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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>aaa || <math>1000 - x</math> || <math>2x - 1000</math>a || <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 $<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 $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></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

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