Problems
  ...the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>. ...if <math>k</math> has an even number of divisors. For how many <math>n \le 1000</math> does there exist an <math>A</math> such that <math>|A| = 620</math>
  8 KB (1,370 words) - 21:52, 27 February 2007
• 2007 iTest Problems
  ...ts+2007</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.
  33 KB (5,177 words) - 21:05, 4 February 2023
• Roman numerals
  1000 - '''M''' ''(mille)''
  865 bytes (140 words) - 13:58, 24 March 2019
• United States of America Computing Olympiad
  The full score is 1000, with each problem having equal weight in the score.
  2 KB (297 words) - 01:41, 21 January 2023
• 2001 AMC 8 Problems
  ...dollar values of each question are shown in the following table (where K = 1000). ...t{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K}
  13 KB (1,994 words) - 13:04, 18 February 2024
• 1985 AIME Problems
  How many of the first <math>1000</math> positive integers can be expressed in the form
  7 KB (1,071 words) - 19:24, 23 February 2024
• 1985 AIME Problems/Problem 2
  ...rac{1}{3} \pi \left(\frac{12}{5}b\right)b^2 = 800\pi</math> so <math>b^3 = 1000</math> and <math>b = 10</math> so <math>a = 24</math>. Then by the [[Pytha
  3 KB (463 words) - 15:10, 4 September 2020
• United Kingdom mathematics competitions
  ...ng cut-off points are awarded bronze, silver or gold certificates. The top 1000 or so contestants are invited to take part in the Junior Mathematical Olymp ...e National Mathematics Summer School are awarded to high-scorers. The nest 1000 in each year group enter the European Kangaroo, a Europe-wide multiple choi
  3 KB (503 words) - 13:53, 11 December 2007
• Mock AIME 6 Pre 2005/Problems
  ...hen <math>\displaystyle 11^{2005}</math> is divided by <math>\displaystyle 1000</math>. ...f(0)f(2006)f(4014)f(6024)f(8036)</math> is divided by <math>\displaystyle 1000</math>.
  7 KB (1,110 words) - 05:15, 31 December 2006
• Mock AIME 4 2006-2007 Problems
  ...y_3)</math>. Find <math>x_3</math> if <math>x_1 = 1</math> and <math>x_2 = 1000</math>. For how many positive integers <math>n < 1000</math> does there exist a regular <math>n</math>-sided polygon such that th
  6 KB (923 words) - 14:17, 16 January 2007
• Mock AIME 4 2006-2007 Problems/Problem 1
  So he needs another <math>1000-321=679</math> digits before he stops. He can accomplish this by writing 1
  1 KB (149 words) - 23:41, 22 April 2010
• Mock AIME 4 2006-2007 Problems/Problem 2
  ...y_3)</math>. Find <math>x_3</math> if <math>x_1 = 1</math> and <math>x_2 = 1000</math>. ...th>C</math> has coordinates <math>(x_0, \frac{2}{3} \ln 1 + \frac{1}{3}\ln 1000) = (x_0, \ln 10)</math> for some <math>x_0</math>. Then the horizontal [[l
  1 KB (214 words) - 15:25, 8 October 2007
• Mock AIME 4 2006-2007 Problems/Problem 6
  For how many positive integers <math>n < 1000</math> does there exist a regular <math>n</math>-sided polygon such that th ...nvex n-gon is <math>\dfrac{n(n-3)}{2}</math>. We need to count the <math>n<1000</math> for which this is a perfect square.
  2 KB (402 words) - 01:41, 31 January 2009
• Mock AIME 4 2006-2007 Problems/Problem 7
  Find the remainder when <math>3^{3^{3^3}}</math> is divided by 1000. ...uch that <math>N\equiv n\pmod{100}</math> so that <math>3^N\equiv 3^n\pmod{1000}</math>.
  1 KB (127 words) - 00:15, 5 January 2010
• Mock AIME 4 2006-2007 Problems/Problem 8
  ...>m > n</math>. Compute the [[remainder]] when <math>m</math> is divided by 1000.
  2 KB (293 words) - 16:20, 8 October 2007
• Mock AIME 4 2006-2007 Problems/Problem 10
  ...choose 3} + \cdots + {2007 \choose 2007}</math></p></center> is divided by 1000. ...so <math>3S \equiv 128-2 \pmod{1000} \Rightarrow S\equiv \boxed{042}\pmod{1000}</math>.
  4 KB (595 words) - 12:14, 25 November 2023
• 2020 AMC 10A Problems
  ...ny <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisi ...right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>not divisible by <math>3</math>? (Recall that <mat
  13 KB (1,968 words) - 18:32, 29 February 2024
• Carmichael function
  Evaluate <math>2009^{2009}\pmod{1000}</math>.
  2 KB (258 words) - 11:56, 1 August 2022
• Mock AIME 1 Pre 2005 Problems
  ...inct digits. Compute the remainder when <math>S</math> is divided by <math>1000</math>. When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a remainder of <math>N</math> is obtained. Determine the value of <
  6 KB (1,100 words) - 22:35, 9 January 2016
• Mock AIME 2 Pre 2005/Problems
  ...Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. (Bridge is a card game played with the standard <math>52-</math>car is divided by <math>1000</math>.
  6 KB (990 words) - 15:23, 11 November 2009
• Mock AIME 3 Pre 2005 Problems
  .... Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. (Repeated digits are allowed.) .... Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>.
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 2
  .... Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. (Repeated digits are allowed.) ...e <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>.
  950 bytes (137 words) - 10:16, 29 November 2019
• Mock AIME 3 Pre 2005 Problems/Problem 5
  .... Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. ...hus, the answer is <math>a_{10} + b_{10} + c_{10} \equiv \boxed{936} \pmod{1000}</math>. (the real answer is <math>15936</math>.)
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 3 Pre 2005 Problems/Problem 8
  Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>.
  3 KB (520 words) - 12:55, 11 January 2019
• Mock AIME 3 Pre 2005 Problems/Problem 10
  ...pute the remainder obtained when <math>a_{2004}</math> is divided by <math>1000</math> if <math>a_1 = 1</math>. Now we can compute <math>a_n \bmod 1000</math> as follows:
  2 KB (306 words) - 10:36, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 12
  Determine the number of integers <math>n</math> such that <math>1 \le n \le 1000</math> and <math>n^{12} - 1</math> is divisible by <math>73</math>.
  714 bytes (105 words) - 23:59, 24 April 2013
• Mock AIME 3 Pre 2005 Problems/Problem 13
  Determine the remainder obtained when <math>S</math> is divided by <math>1000</math>. ...ac{668}{3}\cdot(2005) = 447115</math>, so <math>W \equiv \boxed{115} \pmod{1000}</math>.
  3 KB (502 words) - 14:53, 19 July 2020
• Mock AIME 4 2005-2006/Problems
  ...ive integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>. ...</math>. Find the remainder when <math>a_{2007}</math> is divided by <math>1000</math>.
  7 KB (1,176 words) - 04:44, 26 February 2007
• Mock AIME 5 2005-2006 Problems/Problem 1
  ...the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.
  2 KB (209 words) - 12:43, 10 August 2019
• 2007 AIME I Problems
  ...how many integer Fahrenheit temperatures between <math>32</math> and <math>1000</math> inclusive does the original temperature equal the final temperature? <math>N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )
  7 KB (1,218 words) - 15:28, 11 July 2022
• 2007 AIME I Problems/Problem 1
  ...ger less than <math>\sqrt{10^6}</math>. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions. ..., or <math>1000^2</math>, the closest multiple of <math>12</math> to <math>1000</math> is <math>996</math> (<math>12*83</math>), so we know that this is th
  1 KB (204 words) - 13:56, 7 February 2023
• 2007 AIME I Problems/Problem 5
  For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature? ...th>6</math> numbers from <math>995</math> onwards, <math>995,\ 997,\ 999,\ 1000</math> work, giving us <math>535 + 4 = \boxed{539}</math> as the solution.
  7 KB (1,076 words) - 00:10, 29 November 2023
• 2007 AIME I Problems/Problem 7
  <math>N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor ) Find the [[remainder]] when <math>N</math> is divided by 1000. (<math>\lfloor{k}\rfloor</math> is the [[floor function|greatest integer]
  2 KB (242 words) - 20:26, 20 April 2023
• 2007 AIME I Problems/Problem 10
  ...with this property. Find the remainder when <math>N</math> is divided by 1000. Complete: <math>31\times60=860\mod{1000}</math>
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2007 AIME I Problems/Problem 11
  ...007} b(p),</math> find the [[remainder]] when <math>S</math> is divided by 1000. ...nly consider the <math>740</math> because we are working with modulo <math>1000</math>.
  3 KB (562 words) - 20:02, 30 December 2023
• 2007 AIME II Problems/Problem 8
  ...ngles determined. Find the [[remainder]] when <math>N</math> is divided by 1000. ...hen <math>x=6</math>. Now you just evaluate <math>-20*36+222*6+222^{2}\mod 1000</math> which is <math>{896}</math>.
  3 KB (399 words) - 21:17, 24 February 2021
• 2007 AIME II Problems
  ...ngles determined. Find the [[remainder]] when <math>N</math> is divided by 1000.
  9 KB (1,435 words) - 01:45, 6 December 2021
• 2006 Cyprus MO/Lyceum/Problems
  <math>\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\
  11 KB (1,672 words) - 10:56, 27 April 2008
• 2006 Cyprus MO/Lyceum/Problem 1
  <math>\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\
  852 bytes (135 words) - 10:54, 27 April 2008
• 2005 AMC 12A Problems/Problem 18
  ...math>1000</math>. How many prime-looking numbers are there less than <math>1000</math>? ...ime-looking}\}</math>. Hence, the number of prime-looking numbers is <math>1000 - (168-3) - 1 - |S_2 \cup S_3 \cup S_5|</math> (note that <math>2,3,5</math
  2 KB (277 words) - 18:15, 25 November 2020
• 2005 AMC 12A Problems/Problem 12
  A [[line]] passes through <math>A\ (1,1)</math> and <math>B\ (100,1000)</math>. How many other points with integer coordinates are on the line and \frac{1000-1}{100-1}=\frac{111}{11},
  1 KB (224 words) - 20:18, 15 July 2021
• 2006 AIME II Problems/Problem 13
  How many integers <math> N </math> less than <math>1000</math> can be written as the sum of <math> j </math> consecutive positive o ...all odd <math>j</math>. Looking at the forms above and the bound of <math>1000</math>, <math>N</math> must be
  4 KB (675 words) - 10:40, 14 July 2022
• 2006 AIME II Problems/Problem 11
  ...ind the [[remainder]] when <math>\sum^{28}_{k=1} a_k </math> is divided by 1000. ...problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
  3 KB (417 words) - 10:07, 12 October 2023
• 2006 AIME II Problems/Problem 7
  ...]]s of [[positive]] [[integer]]s <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a zero digit. ...re <math>\left\lfloor\frac{999}{10}\right\rfloor = 99</math> numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities a
  7 KB (1,114 words) - 03:41, 12 September 2021
• 2006 AIME II Problems
  ...dered pairs of positive integers <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a zero digit. ...h> find the remainder when <math>\sum^{28}_{k=1} a_k </math> is divided by 1000.
  8 KB (1,350 words) - 12:00, 4 December 2022
• 2005 PMWC Problems
  "happy" integers are there between <math>100</math> and <math>1000</math>?
  9 KB (1,449 words) - 20:49, 2 October 2020
• 2000 PMWC Problems
  How many whole numbers less than <math>1000</math> contain at least one <math>2</math> but no <math>3</math>?
  11 KB (1,713 words) - 22:47, 13 July 2023
• 1997 PMWC Problems/Problem T3
  .../math> of the time for each digit (if we include 0), there are <math>\frac{1000}{10} = 100</math> '<tt>9</tt>'s in each place, for a total of <math>300</ma
  1 KB (202 words) - 14:39, 20 April 2014
• 1997 PMWC Problems/Problem I9
  ...th> concentration that the chemist had used? (<math>1</math> litre = <math>1000</math> millilitres) There are <math>1000 \cdot \frac{15}{2} = 7500</math> millilitres of the acid.
  947 bytes (136 words) - 14:26, 20 April 2014
• 2005 PMWC Problems/Problem T1
  ...h>. How many "happy" integers are there between <math>100</math> and <math>1000</math>?
  559 bytes (90 words) - 08:07, 9 October 2007
• Liter
  ...n the [[Système international|metric system]]. One liter is equivalent to 1000 cubic centimeters, or one cubic decimeter. When converting from the metric
  413 bytes (58 words) - 21:28, 28 December 2023
• 1995 AHSME Problems/Problem 4
  ...ng <math>100 = 0.2\cdot P</math>, giving <math>P = \frac{100}{0.2} = \frac{1000}{2} = 500</math>.
  2 KB (232 words) - 22:30, 16 February 2018
• Mass
  ...(1000 grams), tonne (1000 kilograms), kilotonne (1000 tonnes), megatonne (1000 kilotonnes), etc.
  1 KB (188 words) - 22:44, 10 October 2013
• 1997 PMWC Problems
  ...th> concentration that the chemist had used? (<math>1</math> litre = <math>1000</math> millilitres)
  15 KB (2,057 words) - 19:13, 10 March 2015
• 1951 AHSME Problems
  ...bf{(B)}\ \text{A makes 1100 on the deal} \qquad\textbf{(C)}\ \text{A makes 1000 on the deal}</math> ...tbf{(D)}\ \text{A loses 900 on the deal} \qquad\textbf{(E)}\ \text{A loses 1000 on the deal}</math>
  23 KB (3,641 words) - 22:23, 3 November 2023
• 1966 AHSME Problems
  The number of positive integers less than <math>1000</math> divisible by neither <math>5</math> nor <math>7</math> is:
  19 KB (3,159 words) - 22:10, 11 March 2024
• 2002 AMC 12B Problems/Problem 15
  ...htarrow 125a = \overline{bcd}</math>. Since <math>100 \le \overline{bcd} < 1000</math>, from <math>a = 1, \ldots, 7</math> we have <math>7</math> three-dig
  2 KB (395 words) - 15:50, 3 April 2022
• Mock AIME 1 Pre 2005 Problems/Problem 1
  ...inct digits. Compute the remainder when <math>S</math> is divided by <math>1000</math>.
  1 KB (194 words) - 13:44, 5 September 2012
• 2003 AMC 12B Problems/Problem 23
  ...th>1000 \le \frac 1x = \pi n \le 10000</math>. There are <math>\frac{10000-1000}{\pi} \approx \boxed{2900} \Rightarrow \mathrm{(A)}</math> solutions for <m ...he zeros of <math>\sin(x)</math> with <math>x</math> values between <math>(1000, 10000)</math>. We know that the <math>x</math> values of any sine graph is
  2 KB (249 words) - 18:39, 9 February 2023
• 2003 AMC 12B Problems/Problem 17
  <cmath> \frac{1000}{x^5} = 10 </cmath>
  2 KB (329 words) - 13:49, 4 April 2024
• 2003 AMC 12B Problems/Problem 24
  ...=1001</math> and <math>c=1002</math>. This is indeed a solution as <math>a=1000</math> puts <math>P</math> on <math>y=2003-2x</math>, and thus the answer i # <math>\frac{2003+a+b-c}3 = \frac{2002+a}3 \leq \frac{2002+1000}3 < 1001 = b</math>
  7 KB (1,183 words) - 11:47, 15 February 2016
• 1998 AHSME Problems/Problem 24
  ...verline{d_1d_2d_3} = \overline{d_4d_5d_6}</math> (of which there are <math>1000</math> possibilities for <math>\overline{d_1d_2d_3}</math> and <math>10</ma <cmath>|A \cup B| = |A| + |B| - |A \cap B| = 1000 \times 10 + 1000 \times 10 - 10 = 19990 \Rightarrow \mathrm{(C)}</cmath>
  2 KB (330 words) - 10:14, 10 August 2016
• 2008 AMC 10A Problems
  <math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math>
  14 KB (2,138 words) - 15:08, 18 February 2023
• 2008 AMC 12A Problems
  <math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math>
  13 KB (2,025 words) - 13:56, 2 February 2021
• 2008 AMC 12A Problems/Problem 6
  <math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math>
  2 KB (240 words) - 19:53, 4 June 2021
• 2008 AMC 10B Problems
  ...{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000</math> ...(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math>
  12 KB (1,838 words) - 16:52, 7 October 2022
• 2008 AMC 12B Problems
  ...500 \qquad \textbf{(B)}\ 900 \qquad \textbf{(C)}\ 950 \qquad \textbf{(D)}\ 1000 \qquad \textbf{(E)}\ 1900</math>
  14 KB (2,199 words) - 13:43, 28 August 2020
• 2006 Alabama ARML TST Problems
  A bored mathematician has his computer calculate 1000 consecutive terms in the Fibonacci sequence. He notes that the smallest of
  5 KB (769 words) - 20:56, 24 March 2015
• 2008 AMC 12B Problems/Problem 10
  ...500 \qquad \textbf{(B)}\ 900 \qquad \textbf{(C)}\ 950 \qquad \textbf{(D)}\ 1000 \qquad \textbf{(E)}\ 1900</math>
  1 KB (171 words) - 17:21, 3 January 2020
• Mock AIME 1 Pre 2005 Problems/Problem 4
  When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a remainder of <math>N</math> is obtained. Determine the value of < ...400 \cdot 5 + 5} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{801} \pmod{1000}</math>.
  685 bytes (81 words) - 10:51, 11 June 2013
• Mock AIME 1 Pre 2005 Problems/Problem 7
  Find the remainder when <math>N</math> is divided by <math>1000</math>. And <math>N \equiv \boxed{320} \pmod{1000}</math>.
  1 KB (221 words) - 17:27, 23 February 2013
• 2008 AIME I Problems
  ..., where <math>m</math> and <math>n</math> are positive integers, <math>m < 1000</math>, and <math>m</math> is not divisible by the <math>n</math>th power o
  9 KB (1,536 words) - 00:46, 26 August 2023
• 2008 AIME II Problems
  ...rnate in pairs. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...ath> with <math>\overline{BC}\parallel\overline{AD}</math>, let <math>BC = 1000</math> and <math>AD = 2008</math>. Let <math>\angle A = 37^\circ</math>, <m
  7 KB (1,167 words) - 21:33, 12 August 2020
• 2008 AIME I Problems/Problem 7
  There are <math>1000 - 266 - 26 = \boxed{708}</math> sets without a perfect square.
  1 KB (195 words) - 22:59, 2 January 2021
• 2008 AIME I Problems/Problem 12
  ...ars as <math>15k</math>. Then, the distance between them is <math>\frac{4}{1000} \times k\text{km}</math>. Therefore, it takes the car closest to the eye n ...the number of these intervals in an hour is <math>\frac{1}{\frac{4/1000+(4/1000)(\lfloor a/15 \rfloor)}{a}}=\frac{250a}{1+\lfloor a/15 \rfloor}</math>. Now
  4 KB (669 words) - 18:35, 8 October 2023
• 2008 AIME I Problems/Problem 15
  ...h>, where <math>m</math> and <math>n</math> are positive integers, <math>m<1000</math>, and <math>m</math> is not divisible by the <math>n</math>th power o
  6 KB (1,041 words) - 00:54, 1 February 2024
• 2008 AIME II Problems/Problem 1
  ...rnate in pairs. Find the remainder when <math>N</math> is divided by <math>1000</math>. Dividing <math>10100</math> by <math>1000</math> yields a remainder of <math>\boxed{100}</math>.
  4 KB (575 words) - 16:41, 14 April 2024
• 2008 AIME II Problems/Problem 5
  ...ath> with <math>\overline{BC}\parallel\overline{AD}</math>, let <math>BC = 1000</math> and <math>AD = 2008</math>. Let <math>\angle A = 37^\circ</math>, <m label("\(1000\)",(B+C)/2,NE);
  8 KB (1,338 words) - 23:15, 28 November 2023
• 2008 AIME II Problems/Problem 12
  ...e are adjacent. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...10</math> and <math>b=9</math>, we get <math>2310 \equiv \boxed{310} \pmod{1000}</math>.
  10 KB (1,550 words) - 12:58, 15 July 2023
• 2008 AIME II Problems/Problem 15
  ...2),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> ...5, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it is AIME problem. Now take the first criterion, let <math>a<
  4 KB (628 words) - 16:23, 2 January 2024
• 0.999...
  ...ts = 0.9 + 0.09 + 0.009 + \ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots
  3 KB (577 words) - 20:04, 4 February 2023
• 2008 AMC 10B Problems/Problem 7
  ...{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000</math>
  2 KB (323 words) - 20:23, 14 July 2021
• 2008 AMC 10B Problems/Problem 18
  ...(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math>
  875 bytes (139 words) - 20:19, 23 March 2023
• KGS math club/solution 1 1
  And another for a 1000 game series: for i in range(0, 1000):
  3 KB (470 words) - 13:26, 28 June 2008
• 2001 IMO Shortlist Problems/C1
  ...ists of 1000 ones, then 500 twos, then 501 threes, the blocks would be the 1000 ones, the 500 twos, and the 501 threes. Let the number of elements in the i
  4 KB (703 words) - 12:45, 27 November 2017
• 2000 USAMO Problems
  ...r <math>n</math> such that if <math>n</math> squares of a <math>1000\times 1000</math> chessboard are colored, then there will exist three colored squares
  3 KB (495 words) - 19:02, 18 April 2014
• 2000 USAMO Problems/Problem 4
  ...<math>n</math> such that if <math>n</math> squares of a <math>1000 \times 1000</math> chessboard are colored, then there will exist three colored squares ...for <math>n = 1999</math>. We call a row or column ''filled'' if all <math>1000</math> of its squares are colored. Then any of the remaining <math>999</mat
  2 KB (382 words) - 13:37, 4 July 2013
• 2008 iTest Problems
  ...th>2008</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>. ...to biomass for between <math>\textdollar{200}</math> and <math>\textdollar{1000}</math> per year. The energy comes from sawdust, switch-grass, and even lan
  71 KB (11,749 words) - 01:31, 2 November 2023
• 2008 iTest Problems/Problem 80
  the remainder when <math>|r(2008)|</math> is divided by <math>1000</math>. Then <math>|r(2008)| \equiv 2008^2 \equiv \boxed{64} \pmod{1000}</math>.
  3 KB (560 words) - 19:49, 23 November 2018
• Kilogram
  The kilogram is equivalent to 1000 [[gram|grams]]. It is the [[Système_international|SI]] unit for mass, mean
  213 bytes (32 words) - 02:03, 7 December 2008
• Binary
  | <math>2^3</math>|| 8|| 1000
  2 KB (294 words) - 12:55, 3 September 2019
• 1951 AHSME Problems/Problem 5
  ...A\ loses\ 900\ on\ the\ deal }</math> <math>\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal } </math>
  1 KB (207 words) - 12:58, 19 February 2016
• 2001 AMC 10 Problems
  \qquad \textbf{(D) } {1000} \qquad \textbf{(E) } {1000\sqrt{2}} </math>
  14 KB (1,983 words) - 16:25, 2 June 2022
• 2021 GMC 10B
  Find the remainder when <math>3^{1624}+7^{1604}</math> is divided by <math>1000</math>.
  11 KB (1,695 words) - 14:33, 7 March 2022
• 2008 AMC 12B Problems/Problem 20
  170&850&1000\\ 180&900&1000\\
  8 KB (1,241 words) - 17:55, 11 August 2023
• 1985 AJHSME Problems/Problem 2
  &= 1000 - 55\\
  2 KB (282 words) - 13:43, 4 April 2024
• 1986 AJHSME Problems
  A contest began at noon one day and ended <math>1000</math> minutes later. At what time did the contest end?
  14 KB (2,054 words) - 15:41, 8 August 2020
• 1986 AJHSME Problems/Problem 4
  ...absolute changes (e.g <math>1001</math> is much closer relatively to <math>1000</math> than <math>2</math> is to <math>1</math>).
  733 bytes (97 words) - 02:17, 24 November 2017
• 1986 AJHSME Problems/Problem 5
  A contest began at noon one day and ended <math>1000</math> minutes later. At what time did the contest end? ...ght is <math>720</math> minutes away, we know that the contest ended <math>1000 - 720 = 280</math> minutes after midnight. The highest multiple of 60 that
  1 KB (170 words) - 13:47, 21 February 2019
• 1987 AJHSME Problems/Problem 5
  ...ac{4}{10} \cdot \frac{22}{100} = \frac{4\cdot 22}{10\cdot 100} = \frac{88}{1000} = 0.088\rightarrow \boxed{\text{A}}</math>
  638 bytes (77 words) - 23:52, 4 July 2013
• 2009 AMC 12A Problems
  How many positive integers less than <math>1000</math> are <math>6</math> times the sum of their digits?
  13 KB (2,105 words) - 13:13, 12 August 2020
• 2009 AMC 12A Problems/Problem 12
  How many positive integers less than <math>1000</math> are <math>6</math> times the sum of their digits?
  3 KB (481 words) - 20:06, 17 December 2017
• 2009 AIME I Problems
  ...is on <math>\overline{AB}</math> so that <math>\frac {AM}{AB} = \frac {17}{1000}</math> and point <math>N</math> is on <math>\overline{AD}</math> so that < How many positive integers <math>N</math> less than <math>1000</math> are there such that the equation <math>x^{\lfloor x\rfloor} = N</mat
  7 KB (1,152 words) - 02:24, 23 July 2021
• 2009 AIME I Problems/Problem 6
  How many positive integers <math>N</math> less than <math>1000</math> are there such that the equation <math>x^{\lfloor x\rfloor} = N</mat ...or}</math> would be at least <math>3125</math> which is greater than <math>1000</math>.
  4 KB (595 words) - 16:38, 15 February 2021
• 2009 AIME I Problems/Problem 8
  ...se differences. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0</math>. Evaluating <math>N \bmod {1000}</math> yields:
  3 KB (396 words) - 11:42, 23 January 2023
• 2009 AIME I Problems/Problem 4
  ...is on <math>\overline{AB}</math> so that <math>\frac {AM}{AB} = \frac {17}{1000}</math> and point <math>N</math> is on <math>\overline{AD}</math> so that < ...der from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2009}{17}MN</math>. It is not difficu
  7 KB (1,117 words) - 00:23, 9 January 2023
• 1989 AJHSME Problems
  <math>\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} =</math>
  13 KB (1,765 words) - 11:55, 22 November 2023
• 2009 AIME II Problems
  ...re consecutive. Find the remainder when <math>m</math> is divided by <math>1000</math>. ...=2000</math>. Find the remainder when <math>m-n</math> is divided by <math>1000</math>.
  8 KB (1,366 words) - 21:33, 3 January 2021
• 2009 AIME II Problems/Problem 9
  ...=2000</math>. Find the remainder when <math>m-n</math> is divided by <math>1000</math>. 1000 + 994 + \cdots + 10 + 4
  5 KB (845 words) - 15:45, 28 December 2020
• 2009 AIME II Problems/Problem 6
  ...re consecutive. Find the remainder when <math>m</math> is divided by <math>1000</math>. ...0\choose 5} = 2002 - 252 = 1750</math>, and the answer is <math>1750 \bmod 1000 = \boxed{750}</math>.
  5 KB (921 words) - 00:15, 11 December 2022
• 2000 AMC 8 Problems
  ...(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000</math>
  15 KB (2,165 words) - 03:32, 13 April 2024
• 2009 AIME II Problems/Problem 7
  ...e number of <math>2</math>s in the summation is clearly greater than <math>1000</math>, dividing by <math>10</math> will yield a number greater than <math>
  8 KB (1,312 words) - 16:23, 30 March 2024
• 2009 AIME II Problems/Problem 13
  ...twelve chords. Find the remainder when <math>n</math> is divided by <math>1000</math>. ...^{12}\cdot \dfrac{14}{2^{13}} = 7\cdot 2^{12} = \boxed{672} \; \text{(mod }1000\text{)}</math>
  8 KB (1,266 words) - 20:27, 10 December 2023
• Olimpiada de Mayo
  .../math> and end in <math>388</math>, and this number is <math>97000=97\cdot 1000=388\cdot 25</math>. <cmath>\begin{eqnarray*} 1000 & < & 3880 + 97k < 100000 \\
  4 KB (668 words) - 14:52, 17 August 2020
• 1989 AJHSME Problems/Problem 2
  <math>\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}=</math> ...+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\
  522 bytes (53 words) - 23:57, 4 July 2013
• 1990 AJHSME Problems
  ...A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990</math> label("$500$",(-1,4),W); label("$1000$",(-1,8),W); label("$1500$",(-1,12),W);
  15 KB (2,059 words) - 15:03, 6 October 2021
• 1990 AJHSME Problems/Problem 16
  ...A)}\ -990 \qquad \text{(B)}\ -10 \qquad \text{(C)}\ 990 \qquad \text{(D)}\ 1000 \qquad \text{(E)}\ 1990</math> In the middle, we have <math>\cdots + 1010-1000+990 -\cdots </math>.
  1 KB (181 words) - 21:44, 1 June 2020
• 1990 AJHSME Problems/Problem 20
  ...1000}</math>, and for the incorrect data the mean is <math>\frac{S+980000}{1000}</math>. The difference is <math>882, or \rightarrow \boxed{\text{A}}</mat
  1 KB (160 words) - 12:26, 13 June 2023
• 1990 AJHSME Problems/Problem 23
  label("$500$",(-1,4),W); label("$1000$",(-1,8),W); label("$1500$",(-1,12),W);
  1 KB (214 words) - 22:03, 11 July 2009
• 1991 AJHSME Problems/Problem 4
  ...93+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\ &\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\
  595 bytes (66 words) - 23:53, 8 October 2014
• 1991 AJHSME Problems/Problem 7
  ...0)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \\
  1,005 bytes (96 words) - 16:34, 24 February 2024
• 1992 AJHSME Problems
  <math>\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900</math>
  17 KB (2,346 words) - 13:36, 19 February 2020
• 1993 AJHSME Problems
  <math>1000\times 1993 \times 0.1993 \times 10 = </math> ...(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000</math>
  14 KB (1,797 words) - 11:13, 28 December 2022
• 2010 AMC 12A Problems/Problem 7
  Note: The fact that <math>1\text{ L}=1000\text{ cm}^3</math> doesn't matter since only the ratios are important.
  1 KB (198 words) - 18:08, 28 June 2021
• 2010 AMC 12A Problems/Problem 6
  ...32</math> is <math>1000</math>. However, the only palindrome between <math>1000</math> and <math>1032</math> is <math>1001</math>, which means that <math>x
  2 KB (320 words) - 04:51, 21 January 2023
• Mock AIME 1 Pre 2005 Problems/Problem 11
  Determine the remainder obtained when <math>S</math> is divided by <math>1000</math>. To find <math>2\cdot3^{1001} \pmod{1000}</math>, we notice that <math>3^{\phi{500}}\equiv 3^{200}\equiv 1 \pmod{500
  2 KB (272 words) - 10:51, 2 July 2015
• 2010 AMC 10B Problems
  A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability
  12 KB (1,817 words) - 22:44, 22 December 2020
• 2010 AMC 12B Problems
  A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability
  12 KB (1,845 words) - 13:00, 19 February 2020
• Mock AIME 1 2010 Problems
  ...that no two overlap, find the remainder when <math>N</math> is divided by 1000.
  7 KB (1,297 words) - 01:29, 25 November 2016
• 2010 AIME I Problems
  ...\times \underbrace{99\cdots9}_{\text{999 9's}}</math> is divided by <math>1000</math>. ...5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  8 KB (1,243 words) - 21:58, 10 August 2020
• 2010 AIME I Problems/Problem 2
  ...\times \underbrace{99\cdots9}_{\text{999 9's}}</math> is divided by <math>1000</math>. ...ion is congruent to <math>- 1\times9\times99 = - 891\equiv\boxed{109}\pmod{1000}</math>.
  1 KB (167 words) - 17:46, 30 April 2023
• 2010 AIME I Problems/Problem 7
  ...5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  2 KB (255 words) - 17:03, 9 August 2018
• 2010 AIME I Problems/Problem 10
  ...iv 0\ (\textrm{mod}\ 100)</math>. Thus, we have <math>2010-10a_1-a_0 \geq 1000</math> always. *If <math>1000 \leq 2010 - 10a_1 - a_0 < 2000</math>, then there are 2 valid choices for <
  7 KB (1,147 words) - 21:58, 23 January 2024
• 2010 AIME I Problems/Problem 14
  ...0</math> (alternatively, use binary search to get to this, with <math>n\le 1000</math>). Manually checking shows that <math>f(109) = 300</math> and <math>f ...n the middle of 0 and 100, let <math>k=50</math>, so <math>50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500</math>, and <math>n = 110</math>. <math>f(110) = 3
  4 KB (739 words) - 22:09, 25 November 2023
• Mock AIME 2 2010 Problems
  ...ath> distinct values, find the remainder when <math>K</math> is divided by 1000.
  7 KB (1,150 words) - 09:10, 8 October 2018
• Mock AIME 2 2010 Answer Key
  ...ath> distinct values, find the remainder when <math>K</math> is divided by 1000. ...the last three nonzero digits, we simply wish to find <math>12^{16} \pmod{1000}</math>. To make this easier, we will find it modulo 8 and modulo 125 and u
  36 KB (6,214 words) - 20:22, 13 July 2023
• 2010 AIME II Problems
  ...s are the same. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  8 KB (1,246 words) - 21:58, 10 August 2020
• 2010 AIME II Problems/Problem 1
  ...s are the same. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...visible by <math>36</math>. Thus, <math>N = 8640 \equiv \boxed{640} \pmod {1000}</math>.
  1 KB (176 words) - 12:54, 28 October 2021
• Getting Started With Python Programming
  ...can see that we can compute with some pretty big numbers such as <math>2^{1000}</math> as illustrated below. '''Find the sum of all the positive multiples of 3 below 1000.'''
  13 KB (2,299 words) - 12:24, 2 March 2024
• Scholarships by State
  ...ation Scholarship Program - The Arizona Private School Association awards $1000 Scholarships to '''every''' public, private and charter High School in Ariz
  9 KB (1,107 words) - 15:32, 16 July 2010
• 2010 AMC 12B Problems/Problem 11
  A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability
  2 KB (354 words) - 19:54, 10 March 2024
• CIA Undergraduate Scholar Program
  * 1500 SAT (1000 Math & Critical Reading, 500 Writing) or 21 ACT scores or higher (high scho
  2 KB (228 words) - 11:04, 23 July 2010
• 2006 AMC 8 Problems
  ...d with water to a depth of <math> 37 </math> cm. A rock with volume <math> 1000 \text{ cm}^3 </math> is then placed in the aquarium and completely submerge
  16 KB (2,215 words) - 19:18, 10 April 2024
• 1997 AJHSME Problems
  <math>\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} = </math>
  12 KB (1,702 words) - 12:35, 6 November 2022
• 1998 AJHSME Problems
  <math>100\times 19.98\times 1.998\times 1000=</math> <math>\text{(A)}\ 600 \qquad \text{(B)}\ 800 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 2000 \qquad \text{(E)}\ 3000</math>
  14 KB (1,920 words) - 19:31, 31 January 2024
• 1958 AHSME Problems
  With \$ <math> 1000</math> a rancher is to buy steers at \$ <math>25</math> each and cows at \$
  25 KB (3,872 words) - 14:21, 20 February 2020
• 2011 AMC 10A Problems
  Seven students count from 1 to 1000 as follows: ...oup of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, ..., 997, 999, 1000.
  13 KB (1,903 words) - 18:09, 19 April 2021
• 1993 AHSME Problems
  ...chedule is 7 work-days followed by 3 rest-days. On how many of their first 1000 days do both have rest-days on the same day?
  20 KB (2,814 words) - 08:15, 27 June 2021
• 2011 AMC 10A Problems/Problem 19
  ...d <math>600</math>. We list perfect squares from <math>400</math> to <math>1000</math>.
  3 KB (545 words) - 20:54, 21 August 2023
• 2011 AMC 10A Problems/Problem 23
  Seven students count from 1 to 1000 as follows: ...p of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
  6 KB (926 words) - 23:38, 8 April 2024
• 2011 AMC 10B Problems/Problem 23
  ...\pmod{1000},</math> we know that <math>2011^{2011} \equiv 11^{2011} \pmod{1000}.</math> ...er than <math>2</math> and so is equivalent to <math>0</math> modulo <math>1000,</math> which means we can ignore it. We have:
  9 KB (1,287 words) - 20:37, 20 August 2023
• 2001 AMC 10 Problems/Problem 20
  \qquad \textbf{(D) } {1000} \qquad \textbf{(E) } {1000\sqrt{2}} </math>
  3 KB (460 words) - 09:59, 1 March 2024
• 2011 AIME I Problems
  ...ts in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000.
  10 KB (1,634 words) - 22:21, 28 December 2023
• 2002 AMC 8 Problems
  ...(B)}\ 500 \qquad \text{(C)}\ 625 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1000</math>
  15 KB (2,102 words) - 09:58, 5 May 2024
• 2011 AIME I Problems/Problem 11
  ...ts in <math>R</math>. Find the remainder when <math>S</math> is divided by 1000. ...pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation:
  11 KB (1,668 words) - 22:10, 24 February 2023
• 2011 AIME II Problems/Problem 5
  ...>\dfrac{1-k^{2011}}{1-k^{6033}}</math>. Since <math>k^{6033} = \dfrac{729}{1000}</math>, plugging all the values in gives <math>\boxed{542}</math>.
  3 KB (441 words) - 21:32, 20 January 2024
• 2011 AIME II Problems/Problem 7
  ...he requirement. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...r is <math>\binom{15}{5} = 3003</math>, take the remainder when divided by 1000 to get the answer: <math>\boxed{003}</math>.
  3 KB (605 words) - 10:30, 12 January 2024
• 2011 AIME II Problems
  ...he requirement. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...ositive integers. Find the remainder when <math>m + n</math> is divided by 1000.
  8 KB (1,301 words) - 08:43, 11 October 2020
• 2011 AIME II Problems/Problem 14
  ...\leq 30</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...8=1440</math>. Thus, the remainder when <math>N</math> is divided by <math>1000</math> is <math>\boxed{440}.</math>
  10 KB (1,581 words) - 22:09, 27 August 2023
• 2011 AIME II Problems/Problem 10
  ...ositive integers. Find the remainder when <math>m + n</math> is divided by 1000. ...= \frac{4050}{7} = (OP)^2</math>; <math>4050 + 7 \equiv \boxed{057} \pmod{1000}</math>.
  11 KB (1,720 words) - 03:12, 18 December 2023
• 1999 AMC 8 Problems
  <math>\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450</ma
  17 KB (2,394 words) - 19:51, 8 May 2023
• 2001 AMC 8 Problems/Problem 17
  ...dollar values of each question are shown in the following table (where K = 1000). ...t{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K}
  2 KB (241 words) - 18:16, 10 January 2024
• 2003 AMC 10B Problems
  Al, Betty, and Clare split <math> \ </math><math>1000</math> among them to be invested in different ways. Each begins with a diff
  15 KB (2,166 words) - 21:17, 16 February 2021
• 1999 AHSME Problems/Problem 11
  ...<math>1000</math> through <math>1999</math> inclusive cost <math>0.08\cdot 1000 = 80.00</math>.
  1 KB (200 words) - 15:22, 11 July 2022
• 1993 AJHSME Problems/Problem 4
  <math>1000\times 1993 \times 0.1993 \times 10 = </math> <math>1000\times10=10^4\\
  448 bytes (45 words) - 00:10, 5 July 2013
• 1998 AJHSME Problems/Problem 7
  <math>100\times 19.98\times 1.998\times 1000=</math>
  508 bytes (51 words) - 00:28, 5 July 2013
• 1998 AJHSME Problems/Problem 15
  <math>\text{(A)}\ 600 \qquad \text{(B)}\ 800 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 2000 \qquad \text{(E)}\ 3000</math>
  1 KB (173 words) - 20:55, 11 May 2021
• 2003 AMC 10B Problems/Problem 12
  Al, Betty, and Clare split <math>\textdollar 1000</math> among them to be invested in different ways. Each begins with a diff <cmath>a+b+c=1000</cmath>
  2 KB (323 words) - 01:11, 2 January 2023
• 1984 AHSME Problems
  <math> \frac{1000^2}{252^2-248^2} </math> equals <math> \mathrm{(A) \ }62500 \qquad \mathrm{(B) \ }1000 \qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ }250 \qquad \mathrm{(E) \
  13 KB (1,879 words) - 14:00, 19 February 2020
• 1984 AHSME Problems/Problem 1
  <math> \frac{1000^2}{252^2-248^2} </math> equals ...2^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{2000} </math>.
  685 bytes (80 words) - 12:48, 5 July 2013
• 1999 AMC 8 Problems/Problem 9
  <math>\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450</ma
  1 KB (178 words) - 13:22, 27 February 2022
• Imperial system
  *1 '''inch''' (in) is equal to 1000 ''thou'' (25.4mm)
  2 KB (233 words) - 01:09, 20 June 2011
• 1992 AJHSME Problems/Problem 8
  <math>\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900</math> ...hem at <math>\$0.25</math> each gives <math>250/0.25= \boxed{\textbf{(C)}\ 1000}</math>.
  700 bytes (96 words) - 00:12, 3 July 2017
• 2000 AMC 8 Problems/Problem 16
  ...(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000</math> ...ac{1000}{25}=40</math> meters. The perimeter <math>P</math> is <math>\frac{1000}{10}=100</math> meters. Since <math>P_{rect} = 2L + 2W</math>, we plug valu
  948 bytes (143 words) - 20:05, 15 April 2023
• 1967 AHSME Problems
  If one uses only the tabular information <math>10^3=1000</math>, <math>10^4=10,000</math>, <math>2^{10}=1024</math>, <math>2^{11}=20
  20 KB (3,108 words) - 14:14, 20 February 2020
• 2003 AMC 8 Problems
  How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
  16 KB (2,236 words) - 12:02, 19 February 2024
• 1997 AJHSME Problems/Problem 1
  <math>\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} = </math> <math>\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000}</math>
  625 bytes (63 words) - 21:02, 27 October 2016
• 1997 AJHSME Problems/Problem 6
  ...th>ths place. The ratio of these two numbers is <math>\frac{100}{\frac{1}{1000}} = 100\times1000 = 100,000</math>, and the answer is <math>\boxed{\textbf{
  1,012 bytes (143 words) - 00:26, 5 July 2013
• 2002 AMC 8 Problems/Problem 13
  ...\text{(B)}\ 500\qquad\text{(C)}\ 625\qquad\text{(D)}\ 750\qquad\text{(E)}\ 1000 </math> \boxed{\text{(E)}\ 1000}</math>.
  761 bytes (120 words) - 10:07, 18 September 2021
• 2009 AMC 8 Problems
  How many whole numbers between 1 and 1000 do not contain the digit 1?
  18 KB (2,551 words) - 18:46, 27 February 2024
• 1996 AJHSME Problems/Problem 11
  ...>100</math> as the reciprocal, and <math>x=0.001</math>, which gives <math>1000</math> as the reciprocal. Thus, <math>D</math> will be huge, and this elim
  2 KB (253 words) - 00:24, 5 July 2013
• 1996 AJHSME Problems/Problem 19
  In the second school, <math>2500 \cdot 40\% = 2500 \cdot 0.40 = 1000</math> students prefer tennis. In total, <math>440 + 1000 = 1440</math> students prefer tennis out of a total of <math>2000 + 2500 =
  2 KB (223 words) - 00:25, 5 July 2013
• Primitive Pythagorean Triple
  280 960 1000 [7 - 24 - 25] 352 936 1000 [44 - 117 - 125]
  55 KB (3,565 words) - 11:05, 21 May 2020
• 2009 AMC 8 Problems/Problem 22
  How many whole numbers between 1 and 1000 do '''not''' contain the digit 1? ...math> such numbers. However, we overcounted by one; 0 is not between 1 and 1000, so there are <math>\boxed{\textbf{(D)}\ 728}</math> numbers.
  2 KB (244 words) - 23:26, 10 March 2024
• 1992 AJHSME Problems/Problem 16
  ...cylinder that has a volume of double the volume is <math>\pi (10)^2(10) = 1000\pi \rightarrow \boxed{(B)}</math>.
  2 KB (266 words) - 00:09, 5 July 2013
• 1993 AJHSME Problems/Problem 17
  ...(B)}\ 500 \qquad \text{(C)}\ 550 \qquad \text{(D)}\ 600 \qquad \text{(E)}\ 1000</math>
  1 KB (187 words) - 09:35, 1 December 2020
• 2011 AMC 8 Problems/Problem 2
  ...e feet. Since <math>1000 > 900,</math> Makenna's garden is larger by <math>1000-900=100</math> square feet. <math>\Rightarrow \boxed{ \textbf{(E)}\ \text{M
  1 KB (155 words) - 12:35, 13 November 2016
• 2003 AMC 8 Problems/Problem 19
  How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors? ..., 600, 900, 1200, 1500, 1800, 2100</math>. The number of multiples between 1000 and 2000 is <math>\boxed{\textbf{(C)}\ 3}</math>.
  737 bytes (88 words) - 00:47, 5 July 2013
• 2006 AMC 8 Problems/Problem 21
  ...led with water to a depth of <math>37</math> cm. A rock with volume <math>1000\text{cm}^3</math> is then placed in the aquarium and completely submerged. ...math>cm for every <math>100 \cdot 40 = 4000\text{cm}^2</math>. Since <math>1000</math> is <math>\frac{1}{4}</math> of <math>4000</math>, the water will ris
  909 bytes (138 words) - 19:43, 15 April 2023
• Mathtime Version 1 Issue 2
  ...er value of <math>n</math> such that <math>75^n</math> has less than <math>1000</math> distinct positive divisors? (Proposed by djmathman)
  15 KB (2,444 words) - 21:46, 1 January 2012
• 2012 AMC 10A Problems/Problem 25
  ...es in the choices, we find that when <math>n=10</math>, <math>p=\frac{512}{1000}>\frac{1}{2}</math>, when <math>n=9</math>, <math>p=\frac{343}{729}<\frac{1
  13 KB (2,133 words) - 01:22, 6 February 2024
• 2012 AMC 10A Problems/Problem 16
  ...from the answers starting with the smallest answer, if they had run <math>1000</math> seconds, they would have run <math>4400, 4800, 5000</math> meters, r ...ed to be multiples of 500. It is now easy to see that <math>0.2t=500, 0.4t=1000, 0.6t=1500</math>, so <math>t=\boxed{\textbf{(C)}\ 2,500}</math>.
  4 KB (718 words) - 12:52, 27 November 2019
• 1991 AHSME Problems/Problem 17
  How many years in the millenium between 1000 and 2000 have properties (a) and (b)?
  1 KB (195 words) - 12:38, 13 December 2016
• 1992 AHSME Problems/Problem 21
  the 99-term sequence <math>(a_1,...,a_{99})</math> is 1000, what is the Cesáro sum of the 100-term sequence \text{(C) } 1000\quad
  2 KB (359 words) - 22:53, 17 April 2022
• 2012 AMC 12B Problems/Problem 14
  ...nardo. The winner is the last person who produces a number less than <math>1000</math>. Let <math>N</math> be the smallest initial number that results in a Thus, <math>950<16x+700<1000</math>. Then, <math>16x>250 \implies x \geq 16</math>.
  3 KB (463 words) - 09:46, 24 April 2024
• 2012 AMC 10B Problems
  The point in the xy-plane with coordinates <math>(1000, 2012)</math> is reflected across the line <math>y = 2000</math>. What are ...extbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012) </math>
  18 KB (2,350 words) - 18:48, 9 July 2023
• 2012 AMC 10B Problems/Problem 3
  The point in the <math>xy</math>-plane with coordinates <math>(1000, 2012)</math> is reflected across the line <math>y=2000</math>. What are t ...extbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012) </math>
  1,021 bytes (147 words) - 20:43, 15 February 2024
• 2012 AMC 12B Problems
  ...nardo. The winner is the last person who produces a number less than <math>1000</math>. Let <math>N</math> be the smallest initial number that results in a
  20 KB (2,681 words) - 09:47, 29 June 2023
• 2012 AMC 10B Problems/Problem 12
  ...an Theorem, <math>DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}</math>. <math>31^2 = 961</math>, and <math>32^2 = 1024</math>. Thus, <math>\sqrt {1000}</math> must be between <math>31</math> and <math>32</math>. The answer is
  2 KB (293 words) - 20:22, 3 September 2021
• 2012 AIME I Problems/Problem 10
  ...the tenth smallest element of <math>\mathcal{T}</math> is divided by <math>1000</math>. ...esponding element in <math>\mathcal{T}</math> is <math>\frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}</math>
  8 KB (1,338 words) - 02:03, 25 November 2023
• 2015 AIME I Problems
  Consider all 1000-element subsets of the set <math> \{ 1, 2, 3, ... , 2015 \} </math>. From ...</math>. Find the number of values of <math>n</math> with <math>2\le n \le 1000</math> for which <math>A(n)</math> is an integer.
  10 KB (1,615 words) - 21:48, 13 January 2024
• 2012 AIME I Problems
  ...the tenth smallest element of <math>\mathcal{T}</math> is divided by <math>1000</math>. ...of such jumps. Find the remainder when <math>M</math> is divided by <math>1000.</math>
  10 KB (1,617 words) - 14:49, 2 June 2023
• 2012 AIME I Problems/Problem 11
  ...of such jumps. Find the remainder when <math>M</math> is divided by <math>1000.</math>
  4 KB (798 words) - 20:32, 24 January 2021
• Mock AIME II 2012 Problems/Problem 1
  ...ime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>. ...equiv 17\pmod{1000}</math> to give us <math>m+n=35+144+17=\boxed{196}\pmod{1000}</math>.
  1 KB (156 words) - 09:53, 13 March 2013
• 1972 IMO Problems/Problem 1
  ...< 10 \cdot 100 = 1000</math>. (There are fewer attainable sums.) As <math>1000 < 1022</math>, the [[Pigeonhole Principle]] implies there are two distinct
  1 KB (233 words) - 16:25, 10 November 2023
• 2012 AIME II Problems
  ...<math>S</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. Find the number of positive integers <math>n</math> less than <math>1000</math> for which there exists a positive real number <math>x</math> such th
  7 KB (1,228 words) - 12:16, 13 March 2020
• 2012 AIME II Problems/Problem 7
  ...<math>S</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...the first 165; for 13 spaces, there are <math>\binom{13}{8}=13 \cdot 99 > 1000</math>, so we now know that <math>N</math> has exactly 13 spaces, so the <m
  2 KB (345 words) - 14:55, 17 October 2020
• 2012 AIME II Problems/Problem 10
  Find the number of positive integers <math>n</math> less than <math>1000</math> for which there exists a positive real number <math>x</math> such th ...1 + \frac{30}{31}</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for <math>n</math> is equ
  3 KB (541 words) - 13:52, 7 May 2023
• 2012 AIME II Problems/Problem 14
  ...er arrangement. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  6 KB (1,098 words) - 14:18, 7 July 2023
• Mock AIME II 2012 Problems
  ...ime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>.
  7 KB (1,274 words) - 21:16, 8 March 2021
• 2017 AMC 8 Problems
  An integer between <math>1000</math> and <math>9999</math>, inclusive, is chosen at random. What is the p
  12 KB (1,771 words) - 21:13, 20 January 2024
• Mock AIME I 2012 Problems
  ...itive integers. Find the remainder when <math>m</math> is divided by <math>1000</math>. ...l list of 50 numbers. Find the remainder when <math>N</math> is divided by 1000.
  7 KB (1,309 words) - 11:13, 8 April 2012
• Mock AIME I 2012 Problems/Problem 4
  ...itive integers. Find the remainder when <math>m</math> is divided by <math>1000</math>. ...5, 3^{31}\right)= 1</math>. So then it suffices to evaluate <math>b_5 \mod 1000</math>.
  1 KB (250 words) - 17:33, 7 April 2012
• Mock AIME I 2012 Problems/Problem 5
  ...l list of 50 numbers. Find the remainder when <math>N</math> is divided by 1000.
  2 KB (335 words) - 17:47, 7 April 2012
• Mock AIME I 2012 Problems/Problem 9
  ...o the desired calculation. We have then that <math>\lfloor 1000P \rfloor = 1000\cdot \frac{1}{8} = \boxed{125}</math>.
  2 KB (285 words) - 17:40, 7 April 2012
• Mock AIME I 2012 Problems/Problem 13
  ...the remainder when the sum of the elements of <math>S</math> is divided by 1000. ...\sum_{k=1}^{499} 2k+1 \right) + \left(\sum_{k=0}^{4} 2^{2k+1} \right) \mod 1000 = \boxed{681}</math>.
  2 KB (445 words) - 17:43, 7 April 2012
• 1974 AHSME Problems/Problem 9
  ...od of <math> 8 </math>. Since <math> 1000\equiv8(\bmod\,8) </math>, <math> 1000 </math> is in the same column as <math> 8 </math>, which is the second colu
  2 KB (222 words) - 12:42, 5 July 2013
• 1952 AHSME Problems
  ...loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \
  23 KB (3,556 words) - 15:35, 30 December 2023
• 1955 AHSME Problems
  ...es } \textdollar{ 100}\qquad\textbf{(C)}\ \text{Mr. A makes } \textdollar{ 1000}\\ \textbf{(D)}\ \text{Mr. B loses } \textdollar{ 100}\qquad\textbf{(E)}\ \
  22 KB (3,509 words) - 21:29, 31 December 2023
• 1971 Canadian MO Problems/Problem 7
  ...f <math>n</math> such that <math>n/m</math> is an integer are multiples of 1000. It is not hard to show that these are all acceptable values.
  1 KB (247 words) - 16:47, 10 June 2020
• 2020 Mock Combo AMC 10
  ...th>S</math> be the sum of all <math>n</math> such that <math>1 \leq n \leq 1000</math> and <math>40 | r_n</math>. <cmath>1 + 11 + 11^2 + 11^3 + … + 11^{2019}</cmath>by <math>1000</math>.
  15 KB (2,452 words) - 03:03, 4 July 2020
• 2012 AMC 8 Problems
  ...05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000 </math> How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
  13 KB (1,835 words) - 08:51, 8 March 2024
• 2012 AMC 8 Problems/Problem 2
  ...05in}800\qquad\textbf{(D)}\hspace{.05in}900\qquad\textbf{(E)}\hspace{.05in}1000 </math>
  947 bytes (139 words) - 21:22, 29 June 2022
• 2012 AMC 8 Problems/Problem 10
  How many 4-digit numbers greater than 1000 are there that use the four digits of 2012? ..., since all of the valid 4-digit number will always be greater than <math> 1000 </math>. The best way to solve this problem is by using casework.
  2 KB (403 words) - 23:53, 10 January 2024
• 2013 AMC 12A Problems/Problem 21
  Since <math>1000 < 2012 < 10000</math>, we know <math>3 < \log(2012) < 4</math>. This gives ...igit numbers, and 1014 4 digit numbers that increase the expression. <math>1000< 90 \cdot 1 + 900 \cdot 2 + 1014 \cdot 3< 10000</math>, so it increases bet
  6 KB (910 words) - 15:30, 30 April 2024
• 2013 AMC 10B Problems/Problem 25
  Notice that there are exactly <math>1000-100=900=5^2\cdot 6^2</math> possible values of <math>N</math>. This means, ...k. Furthermore, when <math>b_1=5</math>, <math>216b_1</math> exceeds <math>1000</math> which is not possible as <math>N</math> is a three digit number, thu
  10 KB (1,623 words) - 15:44, 31 August 2022
• 2013 AMC 10B Problems/Problem 18
  ...ath>. How many integers less than <math>2013</math> but greater than <math>1000</math> have this property?
  3 KB (480 words) - 22:23, 26 March 2023
• 2013 AMC 12B Problems
  <math>\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400</math>
  16 KB (2,459 words) - 02:46, 30 January 2021
• 2013 AMC 12B Problems/Problem 2
  <math>\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400</math>
  1 KB (160 words) - 17:02, 27 January 2015
• 2013 AMC 10B Problems
  <math> \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 </math> ...ath>. How many integers less than <math>2013</math> but greater than <math>1000</math> share this property?
  12 KB (1,926 words) - 21:54, 6 October 2022
• 2013 AIME I Problems
  ...ainder when the smallest possible sum <math>m+n</math> is divided by <math>1000</math>.
  9 KB (1,580 words) - 13:07, 24 February 2024
• 2013 AIME I Problems/Problem 8
  ...he remainder when the smallest possible sum <math>m+n</math> is divided by 1000. ...<math>m+n</math>, which is <math>5368 + 3 = 5371 \equiv \boxed{371} \pmod{1000}</math>.
  4 KB (674 words) - 20:07, 24 January 2021
• 1991 AHSME Problems
  How many years in the millenium between 1000 and 2000 have properties (a) and (b)?
  16 KB (2,451 words) - 04:27, 6 September 2021
• 2013 AIME II Problems
  <cmath>\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.</cmath> ...hat the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer.
  8 KB (1,402 words) - 12:17, 13 March 2020
• 2013 AIME II Problems/Problem 1
  ...a day. The ratio of normal to metric minutes in a day is <math>\frac{1440}{1000}</math>, which simplifies to <math>\frac{36}{25}</math>. This means that ev ...ivalent of <math>\frac{11}{40}</math> of a day is <math>\frac{11}{40}\cdot 1000=275</math> metric minutes, which is equivalent to 2 metric hours and 75 met
  2 KB (337 words) - 19:03, 3 September 2023
• 2013 AIME II Problems/Problem 2
  <cmath>\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.</cmath> ...math>. The largest <math>a</math> such that <math>2^a</math> divides <math>1000</math> is <math>3</math>, so we only need to check <math>1</math>,<math>2</
  2 KB (415 words) - 23:19, 10 January 2024
• 2013 AIME II Problems/Problem 6
  ...hat the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer. ...0N</math> within the difference of the two squares. Examine when <math>n^2=1000</math>. Then, <math>n=10\sqrt{10}</math>. One example way to estimate <math
  5 KB (826 words) - 09:29, 1 January 2023
• 2013 AIME II Problems/Problem 14
  ...ainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>.
  4 KB (637 words) - 08:59, 30 September 2023
• 2013 AIME II Problems/Problem 9
  ...fferent tiling. Find the remainder when <math>N</math> is divided by <math>1000</math>. Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of <math>\boxed{106}</math>.
  6 KB (1,051 words) - 14:14, 21 November 2023
• 2013 AIME II Problems/Problem 11
  ...stant function. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  3 KB (574 words) - 01:22, 1 February 2024
• Mock AIME 6 2006-2007 Problems
  ...math>4</math>. Find the remainder when <math>T</math> is divided by <math>1000</math>. Alvin, Simon, and Theodore are running around a <math>1000</math>-meter circular track starting at different positions. Alvin is runn
  7 KB (1,173 words) - 21:04, 7 December 2018
• 2014 AIME II Problems/Problem 10
  ...is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>. ...3R^2}{4} \equiv 3 \cdot 1007^2 \equiv 3 \cdot 7^2 \equiv \boxed{147} \pmod{1000}</cmath>
  6 KB (1,045 words) - 13:08, 21 January 2024
• Northeastern WOOTers Mock AIME I Problems
  ...<math>T</math>. Find the remainder when <math>T</math> is divided by <math>1000</math>. ...e his earnings. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  8 KB (1,336 words) - 09:10, 30 May 2020
• ASIA TEAM Problems
  How many of the lily pads numbered from <math>1</math> to <math>1000</math> inclusive have an even order? ...of all such <math>n</math>. Find the remainder when n is divided by <math>1000</math>.
  10 KB (1,710 words) - 23:23, 10 January 2020
• Mock AIME I 2015 Problems
  ...mainder when <cmath>k_1+k_2+k_3+\cdots+k_{100}</cmath> is divided by <math>1000</math>. ...th>q</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>.
  9 KB (1,463 words) - 14:48, 12 February 2017
• 1966 AHSME Problems/Problem 29
  The number of positive integers less than <math>1000</math> divisible by neither <math>5</math> nor <math>7</math> is: ...ositive integers divisible be <math>7</math> or <math>5</math> under <math>1000</math> is <math>341-28=313</math>. We can conclude that the number of posit
  1 KB (167 words) - 08:44, 1 January 2024
• 2016 AMC 10A Problems
  For what value <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?
  14 KB (2,104 words) - 22:26, 16 September 2022
• 1962 AHSME Problems/Problem 40
  So.. we have the sum to be <math>\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}</math>...
  2 KB (294 words) - 17:01, 28 June 2023
• 2013 AMC 8 Problems
  What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>?
  15 KB (2,162 words) - 20:05, 8 May 2023
• 2013 AMC 8 Problems/Problem 9
  ...ence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that <math>2^{10}=1024</math>.
  923 bytes (144 words) - 17:56, 5 May 2022
• 2013 AMC 8 Problems/Problem 3
  What is the value of <math>4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)</math>? ...6 - 7 + \ldots + 1000 &= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\
  798 bytes (103 words) - 01:01, 26 December 2022
• 2015 AMC 10B Problems/Problem 16
  <math>\textbf{(A) } \dfrac{9}{1000} \qquad\textbf{(B) } \dfrac{1}{90} \qquad\textbf{(C) } \dfrac{1}{80} \qquad
  2 KB (300 words) - 21:24, 26 December 2020
• 1994 AHSME Problems
  ...f{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}</math>
  14 KB (2,124 words) - 13:39, 19 February 2020
• 1952 AHSME Problems/Problem 14
  ...loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \
  1 KB (173 words) - 20:09, 19 December 2017
• 2014 AMC 10A Problems
  ...has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>. What is <math>k</math>?
  15 KB (2,190 words) - 15:21, 22 December 2020
• 2014 AMC 10A Problems/Problem 20
  ...has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>. What is <math>k</math>?<!-- don't remove the following tag, for Po ...h>04</math>. We want to make the numbers sum to 1000, so <math>7+4+(k-2) = 1000</math>. Solving, we get <math>k = 991</math>, meaning the answer is <math>\
  3 KB (534 words) - 21:30, 9 May 2023
• 2014 AMC 10A Problems/Problem 24
  Clearly, <math>\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}</math>. This means that there are <math>999-3=996</math> skipped ...h>7000</math>, we have to decrease <math>4</math> times leading to <math>n=1000-4=996</math>. We plug it in to <math>n^2+4n</math> getting <math>996\cdot10
  6 KB (893 words) - 21:33, 4 May 2024
• 1994 AHSME Problems/Problem 5
  ...f{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}</math> ...4>1000</math>, our answer is <math>\boxed{\textbf{(E)}\ \text{greater than 1000}}</math>.
  1 KB (179 words) - 17:26, 9 January 2021
• 2000 PMWC Problems/Problem I3
  How many whole numbers less than <math>1000</math> contain at least one <math>2</math> but no <math>3</math>?
  149 bytes (22 words) - 17:16, 20 April 2014
• 2014 AMC 12A Problems
  ...has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>. What is <math>k</math>? ...,y)\in P</math> with integer coordinates is it true that <math>|4x+3y|\leq 1000</math>?
  12 KB (1,863 words) - 19:04, 11 April 2024
• 1967 AHSME Problems/Problem 26
  If one uses only the tabular information <math>10^3=1000</math>, <math>10^4=10,000</math>, <math>2^{10}=1024</math>, <math>2^{11}=20 Since <math>1024</math> is greater than <math>1000</math>.
  1 KB (131 words) - 01:40, 16 August 2023
• 2014 AMC 12A Problems/Problem 25
  ...,y)\in P</math> with integer coordinates is it true that <math>|4x+3y|\leq 1000</math>? ...<math>m = 4x+3y \in \mathbb Z</math>, <math>\left\lvert m \right\rvert \le 1000</math>. Put <math>m = 25k</math> to obtain
  4 KB (661 words) - 16:18, 2 September 2022
• 2014 AIME II Problems
  ...is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>.
  8 KB (1,410 words) - 00:04, 29 December 2021
• 2014 AIME I Problems
  ...ion in lowest terms, the numerator and the denominator have a sum of <math>1000</math>.
  9 KB (1,472 words) - 13:59, 30 November 2021
• 2014 AIME I Problems/Problem 3
  ...fraction in lowest terms, the numerator and the denominator have a sum of 1000. ...Euclidean algorithm <math>\gcd(1000,1000-x)=1</math>, so we want <math>x<y=1000-x.</math> Thus the <math>400</math> relatively prime numbers can generate <
  6 KB (965 words) - 14:17, 27 December 2023
• 2014 AIME I Problems/Problem 8
  <math>200bd+100c^2= 200b \equiv 100b \pmod{1000}</math> <math>200bd+100c^2+100c= 1000b+600 \equiv600\equiv 100b \pmod{1000}</math>
  12 KB (1,962 words) - 08:40, 4 November 2022
• 1993 AHSME Problems/Problem 18
  ...chedule is 7 work-days followed by 3 rest-days. On how many of their first 1000 days do both have rest-days on the same day? ...ys they have in the first 20 days, and multiply that number by <math>\frac{1000}{20}=50</math> to determine the total number of coinciding rest days. Thes
  2 KB (294 words) - 16:49, 9 September 2020
• 1992 AHSME Problems
  the 99-term sequence <math>(a_1,...,a_{99})</math> is 1000, what is the Cesáro sum of the 100-term sequence \text{(C) } 1000\quad
  16 KB (2,548 words) - 13:40, 19 February 2020
• 1968 AHSME Problems/Problem 26
  ...that <math>S = 2(1+2+3+...+N) = N(N +1)</math>. It follows that <math>N = 1000</math>, so the sum of the digits of <math>N</math> is <math>\fbox{E}</math>
  609 bytes (93 words) - 01:53, 16 August 2023
• 1958 AHSME Problems/Problem 32
  With &#036; <math>1000</math> a rancher is to buy steers at &#036; <math>25</math> each and cows a
  768 bytes (117 words) - 06:24, 3 October 2014
• 1978 AHSME Problems
  If <math>a = 1,~ b = 10, ~c = 100</math>, and <math>d = 1000</math>, then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</ma
  15 KB (2,432 words) - 01:06, 22 February 2024
• 1976 AHSME Problems
  ...roduct <math>2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}</math> is greater than <math>1000</math>?
  17 KB (2,835 words) - 14:36, 8 September 2021
• 1972 AHSME Problems
  When the number <math>2^{1000}</math> is divided by <math>13</math>, the remainder in the division is
  17 KB (2,732 words) - 13:54, 20 February 2020
• 2007 iTest Problems/Problem 29
  ...ts+2007</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>. ...} = 2015028</math>. The remainder when <math>S</math> is divided by <math>1000</math> is <math>\boxed{28}</math>.
  601 bytes (83 words) - 19:16, 10 June 2018
• Mock AIME 5 2005-2006 Problems/Problem 11
  ...if <math>k</math> has an even number of divisors. For how many <math>n \le 1000</math> does there exist an <math>A</math> such that <math>|A| = 620</math>
  786 bytes (131 words) - 21:19, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 13
  ...f <math>S</math>. Find the remainder upon dividing <math>A</math> by <math>1000</math>.
  539 bytes (83 words) - 21:20, 8 October 2014
• 1961 AHSME Problems
  He gives the salesman a <math>1000</math> m.u bill, and receives, in change, <math>340</math> m.u. The base <m
  18 KB (2,905 words) - 18:33, 5 April 2023
• UNM-PNM Statewide High School Mathematics Contest
  Approximately 1000 students compete in this contest annually.
  2 KB (225 words) - 03:42, 12 January 2019
• 1993 UNCO Math Contest II Problems
  A collection of <math>25</math> consecutive positive integers adds to <math>1000.</math> What are the smallest and largest integers in this collection?
  4 KB (607 words) - 15:40, 20 October 2014
• 1993 UNCO Math Contest II Problems/Problem 5
  A collection of <math>25</math> consecutive positive integers adds to <math>1000.</math> What are the smallest and largest integers in this collection? The thirteenth integer is the average, which is <math>\frac{1000}{25}=40</math>. So, the largest integer is 12 larger, which is <math>40+12=
  1,003 bytes (141 words) - 01:05, 20 January 2023
• 1998 CEMC Gauss (Grade 7) Problems
  The value of <math>\frac{1998 - 998}{1000}</math> is <math>\text{(A)}\ 1 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 0.1 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 0.001</math
  10 KB (1,590 words) - 16:43, 29 January 2021
• 2016 AMC 12A Problems
  For what value of <math>x</math> does <math>10^x \cdot 100^{2x} = 1000^5</math>?
  15 KB (2,348 words) - 17:20, 19 January 2024
• 2014 AMC 8 Problems/Problem 11
  ...This is not recommended in more complicated problems (e.g. Jill's house is 1000 blocks east and 400 blocks north of Jack's house).
  2 KB (352 words) - 21:47, 6 January 2024
• 1956 AHSME Problems/Problem 4
  ...man currently earns <math>4000 \cdot \frac{5}{1000} + 3500 \cdot \frac{4}{1000} = 340</math> dollars. So, we need to find the value of <math>x</math> such <cmath>2500 \cdot \frac{x}{1000} = 160.</cmath>
  800 bytes (106 words) - 17:11, 14 March 2023
• 1998 CEMC Gauss (Grade 7) Problems/Problem 1
  The value of <math>\frac{1998-998}{1000}</math> is <math> \text{ (A) }\ 1\qquad\text{ (B) }\ 1000\qquad\text{ (C) }\ 0.1\qquad\text{ (D) }\ 10\qquad\text{ (E) }\ 0.001 </mat
  178 bytes (24 words) - 20:59, 20 January 2015
• 2015 AMC 10A Problems
  ...o represent <math>10</math> through <math>15</math>. Among the first <math>1000</math> positive integers, there are <math>n</math> whose hexadecimal repres
  12 KB (1,897 words) - 22:45, 18 March 2024
• 2015 AMC 10A Problems/Problem 18
  ...o represent <math>10</math> through <math>15</math>. Among the first <math>1000</math> positive integers, there are <math>n</math> whose hexadecimal repres Notice that <math>1000</math> is <math>3E8</math> when converted to hexadecimal (<math>3 \cdot 16^
  3 KB (455 words) - 07:19, 31 March 2023
• 2015 AMC 10B Problems
  <math>\textbf{(A) } \dfrac{9}{1000} \qquad\textbf{(B) } \dfrac{1}{90} \qquad\textbf{(C) } \dfrac{1}{80} \qquad
  14 KB (2,247 words) - 11:38, 26 March 2024
• 2015 AIME II Problems/Problem 1
  ...nother integer. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  3 KB (461 words) - 08:44, 17 December 2023
• 2015 AIME I Problems/Problem 8
  ...hen the units digit of <math>N</math> is clearly less than 4. But as <math>1000 < 200 + 864 < N < 600 + 864 = 1464</math>, the sum of the thousands digit a ...the real test, I immediately noticed that <math>n</math> must be less than 1000 (AIME problem) and that <math>n</math> must be a three-digit number. Theref
  10 KB (1,797 words) - 14:06, 21 January 2024
• 2015 AIME I Problems/Problem 12
  Consider all 1000-element subsets of the set <math>\{1, 2, 3, ... , 2015\}</math>. From each ...e desired mean. Then because <math>\dbinom{2015}{1000}</math> subsets have 1000 elements and <math>\dbinom{2015 - i}{999}</math> have <math>i</math> as the
  5 KB (683 words) - 23:01, 9 August 2021
• 2015 AIME I Problems/Problem 14
  ...</math>. Find the number of values of <math>n</math> with <math>2\le n \le 1000</math> for which <math>A(n)</math> is an integer. ...ath>a(n) = \left\lfloor \sqrt n \right\rfloor</math>. For <math>2\le n \le 1000</math>, we have <math>1\le a(n)\le 31</math>.
  6 KB (1,079 words) - 18:58, 5 August 2023
• 2015 AIME II Problems
  ...nother integer. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  8 KB (1,326 words) - 19:15, 13 January 2024
• 2017 AIME I Problems
  <cmath>\sum_{n=1}^{2017} d_n</cmath>is divided by <math>1000</math>. ...\leq 6</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.
  7 KB (1,163 words) - 16:43, 2 June 2022
• 2015 AMC 8 Problems/Problem 10
  How many integers between <math>1000</math> and <math>9999</math> have four distinct digits? ...\times 8 \times 7=\boxed{\textbf{(B) }4536}</math> integers between <math>1000</math> and <math>9999</math> with four distinct digits.
  1 KB (165 words) - 11:22, 23 November 2023
• 2015 AMC 8 Problems
  How many integers between <math>1000</math> and <math>9999</math> have four distinct digits?
  16 KB (2,322 words) - 14:04, 2 February 2024
• 2016 AMC 12A Problems/Problem 25
  ...the last number Bernardo writes for <math>k=6</math> is <math>\left(500000+1000\right)^2</math>, the last for <math>k = 8</math> will be <math>\left(500000 <cmath>f(6) =251000 =250000 +1000</cmath>
  8 KB (1,224 words) - 18:28, 16 November 2022
• 2019 AMC 10A Problems/Problem 2
  We can clearly see that <math>20! \equiv 15! \equiv 0 \pmod{1000}</math>, so <math>20! - 15! \equiv 0 \pmod{100}</math> meaning that the las
  1 KB (182 words) - 12:06, 26 December 2023
• 2016 AMC 10B Problems
  ...e times as many sets of twins as sets of triplets. How many of these <math>1000</math> babies were in sets of quadruplets?
  12 KB (1,868 words) - 17:50, 30 December 2023
• 2016 AMC 10B Problems/Problem 13
  ...e times as many sets of twins as sets of triplets. How many of these <math>1000</math> babies were in sets of quadruplets? 2a + 3b + 4c & = 1000 \\
  2 KB (306 words) - 21:57, 18 April 2022
• 2016 AMC 12B Problems/Problem 22
  For a certain positive integer <math>n</math> less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\over ...tors of <math>999999</math> and <math>9999</math> that are less than <math>1000</math>, we see that <math>n = 297</math> is a solution, so the answer is <m
  4 KB (542 words) - 19:29, 31 October 2021
• 2016 AMC 12B Problems
  For a certain positive integer <math>n</math> less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\over
  14 KB (2,037 words) - 19:09, 29 July 2023
• 2016 AIME II Problems/Problem 9
  ...nteger <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. ...= \max \{x : r^x < 700\}</math>, and define <math>j(r) = \min \{x : r^x > 1000\}</math>. We are looking for values of <math>r</math> such that <math>j(r)
  6 KB (983 words) - 01:18, 2 February 2023
• 2016 AIME II Problems/Problem 11
  ...h> positive divisors. Find the number of positive integers less than <math>1000</math> that are neither <math>7</math>-nice nor <math>8</math>-nice. ...perfect kth power. By PIE, the number of positive integers less than <math>1000</math> that are either <math>1 \pmod 7</math> or <math>1\pmod 8</math> is <
  2 KB (368 words) - 20:09, 20 February 2021
• 2016 AIME II Problems
  ...nteger <math>k</math> such that <math>c_{k-1}=100</math> and <math>c_{k+1}=1000</math>. Find <math>c_k</math>. ...h> positive divisors. Find the number of positive integers less than <math>1000</math> that are neither <math>7</math>-nice nor <math>8</math>-nice.
  8 KB (1,312 words) - 21:16, 3 March 2021
• Mock AIME I 2015 Problems/Problem 8
  ...ath>q</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>. ...dy relatively prime; hence, <math>p+q=1005+2014=3019\equiv\boxed{019}\pmod{1000}.</math>
  2 KB (310 words) - 19:46, 27 March 2016
• Gmaas
  ...blue that shines with the brightness of lightning. EDIT: it's <math>1000^{1000}</math> times brighter than lightning. Gmaasian blue occurs when a meteor h 209. Gmaas is a supreme overlord who must be given<cmath>1000^{1,000,000,000,000,000,000,000^{1,000,000,000,000,000,000,000}}</cmath>mine
  69 KB (11,805 words) - 20:49, 18 December 2019
• 1954 AHSME Problems/Problem 15
  <math>\log(1000)=\log(125)+\log(8)\implies 3=\log(125)+\log(2^3)\implies \log(125)=3-3\log(
  441 bytes (54 words) - 01:26, 28 February 2020
• Mock AIME 4 Pre 2005/Problems
  Determine the remainder when <math>1000!</math> is divided by <math>2003</math>. ...termine the remainder obtained when <math>m + n</math> is divided by <math>1000</math>.
  7 KB (1,094 words) - 15:39, 24 March 2019
• Mock AIME 3 2010 Problems
  is divided by 1000.
  14 KB (2,904 words) - 18:24, 16 May 2017
• 2017 AMC 10B Problems/Problem 16
  ...t 9 = 27</math> with two zeros, and <math>1</math> with three zeros <math>(1000)</math>. Finally, we consider the numbers <math>2000</math> to <math>2017</
  2 KB (216 words) - 11:03, 20 November 2023
• 2017 AIME I Problems/Problem 7
  ...\leq 6</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>. <cmath>\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}</cmath>
  8 KB (1,348 words) - 19:51, 29 January 2024
• 2017 AIME I Problems/Problem 13
  ...nd the remainder when <cmath>\sum_{m = 2}^{2017} Q(m)</cmath>is divided by 1000. ...erefore <cmath>\sum_{m = 2}^{2017} Q(m) \equiv \sum_{m = 2}^{17} Q(m) \mod 1000</cmath>
  3 KB (581 words) - 15:36, 18 January 2024
• 2017 AIME I Problems/Problem 11
  ...h>m = 5</math>. Find the remainder when <math>Q</math> is divided by <math>1000</math>. ...refore, <cmath>Q = 9! \cdot \frac{4}{7} = 207360 \equiv \boxed{360} \pmod {1000} .</cmath>
  6 KB (1,019 words) - 12:40, 24 January 2024
• 2017 AIME I Problems/Problem 14
  ...) = 256</math>. Find the remainder when <math>x</math> is divided by <math>1000</math>. We only wish to find <math>x\bmod 1000</math>. To do this, we note that <math>x\equiv 0\bmod 8</math> and now, by
  4 KB (496 words) - 14:35, 3 January 2024
• 2017 AIME I Problems/Problem 3
  <cmath>\sum_{n=1}^{2017} d_n</cmath>is divided by <math>1000</math>.
  944 bytes (148 words) - 23:37, 6 February 2024
• 2020 AMC 12A Problems
  ...ny <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisi
  14 KB (2,073 words) - 15:15, 21 October 2021
• 2020 AMC 12B Problems
  How many positive integers <math>n</math> satisfy<cmath>\frac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor
  16 KB (2,417 words) - 01:03, 28 April 2022
• 2017 AMC 8 Problems/Problem 20
  An integer between <math>1000</math> and <math>9999</math>, inclusive, is chosen at random. What is the p
  1 KB (202 words) - 19:20, 15 April 2023
• Mock AIME 2 2010 Problems/Problem 2
  ...ath> distinct values, find the remainder when <math>K</math> is divided by 1000.
  473 bytes (79 words) - 23:34, 27 January 2018
• 2018 AMC 10A Problems/Problem 25
  <cmath>c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \
  6 KB (1,001 words) - 16:50, 21 January 2024
• 2018 AMC 10B Problems/Problem 21
  ...46k</math> for some positive integer <math>k.</math> Moreover, since <math>1000\leq n\leq9998,</math> we get <math>2\leq k\leq15.</math> As <math>d>323,</m
  5 KB (778 words) - 11:04, 17 June 2023
• 2018 AIME I Problems/Problem 1
  ...r coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>. This gives us <math>2600</math>, and <math>2600\equiv 600 \bmod{1000}.</math>
  10 KB (1,670 words) - 16:38, 15 January 2024
• 1960 AHSME Problems/Problem 8
  <cmath>2+\frac{5}{10} + \frac{2}{100} + \frac{5}{1000} + \frac{2}{10,000} \ldots</cmath>
  2 KB (219 words) - 19:02, 17 May 2018
• Mock AIME I 2015 Problems/Problem 5
  ...mainder when <cmath>k_1+k_2+k_3+\cdots+k_{100}</cmath> is divided by <math>1000</math> ...he total sum as <math>348550</math>. Taking the remainder after dividng by 1000, we get <math>550</math> as our answer.
  3 KB (586 words) - 15:21, 8 May 2018
• 1961 AHSME Problems/Problem 17
  He gives the salesman a <math>1000</math> m.u bill, and receives, in change, <math>340</math> m.u. The base <m If Jones received <math>340</math> m.u. change after paying <math>1000</math> m.u. for something that costs <math>440</math> m.u., then
  1 KB (188 words) - 01:50, 7 June 2018
• 2018 AIME I Problems
  ...r coefficients. Find the remainder when <math>S</math> is divided by <math>1000</math>. ...a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  8 KB (1,284 words) - 14:35, 9 August 2021
• 2018 AIME I Problems/Problem 6
  ...a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...th>120</math>. The answer is <math>12 \cdot 120 = 1440 = \boxed{440} \pmod{1000}</math>.
  7 KB (1,211 words) - 00:23, 20 January 2024
• 2018 AIME II Problems
  Find the sum of all positive integers <math>b < 1000</math> such that the base-<math>b</math> integer <math>36_{b}</math> is a p
  9 KB (1,385 words) - 00:26, 21 January 2024
• 2018 AIME I Problems/Problem 12
  ...test integer <math>n</math> satisfying <math>\left(\frac{2^n}{3}\right)-1 <1000</math> can be plugged in to get the solution of round <math>\left( \frac{2^
  26 KB (4,044 words) - 13:58, 24 January 2024
• 2018 AIME I Problems/Problem 10
  ...<math>A</math>. Find the remainder when <math>n</math> is divided by <math>1000.</math> ...cmath>\binom{14}4+\binom{14}9+\binom{14}{14} = 3004\equiv \boxed{004}\pmod{1000}.</cmath>
  11 KB (1,934 words) - 12:18, 29 March 2024
• 2018 AIME I Problems/Problem 11
  ...h>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>.
  10 KB (1,448 words) - 06:30, 21 April 2024
• Basic Programming With Python
  while i < 1000: ...oop loops 100 times, until <math>i</math> becomes greater than or equal to 1000.
  33 KB (5,277 words) - 22:14, 3 June 2023
• 2018 AIME II Problems/Problem 3
  Find the sum of all positive integers <math>b < 1000</math> such that the base-<math>b</math> integer <math>36_{b}</math> is a p ...ath>27_{b} = 2b + 7</cmath>. It should also be noted that <math>8 \leq b < 1000</math>.
  3 KB (588 words) - 02:28, 23 December 2023
• 2008 iTest Problems/Problem 14
  Note that <math>10^3 = 1000</math> and <math>20^3 = 8000</math>, so check the first few cubes of number
  543 bytes (76 words) - 13:19, 22 June 2018
• 2008 iTest Problems/Problem 39
  ...th>2008</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>. ...S = 2008</math>, and the remainder when <math>S</math> is divided by <math>1000</math> is <math>\boxed{8}</math>.
  4 KB (562 words) - 19:49, 22 August 2023
• 1973 AHSME Problems/Problem 2
  ...be itself, and cube it to find the volume of that cube. So there are <math>1000-512=\boxed{\textbf{(C) } 488}</math> cubes that have at least one face pain
  1 KB (209 words) - 13:56, 20 February 2020
• 2008 iTest Problems/Problem 56
  ...to biomass for between <math>\textdollar{200}</math> and <math>\textdollar{1000}</math> per year. The energy comes from sawdust, switch-grass, and even lan
  7 KB (1,092 words) - 19:05, 17 December 2021
• 2008 iTest Problems/Problem 66
  ...th> easily by using a calculator, but if we do not, note that <math>1732 < 1000\sqrt{3} < 1733,</math> making <math>\lfloor 1000p \rfloor = 2000 - 1733 = \
  2 KB (359 words) - 09:39, 8 August 2018
• 2000 JBMO Problems/Problem 1
  0 &= (x+y)^3 - 1000 + x^3 + 3x^2y - 3x^2y + 3xy^2 - 3xy^2 + y^3 - 1000 + 30xy \\ &= 2[(x+y)^3 - 1000] - 3x^2y - 3xy^2 + 30xy.
  2 KB (313 words) - 16:17, 24 March 2024
• 2018 AMC 8 Problems/Problem 21
  ...tions: <math>6x+2 = 9y+5 = 11z+7</math>, and we know <math>100 \leq 11z+7 <1000</math>, it gives us <math>9 \leq z \leq 90</math>, which is the range of th
  4 KB (605 words) - 12:31, 23 December 2023
• 2008 iTest Problems/Problem 86
  ...t}}c@{\hspace{3pt}}c@{\hspace{3pt}}c}a^2+b^2&=&c^2+d^2&=&2008,\\ ac&=&bd&=&1000.\end{array} </cmath> ...ath>a^2 + b^2 = 2008</math>, <math>a^2 b^2 = 1000000</math>, so <math>ab = 1000</math>. Thus, <math>a^2 + 2ab + b^2 = 4008</math>, so <math>a+b = \sqrt{400
  1 KB (173 words) - 21:36, 21 November 2018
• 1978 IMO Problems/Problem 1
  We have <math>1978^m\equiv 1978^n\pmod {1000}</math>, or <math>978^m-978^n=1000k</math> for some positive integer <math>
  2 KB (286 words) - 00:21, 21 October 2021
• 2019 AMC 10C Problems
  ...and <math>d</math> are integers no more than 1000 and no less than <math>-1000</math>, is <math>7</math>. ...and <math>d</math> are integers no more than 1000 and no less than <math>-1000</math>, is <math>420</math>.
  12 KB (1,917 words) - 12:14, 29 November 2021
• 2019 AIME II Problems/Problem 8
  ...{3}i</math>. Find the remainder when <math>f(1)</math> is divided by <math>1000</math>. ...}{2} = 2015</math>, so <math>f(1) = 4038 + 2015 = 6053 \implies f(1) \pmod{1000} = \boxed{053}</math>.
  4 KB (706 words) - 22:18, 28 December 2023
• 2020 AMC 10B Problems
  How many positive integers <math>n</math> satisfy<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?
  16 KB (2,497 words) - 21:14, 12 November 2023
• 2019 USAMO Problems
  ...n</math>. Given this information, determine all possible values of <math>f(1000)</math>.
  3 KB (495 words) - 13:47, 22 November 2023
• 2019 USAMO Problems/Problem 1
  ...n</math>. Given this information, determine all possible values of <math>f(1000)</math>. By injectivity, <math>f(1000)</math> is not odd.
  3 KB (538 words) - 18:04, 5 April 2021
• 2019 AIME I Problems
  ...substitutions). Find the remainder when <math>n</math> is divided by <math>1000</math>.
  8 KB (1,331 words) - 06:57, 4 January 2021
• 2019 AMC 10B Problems/Problem 3
  ...seniors, then <cmath>\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500</cmath> ...mber of non-seniors. Then <cmath>\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234</cmath>
  3 KB (484 words) - 16:57, 16 July 2023
• 2019 AIME II Problems
  ...ese conditions. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...rt3i</math>. Find the remainder when <math>f(1)</math> is divided by <math>1000</math>.
  7 KB (1,254 words) - 14:45, 21 August 2023
• 2019 AIME I Problems/Problem 4
  ...substitutions). Find the remainder when <math>n</math> is divided by <math>1000</math>. ...11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)</math>. When taking modulo <math>1000</math>, the last term goes away. What is left is <math>1+11^2=\boxed{122}</
  4 KB (551 words) - 10:32, 5 February 2022
• 2019 AIME II Problems/Problem 5
  ...ese conditions. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...t the people, and the remainder when <math>2520</math> is divided by <math>1000</math> is <math>\boxed{520}</math>. ~[[User:emerald_block|emerald_block]]
  8 KB (1,329 words) - 17:00, 27 January 2024
• Gmaasology
  c. 3000-1000 BC: Some Egyptologists suggest that the goddess Bastet, who was half human
  15 KB (2,297 words) - 03:29, 10 March 2024
• 2019 Mock AMC 10B Problems/Problem 25
  ...math>\sum_{n=0}^{200} \sum_{k=0}^{200} S_{n, k}</math> is divided by <math>1000</math>. And, this gives us <math>2^{201}-1\equiv\boxed{751}\pmod{1000}</math>.
  7 KB (1,148 words) - 17:09, 22 September 2019
• Mock AIME 2 Pre 2005 Problems
  .... Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. (Bridge is a game played with the standard <math>52</math>-card dec ...he expression <cmath>2004^{2003^{2002^{2001}}}</cmath> is divided by <math>1000</math>.
  6 KB (1,052 words) - 13:52, 9 June 2020
• Gods Fighting Sim
  *Life Span: 1000 yrs (immortal) Sight: 1000% (anything in line of sight)
  5 KB (832 words) - 17:10, 9 August 2019
• 2020 AIME I Problems/Problem 7
  size(1000, 100); size(1000, 100);
  7 KB (1,023 words) - 23:56, 13 February 2023
• 1972 AHSME Problems/Problem 31
  When the number <math>2^{1000}</math> is divided by <math>13</math>, the remainder in the division is ...>. However, we find that <math>1000 \equiv 4 \pmod{12}</math>, so <math>2^{1000} \equiv 2^4 = 16 \equiv 3 \pmod{13}</math>, so the answer is <math>\boxed{\
  527 bytes (69 words) - 14:10, 23 June 2021
• Mock AMC 10B Problems
  <math>\mathrm{(A) \ } 100.5\qquad \mathrm{(B) \ } 1000.5\qquad \mathrm{(C) \ } 10,000.5\qquad \mathrm{(D) \ } 100,000.5\qquad \mat ...math>\sum_{n=0}^{200} \sum_{k=0}^{200} S_{n, k}</math> is divided by <math>1000</math>.
  13 KB (2,059 words) - 02:59, 21 January 2021
• 2023 AIME I Problems/Problem 10
  ...rfloor</cmath> is an integer strictly between <math>-1000</math> and <math>1000</math>. For that unique <math>a</math>, find <math>a+U</math>. ...h>UB \in \left[ - 1000, 1000 \right]</math> or <math>LB \in \left[ - 1000, 1000 \right]</math>.
  10 KB (1,578 words) - 09:48, 24 April 2024
• GMAAS
  ...blue that shines with the brightness of lightning. EDIT: it's <math>1000^{1000}</math> times brighter than lightning. Almighty Gmaasian blue occurs when a 209. Almighty Gmaas is a supreme overlord who must be given<cmath>1000^{1,000,000,000,000,000,000,000^{1,000,000,000,000,000,000,000}}</cmath>mine
  85 KB (13,954 words) - 17:25, 22 March 2024
• 2019 AMC 10C Problems/Problem 23
  <math>42\cdot 31^{2018} \pmod {1000} </math> <math>42 \cdot (3 \cdot 10 +1 )^{2018} \pmod {1000}</math>
  1 KB (209 words) - 19:09, 27 January 2020
• 2019 AMC 10C Problems/Problem 22
  ...>8*25=200</math> numbers said by all chicks, so our answer is <math>4\cdot{1000/200}+1=\boxed{21}</math>
  1 KB (232 words) - 20:16, 5 November 2022
• 1978 AHSME Problems/Problem 4
  If <math>a = 1,~ b = 10, ~c = 100</math>, and <math>d = 1000</math>, then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</ma
  549 bytes (86 words) - 12:00, 13 February 2021
• 2000 SMT/Team Problems
  Find the sum of all integers from 1 to 1000 inclusive which contain at least one 7 in their digits, i.e. find 7 + 17 +
  4 KB (618 words) - 13:33, 21 January 2020
• 2020 AMC 10A Problems/Problem 6
  ...ny <math>4</math>-digit positive integers (that is, integers between <math>1000</math> and <math>9999</math>, inclusive) having only even digits are divisi
  1 KB (188 words) - 04:01, 30 March 2024
• 2020 AMC 10A Problems/Problem 22
  ...right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>not divisible by <math>3</math>? (Recall that <mat <cmath>\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor</cmath>
  12 KB (1,978 words) - 22:13, 29 September 2023
• 2020 AMC 10A Problems/Problem 24
  Let <math>n</math> be the least positive integer greater than <math>1000</math> for which ...alpha \equiv{14} \pmod{40}</math>. We are given that <math>n=21\alpha -57 >1000</math>, so <math>\alpha \geq 51</math>. Notice that if <math>\alpha =54</ma
  17 KB (2,544 words) - 12:09, 1 September 2023
• 2020 AMC 10B Problems/Problem 24
  How many positive integers <math>n</math> satisfy <cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor <cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath>
  10 KB (1,550 words) - 10:26, 9 March 2024
• 2020 AMC 10B Problems/Problem 12
  ...2</math>, which means that <math>2^{20}</math> is a little more than <math>1000^2=1,000,000</math>. Multiplying it with <math>10^{20}</math>, we get that t (1000 + 24)^2 &= 1024^2 \\
  6 KB (918 words) - 19:09, 22 June 2023
• 2020 AIME II Problems
  ...<math>9</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...eq 2020</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  7 KB (1,199 words) - 16:17, 12 March 2021
• 2023 AIME I Problems/Problem 15
  Find the largest prime number <math>p<1000</math> for which there exists a complex number <math>z</math> satisfying <cmath>1000>p=a^2+b^2\geq 12b^2+b^2=13b^2</cmath>so <math>b^2<81</math>, and <math>b<9<
  8 KB (1,299 words) - 17:37, 3 June 2023
• 2024 AIME II Problems
  ...tient and remainder, respectively, when <math>N</math> is divided by <math>1000</math>. Find <math>Q+R</math>. is divided by 1000.
  8 KB (1,236 words) - 23:11, 12 March 2024
• 2021 JMPSC Accuracy Problems/Problem 15
  ...f(2)+ \cdots + f(100)}{25}\equiv19+191+911+111 \cdot 97 \equiv 11888 \pmod{1000} \rightarrow \boxed{888}.</cmath> ...3</math> digits to get <cmath>20+28(10+99+980+700)+4(444)=\boxed{888} \mod 1000</cmath>
  2 KB (304 words) - 01:19, 12 July 2021
• AMC 12C 2020
  <math>\mathrm{(A) \ } -1000\qquad \mathrm{(B) \ } 1290\qquad \mathrm{(C) \ } 4356\qquad \mathrm{(D) \ } Let <math>N</math> be <math>10^{100^{1000^{10000…}}}</math>. (All the way till the number consisting of <math>100</
  9 KB (1,450 words) - 18:33, 21 April 2020
• AMC 12C 2020 Problems
  ...math>R(1) = \frac{1}{2}.</math> What is <math>R(1) + R(2) + R(3) + … + R(1000)</math>?
  8 KB (1,223 words) - 15:02, 27 November 2022
• AIME 2020(MOCK) Problems
  Let <math>N</math> be <math>112123123412345... (1000 digits)</math>. What is the remainder when <math>N</math> is divided by <ma
  891 bytes (139 words) - 06:45, 23 December 2020
• 2020 AIME II Problems/Problem 5
  ...<math>9</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...ving trouble with this step, note that <math>2^{10} = 1024 \equiv 24 \pmod{1000}</math>) ~ TopNotchMath
  3 KB (414 words) - 22:01, 23 March 2023
• 2020 AIME II Problems/Problem 14
  ...eq 2020</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...dot 2004 \cdot 2003}{3 \cdot 2\cdot 1} \equiv 10 (\mathrm{mod} \hskip .2cm 1000)</cmath>
  7 KB (1,186 words) - 15:31, 5 January 2024
• Sum Product Conjecture
  ...4</math> can be computed by hand, since <math>\tfrac{81}4\cdot 4^2 = 324 < 1000</math>. For the inductive step, write
  6 KB (990 words) - 10:47, 12 June 2020
• 2020 FMC 10A Problems
  ...57) + S(2, 257) + S(3, 257) + ... + S(256, 257)</cmath>is divided by <math>1000</math>. ...of the company. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  14 KB (2,267 words) - 12:49, 9 June 2020
• 2020 FMC 12A Problems
  ...57) + S(2, 257) + S(3, 257) + ... + S(256, 257)</cmath>is divided by <math>1000</math>. Find the value of <math>S</math> modulo <math>1000.</math>
  15 KB (2,388 words) - 13:24, 9 June 2020
• 2021 AIME I Problems/Problem 3
  Find the number of positive integers less than <math>1000</math> that can be expressed as the difference of two integral powers of <m ...math> and <math>b</math>, the value of <math>2^a - 2^b</math> is less than 1000.
  7 KB (1,174 words) - 08:21, 13 May 2023

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