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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><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

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