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Mock AIME 2 Pre 2005 Problems/Problem 8
...he expression <cmath>2004^{2003^{2002^{2001}}}</cmath> is divided by <math>1000</math>. ...te that <math>2004^{2003^{2002^{2001}}} \equiv 4^{2003^{2002^{2001}}}\pmod{1000}</math>. The remainder of the RHS modulo <math>8</math> is trivially zero,

1 KB (188 words) - 12:01, 10 August 2020
2021 AIME I Problems
Find the number of positive integers less than <math>1000</math> that can be expressed as the difference of two integral powers of <m

7 KB (1,182 words) - 14:54, 13 March 2023
2023 AIME II Problems
...s the same forward and backward. Find the greatest integer less than <math>1000</math> that is a palindrome both when written in base ten and when written ...the number of positive integers <math>n</math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer who

8 KB (1,370 words) - 21:34, 28 January 2024
2020 CIME I Problems/Problem 13
...ath>q</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>.

648 bytes (99 words) - 14:51, 22 December 2023
2022 AIME I Problems/Problem 3
draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red);

24 KB (3,832 words) - 20:59, 2 March 2024
2022 AIME I Problems/Problem 12
1000. ...\cdot 41 + 21^2 \pmod{1000} \equiv 1804+441 \pmod{1000} \equiv 2245 \pmod{1000} \equiv \boxed{245}</math>

9 KB (1,471 words) - 16:41, 1 February 2024
2022 AIME I Problems/Problem 13
...}{3333}.</math> Find the remainder when <math>N</math> is divided by <math>1000.</math> ...\bf{6000}</cmath> values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer)

7 KB (1,207 words) - 21:53, 2 January 2024
2022 AIME II Problems
...m{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}</cmath>is divided by <math>1000</math>. ...hat contains sub-collections worth every whole number of cents up to <math>1000</math> cents, let <math>f(a, b, c)</math> be the minimum number of stamps i

8 KB (1,397 words) - 09:18, 15 August 2022
2020 CIME I Problems
...<math>q</math>. Find the remainder when <math>p</math> is divided by <math>1000</math>. ...ath>3</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>.

7 KB (1,188 words) - 18:00, 31 August 2020
2020 CIME I Problems/Problem 7
...<math>q</math>. Find the remainder when <math>p</math> is divided by <math>1000</math>. ...Solving the system of congruences, we get <math>p \equiv \boxed{312}\pmod{1000}</math>. ~rocketsri (based off of official solution)

2 KB (282 words) - 12:05, 2 February 2021
2020 CIME I Problems/Problem 8
...ath>3</math>. Find the remainder when <math>p+q</math> is divided by <math>1000</math>. ...56</math> and <math>q=20</math>. Taking the remainder of <math>\frac{1176}{1000}</math>, we get <math>176</math>.

3 KB (487 words) - 21:23, 30 January 2023
2020 CIME I Problems/Problem 10
...+d_4^4.</cmath> Find the remainder when <math>S</math> is divided by <math>1000</math>.

1 KB (239 words) - 11:47, 2 February 2021
2020 CIME I Problems/Problem 12
...ssed in binary. Find the remainder when <math>S</math> is divided by <math>1000</math>.

505 bytes (80 words) - 16:20, 31 August 2020
2022 AMC 12B Problems
&\{991,992,993,\ldots,1000\}.

15 KB (2,233 words) - 13:02, 10 November 2023
2021 AIME II Problems
...<math>n</math> for which <math>2^n + 5^n - n</math> is a multiple of <math>1000</math>.

8 KB (1,429 words) - 14:31, 26 February 2024
2020 CIME II Problems
The probability a randomly chosen positive integer <math>N<1000</math> has more digits when written in base <math>7</math> than when writte ...ve positive integers. Find the number of positive integers less than <math>1000</math> that are either <math>9</math>-consecutive or <math>11</math>-consec

8 KB (1,298 words) - 18:32, 7 January 2021
2020 CIME II Problems/Problem 4
The probability a randomly chosen positive integer <math>N<1000</math> has more digits when written in base <math>7</math> than when writte ...math>. If <math>k \geq 4</math>, <math>N</math> will be greater than <math>1000</math>, so we only need to consider <math>k \le 3</math>. The number of pos

2 KB (242 words) - 00:49, 19 December 2020
2020 CIME II Problems/Problem 5
...ve positive integers. Find the number of positive integers less than <math>1000</math> that are either <math>9</math>-consecutive or <math>11</math>-consec ...ath>1+2+3+4+5+6+7+8+9+10+11=66</math>, and the largest one less than <math>1000</math> is <math>990</math>. There are <math>\frac{990-66}{11}+1=85</math> o

2 KB (222 words) - 13:13, 6 September 2020
2021 AIME II Problems/Problem 1
&=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\

5 KB (794 words) - 15:36, 30 August 2022
1955 AHSME Problems/Problem 26
...es } \textdollar{ 100}\qquad\textbf{(C)}\ \text{Mr. A makes } \textdollar{ 1000}\\ \textbf{(D)}\ \text{Mr. B loses } \textdollar{ 100}\qquad\textbf{(E)}\ \

1 KB (227 words) - 13:01, 18 November 2020
2013 Mock AIME I Problems
...mainder when the sum of the elements in <math>S</math> is divided by <math>1000</math>.

7 KB (1,149 words) - 17:16, 15 December 2020
2021 MECC Mock AMC 10
...ath>147_{-16}</math> expressed in base <math>10</math> is divided by <math>1000</math>.

12 KB (1,915 words) - 17:38, 29 April 2021
2021 CMC 12A Problems
...</math> such that <cmath>\frac{\tfrac{1}{999}+\tfrac{1}{1001}}{2}=\frac{1}{1000}+\frac{1}{N}</cmath> What is the sum of the digits of <math>N</math>?

5 KB (776 words) - 09:35, 8 August 2023
Mock AIME 2 Pre 2005 Problems/Problem 9
...mine the remainder obtained when <math>a_{1997}</math> is divided by <math>1000</math>.

2 KB (232 words) - 00:22, 1 January 2021
2021 CMC 12A Problems/Problem 4
...</math> such that <cmath>\frac{\tfrac{1}{999}+\tfrac{1}{1001}}{2}=\frac{1}{1000}+\frac{1}{N}.</cmath> What is the sum of the digits of <math>N</math>? ...or <math>\frac{1}{a^{3}-a}=\frac{1}{N}</math>, so <math>N=a^{3}-a=1000^{3}-1000=999,999,000</math> whose digits sum to <math>\boxed{\textbf{(C) } 54}</math

898 bytes (129 words) - 00:41, 25 March 2023
2022 AMC 12A Problems
...{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}</math>

15 KB (2,250 words) - 00:32, 9 March 2024
1998 CEMC Gauss (Grade 7) Problems/Problem 25
...ir, <math>p</math> and <math>q,</math> is a power of 10 (that is, 10, 100, 1000, 10 000 , ...). If <math>p > q</math> , the last digit of <math>p-q</math>

898 bytes (157 words) - 18:59, 7 April 2024
1985 AJHSME Problem 2
We know that <math>10(90) = 900</math> and <math>10(100) = 1000.</math> Quick estimation reveals that this sum is in between these two numb

891 bytes (111 words) - 08:36, 7 January 2023
1985 AJHSME Problem 18
<cmath>9x < 10 \rightarrow 900x < 1000</cmath>

880 bytes (126 words) - 16:03, 4 July 2023
2021 GCIME Problems
...which <math>\frac{H(n)}{n}</math> is the minimum amongst all <math>1<n\leq 1000</math>. ...gers, then what is the remainder when <math>m+n</math> is divided by <math>1000</math>?

4 KB (651 words) - 20:18, 6 March 2021
2021 GCIME Problems/Problem 5
...gers, then what is the remainder when <math>m+n</math> is divided by <math>1000</math>?

1 KB (165 words) - 14:45, 6 March 2021
2021 AIME II Problems/Problem 13
...<math>n</math> for which <math>2^n + 5^n - n</math> is a multiple of <math>1000</math>. Recall that <math>1000</math> divides this expression if <math>8</math> and <math>125</math> both

16 KB (2,240 words) - 23:16, 26 January 2024
2022 AIME I Problems
...ive integers. Find the remainder when <math>p+q</math> is divided by <math>1000.</math> ...}{3333}.</math> Find the remainder when <math>N</math> is divided by <math>1000.</math>

9 KB (1,520 words) - 19:06, 2 January 2023
2021 GMC 12
...<math>n</math>. Find the remainder when <math>n</math> is divided by <math>1000</math>. (Examples: <math>AAAAABCBCBCBCBC</math> and <math>ACCBBBBBACACCAA</

13 KB (2,097 words) - 17:38, 29 April 2021
G285 2021 MC10B
...(B)}\ 820\qquad\textbf{(C)}\ 900\qquad\textbf{(D)}\ 940\qquad\textbf{(E)}\ 1000</math>

4 KB (675 words) - 14:01, 28 May 2021
2021 GMC 12B
Find the remainder when <math>3^{1624}+7^{1604}</math> is divided by <math>1000</math>. ...Find the remainder when the total ways of arrangements is divided by <math>1000</math>.

11 KB (1,691 words) - 18:56, 25 April 2022
2022 AIME II Problems/Problem 10
...m{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}</cmath>is divided by <math>1000</math>. & \equiv \boxed{004}\pmod{1000}\

8 KB (1,098 words) - 23:31, 30 January 2024
2022 AIME II Problems/Problem 14
...hat contains sub-collections worth every whole number of cents up to <math>1000</math> cents, let <math>f(a, b, c)</math> be the minimum number of stamps i ...{c} \rfloor</math> stamps of value <math>c</math>, every value up to <math>1000</math> can be represented.

6 KB (860 words) - 19:05, 9 February 2024
G285 2021 MC-IME II
...\phi(n) \cdot l</math> for some constant <math>l</math>, find <math>l \mod 1000</math>

641 bytes (108 words) - 20:58, 4 July 2021
2023 IOQM Problems
Let <math>n</math> be a positive integer such that <math>1 \leq n \leq 1000</math>. Let <math>M_n</math> be the number of integers in the set ...M_n : 1 \leq n \leq 1000\}</math>, and <math>b = \min\{M_n : 1 \leq n \leq 1000\}</math>.

7 KB (1,225 words) - 03:05, 4 May 2024
G285 2021 Summer Problem Set
...idered unique, find the remainder when <cmath>\sum_{p=1}^{1000}\sum_{k=1}^{1000} S_{p,k}</cmath> is divided by <math>100</math>.

9 KB (1,577 words) - 23:28, 28 June 2021
2022 AMC 10B Problems
&\{991,992,993,\ldots,1000\}.

15 KB (2,224 words) - 13:10, 20 February 2024
2023 AIME I Problems/Problem 7
...tinct. Find the number of extra-distinct positive integers less than <math>1000</math>. ...of our pattern, we know that the numbers from <math>961</math> thru <math>1000</math> will have the same remainders as <math>1</math> thru <math>40</math>

8 KB (1,127 words) - 17:55, 31 October 2023
1976 AHSME Problems/Problem 21
...roduct <math>2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}</math> is greater than <math>1000</math>? ...t positive odd integer value of <math>n</math>: <cmath>2^{\frac{n^2}{7}} > 1000.</cmath>

1 KB (166 words) - 20:54, 19 July 2021
2021 Fall AMC 12B Problems/Problem 18
...the least value of <math>k</math> such that <cmath>|u_k-L| \le \frac{1}{2^{1000}}?</cmath> ...{2^{1000}},</cmath> and we only need to consider <math>\frac12-\frac{1}{2^{1000}} \leq u_k.</math>

5 KB (733 words) - 10:36, 5 November 2022
2021 Fall AMC 10B Problems/Problem 19
<math>f(5) - \sqrt[5]{\frac{7}{9} \cdot 1000} = \sqrt[5]{\frac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3</math>. ...ffect the leading digit of a number. Similarly, <cmath>B(\sqrt[3]{\frac{n}{1000}}) = B(\sqrt[3]{n}).</cmath>

7 KB (1,032 words) - 14:16, 13 November 2023
2021 Fall AMC 12B Problems/Problem 1
...+2+3+4=10</math>, we find that the sum is equal to <cmath>10\cdot(1+10+100+1000)=\boxed{\textbf{(E)} \: 11{,}110}.</cmath>

2 KB (292 words) - 01:48, 30 January 2024
2021 Fall AMC 12A Problems
...math> terms in the expansion of <math>\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?</math>

15 KB (2,452 words) - 19:37, 7 June 2023
2021 ZeMC10 Problems/Problem 3
<math>\frac{9}{10}*\frac{1}{10}*\frac{1}{10}=\frac{9}{1000}</math> <math>\frac{9}{1000} * 3 = \frac{27}{1000}</math>

928 bytes (133 words) - 19:50, 1 September 2022
2021 Fall AMC 12A Problems/Problem 1
<cmath>\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.</cmath>

2 KB (245 words) - 12:23, 21 July 2023
2021 Fall AMC 12A Problems/Problem 12
...math> terms in the expansion of <math>\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?</math> ...3^{\frac{1000-k}{2}}x^k y^{1000-k},</cmath> where <math>k\in\{0,1,2,\ldots,1000\}.</math>

1 KB (182 words) - 22:01, 7 April 2022
2021 Fall AMC 12B Problems
...the least value of <math>k</math> such that <cmath>|u_k-L| \le \frac{1}{2^{1000}}?</cmath>

14 KB (2,191 words) - 19:57, 12 November 2023
2020 CIME II Problems/Problem 9
...ldots)).</cmath>Find the remainder when <math>N</math> is divided by <math>1000</math>. ...\mod 1000</math>. Hence <math>3\cdot 2^{2018}+1\equiv 3\cdot 144+1=433\mod 1000</math>. Thus the answer is 433 as desired.

7 KB (1,282 words) - 03:48, 23 May 2024
Mock AIME 6 2006-2007 Problems/Problem 3
Alvin, Simon, and Theodore are running around a <math>1000</math>-meter circular track starting at different positions. Alvin is runn

929 bytes (140 words) - 10:20, 23 November 2023
2021 GMC 10B Problems/Problem 21
Find the remainder when <math>3^{1624}+7^{1604}</math> is divided by <math>1000</math>. Since <math>\varphi(1000) = 400</math>, we have

930 bytes (102 words) - 12:50, 11 April 2022
2023 AIME I Problems
...tinct. Find the number of extra-distinct positive integers less than <math>1000</math>. ...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>.

7 KB (1,154 words) - 12:54, 20 February 2024
Crusade
...f a battle has a gold limit of 1000 G, then there can only be a maximum of 1000 G spent.

36 KB (5,892 words) - 20:26, 11 September 2022
Mock AIME 6 2006-2007 Problems/Problem 1
...math>4</math>. Find the remainder when <math>T</math> is divided by <math>1000</math>. which is <math>\framebox{751}</math> mod 1000.

464 bytes (68 words) - 10:17, 23 November 2023
2022 AMC 10A Problems/Problem 6
We test <math>x = - 1000</math> and get <math>\left|-1000-2- \text{positive} \right| \ne 3 \implies \boxed{\textbf{(A) } 3-2a}.</math

2 KB (293 words) - 12:13, 25 December 2023
2022 AMC 12A Problems/Problem 8
...{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}</math>

3 KB (345 words) - 12:34, 25 October 2023
2022 AMC 10B Problems/Problem 22
pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0];

7 KB (1,202 words) - 01:15, 10 June 2023
2022 AMC 10B Problems/Problem 6
Factoring each of the sums, we have <cmath>11(10+1), 111(100+1), 1111(1000+1), \ldots</cmath> respectively. With each number factored, there are <math

3 KB (361 words) - 12:18, 20 March 2024
2022 AMC 10B Problems/Problem 8
&\{991,992,993,\ldots,1000\}. &\{991,992,993,\ldots,1000\}.

4 KB (587 words) - 17:40, 11 November 2023
2022 MMATHS Individual Round Problems/Problem 1
...\frac {1010-(-990)}{1012-1010}</math> gives us the answer of <math>\boxed {1000}</math>.

549 bytes (86 words) - 20:58, 18 December 2022
2009 MMATHS Individual Round Problems/Problem 1
...\frac {1010-(-990)}{1012-1010}</math> gives us the answer of <math>\boxed {1000}</math>.

422 bytes (66 words) - 21:05, 18 December 2022
2023 AIME I Problems/Problem 14
...itial position. Find the remainder when <math>N</math> is divided by <math>1000</math>.

11 KB (1,834 words) - 22:01, 4 January 2024
2023 AIME II Problems/Problem 2
...s the same forward and backward. Find the greatest integer less than <math>1000</math> that is a palindrome both when written in base ten and when written ...ath>72</math> to <math>513</math> until we get palindromes less than <math>1000:</math>

1 KB (174 words) - 20:32, 8 June 2023
2023 AIME II Problems/Problem 13
...the number of positive integers <math>n</math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer who For <math>n \leq 1000</math>, because <math>n = 4m</math>, we only need to count feasible <math>m

5 KB (881 words) - 04:52, 19 December 2023
2023 AIME II Problems/Problem 15
...the number of positive integers <math>n</math> less than or equal to <math>1000</math> that satisfy <math>a_n = a_{n+1}.</math> We have <math>1000 = 90 \cdot 11 + 10</math>.

7 KB (1,144 words) - 08:13, 1 February 2024
2013 TSTST Problems
...e maximum possible number of forced terms in a regular sequence with <math>1000</math> terms.

5 KB (873 words) - 08:45, 24 February 2023
2023 AMC 10A Problems/Problem 8
...We get <math>b = \frac{-275}{6}</math>. Plugging in <math>(200, y), \frac{1000}{12}-\frac{550}{12}=y</math>. Simplifying, <math>\frac{450}{12} = \boxed{\t

4 KB (628 words) - 06:57, 28 January 2024
Xosomiennamxsmn
<p><span style="color: #000000; font-weight: 400;">1000 giải Tám 2 chữ số, 100 nghìn đồng mỗi giải.</spa

11 KB (2,230 words) - 23:24, 9 May 2023
2022 SSMO Relay Round 1 Problems
...the product of all possible values of <math>BC</math> is divided by <math>1000</math>.

1 KB (202 words) - 21:55, 31 May 2023
2022 SSMO Relay Round 3 Problems
...\dots,f(21)\}</math>. Find the remainder when <math>N</math> is divided by 1000. ...=T</math>. Find the remainder when <math>f(13)</math> is divided by <math>1000</math>.

1 KB (182 words) - 21:57, 31 May 2023
2022 SSMO Relay Round 4 Problems
...1)^{n}g_n(5),</cmath> find the remainder when <math>S</math> is divided by 1000. ...th>n</math>, find the remainder when <math>f(T)</math> is divided by <math>1000.</math>

1 KB (260 words) - 21:58, 31 May 2023
Mock AIME 3 2006-2007 Problems/Problem 15
...ivided by k it leaves remainder n. Find the remainder when N is divided by 1000.

210 bytes (41 words) - 09:51, 18 June 2023
Xarcade! (c)
...end of April, bpan2021 was the first official user of Xarcade! to get past 1000 XB, beating the previous record holder, LAFILLEDEPARIS, who had around 650

6 KB (929 words) - 14:54, 5 March 2024
Know the Facts: Gmaas
c. 3000-1000 BC: Some Egyptologists suggest that the goddess Bastet, who was half human ...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 hi

88 KB (14,928 words) - 13:54, 29 April 2024
2024 AMC 8 Problems/Problem 23
...cd}(3000, 5000)-1=999</math> times. This partitions the segment into <math>1000</math> congruent pieces that each pass through <math>f(3, 5)</math> cells, ...that the line crossed 7 blocks in this pattern. Such a pattern is repeated 1000 times between <math>(2000,3000)</math> and <math>(5000,8000)</math>, then t

7 KB (1,167 words) - 15:40, 19 May 2024
2023 IOQM/Problem 1
Let <math>n</math> be a [[positive integer]] such that <math>1 \leq n \leq 1000</math>. Let <math>M_n</math> be the number of [[integers]] in the [[set]] ...M_n : 1 \leq n \leq 1000\}</math>, and <math>b = \min\{M_n : 1 \leq n \leq 1000\}</math>.

8 KB (670 words) - 22:07, 27 April 2024
2023 AMC 12A Problems/Problem 12
<math>=8-1+64-27+216-125+512-343+1000-729+1728-1331+2744-2197+4096-3375+5832-4913</math>

5 KB (604 words) - 09:40, 7 April 2024
2024 AMC 8 Problems/Problem 22
...layers with the diameter <math>2</math>, the length should be <math>\frac{1000}{0.015}2\pi\approx 400</math>. If the diameter is seem as <math>4</math>, t

4 KB (686 words) - 22:06, 27 April 2024
Mock AIME 6 2006-2007 Problems/Problem 4
...</math> points are <math>13</math> consecutive vertices of a regular <math>1000-gon</math> (because every triangle out of these <math>13</math> points has

878 bytes (151 words) - 02:42, 3 January 2024
Mock AIME 6 2006-2007 Problems/Problem 7
<math>5, 10, 15, 20, 25,\cdots, 1000</math> gives '''200''' terms

3 KB (442 words) - 20:51, 26 November 2023
Mock AIME 6 2006-2007 Problems/Problem 12
..._k),</cmath> what is the remainder when <math>S</math> is divided by <math>1000</math>? Therefore, <math>S=2007-1=2006 \equiv 6\;(mod\;1000)</math>

1 KB (196 words) - 21:17, 26 November 2023
Divisibility rules/Rule 1 for 7 proof
...3k-1} + 10 a_{3k-2} + a_{3k-3}) \cdots + 1000^2 (100 a_8 + 10 a_7 + a_6) + 1000 (100 a_5 + 10 a_4 + a_3) + (100 a_2 + 10 a_1 + a_0).</math> <math>N = 1000^{k-1} (a_{3k-1}a_{3k-2}a_{3k-3}) \cdots + 1000^2 (a_8a_7a_6) + 1000 (a_5a_4a_3) + (a_2a_1a_0).\\

1 KB (207 words) - 22:17, 25 November 2023
Divisibility rules/Rule 2 for 13 proof
...3k-1} + 10 a_{3k-2} + a_{3k-3}) \cdots + 1000^2 (100 a_8 + 10 a_7 + a_6) + 1000 (100 a_5 + 10 a_4 + a_3) + (100 a_2 + 10 a_1 + a_0).</math> <math>= 1000^{k-1} (a_{3k-1}a_{3k-2}a_{3k-3}) \cdots + 1000^2 (a_8a_7a_6) + 1000 (a_5a_4a_3) + (a_2a_1a_0).</math>

2 KB (241 words) - 11:34, 25 November 2023
1999 OIM Problems/Problem 1
Find all positive integers that are less than 1000 and satisfy the following condition: the cube of the sum of their digits is

329 bytes (51 words) - 15:59, 13 December 2023
2022 SSMO Relay Round 1 Problems/Problem 3
...the product of all possible values of <math>BC</math> is divided by <math>1000</math>.

291 bytes (48 words) - 13:09, 14 December 2023
2022 SSMO Relay Round 3 Problems/Problem 1
...\dots,f(21)\}</math>. Find the remainder when <math>N</math> is divided by 1000.

413 bytes (73 words) - 13:10, 14 December 2023
2022 SSMO Relay Round 3 Problems/Problem 3
...=T</math>. Find the remainder when <math>f(13)</math> is divided by <math>1000</math>.

273 bytes (47 words) - 13:11, 14 December 2023
2022 SSMO Relay Round 4 Problems/Problem 2
...-1)^{n}g_n(5),</cmath>find the remainder when <math>S</math> is divided by 1000.

286 bytes (57 words) - 13:11, 14 December 2023
2022 SSMO Relay Round 4 Problems/Problem 3
...th>n</math>, find the remainder when <math>f(T)</math> is divided by <math>1000.</math>

255 bytes (44 words) - 13:11, 14 December 2023
2023 SSMO Team Round Problems
...ath>0</math> as your answer. If you believe the sum is infinity, put <math>1000</math> as your answer.

8 KB (1,238 words) - 02:14, 3 January 2024
2023 SSMO Team Round Problems/Problem 6
...ath>0</math> as your answer. If you believe the sum is infinity, put <math>1000</math> as your answer.

353 bytes (60 words) - 22:25, 15 December 2023
2002 AMC 12P Problems
\text{(E) }\frac{1}{1000} \text{(A) }1000

10 KB (1,606 words) - 01:46, 31 December 2023
2002 AMC 12P Problems/Problem 14
...in the imaginary part. Therefore, the sum becomes <math>(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}</math>.

2 KB (313 words) - 20:07, 10 March 2024
2002 AMC 12P Problems/Problem 15
\text{(E) }\frac{1}{1000} ...th> marbles remaining that match that color. Therefore, <math>P_s = \frac {1000}{2001}</math>.

1 KB (191 words) - 19:16, 10 March 2024
2002 AMC 12P Problems/Problem 20
\text{(A) }1000

784 bytes (103 words) - 12:30, 28 March 2024
2024 AMC 8 Problems/Problem 15
...F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\u ...B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U}~\underline{G}) + \underline{B}~\underline{U}~\u

4 KB (559 words) - 01:46, 18 April 2024
Problems Collection
Find the remainder when <math>\min{x}</math> is divided by <math>1000</math>. ...the second peg. Find the remainder when <math>x</math> is divided by <math>1000</math>.

43 KB (7,006 words) - 14:24, 19 February 2024
2024 AIME II Problems/Problem 7
...tient and remainder, respectively, when <math>N</math> is divided by <math>1000</math>. Find <math>Q+R</math>. ...-1)</math>, or <math>d-1</math>. Thus, <math>1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}</math>. We can case

5 KB (777 words) - 03:57, 10 February 2024
2024 AIME II Problems/Problem 13
is divided by 1000. ...^2 = \big| -65 - 64i \big|^2 = 65^2 + 64^2 = 8321 \equiv \boxed{321} \pmod{1000}.

5 KB (771 words) - 10:36, 19 April 2024
Mock AIME 2 Pre 2005 Problems/Problem 7
Given \(\binom{26}{13} \mod 1000 = 576\): ...76 \equiv 6 \times 331776 \equiv 6 \times 776 \equiv 4560 \equiv 560 \pmod{1000}

543 bytes (64 words) - 16:34, 24 May 2024
2023 CMO Problems/Problem 4
...y to see that <math>|B| \leq 100</math>. Since <math>\sum_{i=1}^{2023} a_i>1000</math>, there is a solution!

2 KB (283 words) - 05:36, 25 May 2024
2023 CMO(CHINA) Problems/Problem 4
...y to see that <math>|B| \leq 100</math>. Since <math>\sum_{i=1}^{2023} a_i>1000</math>, there is a solution!

2 KB (283 words) - 06:05, 25 May 2024

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