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• 2020 AMC 10B Problems/Problem 24
  How many positive integers <math>n</math> satisfy <cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor <cmath>\dfrac{n+1000}{70} = \sqrt{n} </cmath>
  10 KB (1,550 words) - 10:26, 9 March 2024
• 2020 AMC 10B Problems/Problem 12
  ...2</math>, which means that <math>2^{20}</math> is a little more than <math>1000^2=1,000,000</math>. Multiplying it with <math>10^{20}</math>, we get that t (1000 + 24)^2 &= 1024^2 \\
  6 KB (918 words) - 19:09, 22 June 2023
• 2020 AIME II Problems
  ...<math>9</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...eq 2020</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  7 KB (1,199 words) - 16:17, 12 March 2021
• 2023 AIME I Problems/Problem 15
  Find the largest prime number <math>p<1000</math> for which there exists a complex number <math>z</math> satisfying <cmath>1000>p=a^2+b^2\geq 12b^2+b^2=13b^2</cmath>so <math>b^2<81</math>, and <math>b<9<
  8 KB (1,299 words) - 17:37, 3 June 2023
• 2024 AIME II Problems
  ...tient and remainder, respectively, when <math>N</math> is divided by <math>1000</math>. Find <math>Q+R</math>. is divided by 1000.
  8 KB (1,236 words) - 23:11, 12 March 2024
• 2021 JMPSC Accuracy Problems/Problem 15
  ...f(2)+ \cdots + f(100)}{25}\equiv19+191+911+111 \cdot 97 \equiv 11888 \pmod{1000} \rightarrow \boxed{888}.</cmath> ...3</math> digits to get <cmath>20+28(10+99+980+700)+4(444)=\boxed{888} \mod 1000</cmath>
  2 KB (304 words) - 01:19, 12 July 2021
• AMC 12C 2020
  <math>\mathrm{(A) \ } -1000\qquad \mathrm{(B) \ } 1290\qquad \mathrm{(C) \ } 4356\qquad \mathrm{(D) \ } Let <math>N</math> be <math>10^{100^{1000^{10000…}}}</math>. (All the way till the number consisting of <math>100</
  9 KB (1,450 words) - 18:33, 21 April 2020
• AMC 12C 2020 Problems
  ...math>R(1) = \frac{1}{2}.</math> What is <math>R(1) + R(2) + R(3) + … + R(1000)</math>?
  8 KB (1,223 words) - 15:02, 27 November 2022
• AIME 2020(MOCK) Problems
  Let <math>N</math> be <math>112123123412345... (1000 digits)</math>. What is the remainder when <math>N</math> is divided by <ma
  891 bytes (139 words) - 06:45, 23 December 2020
• 2020 AIME II Problems/Problem 5
  ...<math>9</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...ving trouble with this step, note that <math>2^{10} = 1024 \equiv 24 \pmod{1000}</math>) ~ TopNotchMath
  3 KB (414 words) - 22:01, 23 March 2023
• 2020 AIME II Problems/Problem 14
  ...eq 2020</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...dot 2004 \cdot 2003}{3 \cdot 2\cdot 1} \equiv 10 (\mathrm{mod} \hskip .2cm 1000)</cmath>
  7 KB (1,186 words) - 15:31, 5 January 2024
• Sum Product Conjecture
  ...4</math> can be computed by hand, since <math>\tfrac{81}4\cdot 4^2 = 324 < 1000</math>. For the inductive step, write
  6 KB (990 words) - 10:47, 12 June 2020
• 2020 FMC 10A Problems
  ...57) + S(2, 257) + S(3, 257) + ... + S(256, 257)</cmath>is divided by <math>1000</math>. ...of the company. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  14 KB (2,267 words) - 12:49, 9 June 2020
• 2020 FMC 12A Problems
  ...57) + S(2, 257) + S(3, 257) + ... + S(256, 257)</cmath>is divided by <math>1000</math>. Find the value of <math>S</math> modulo <math>1000.</math>
  15 KB (2,388 words) - 13:24, 9 June 2020
• User:Vqbc/Testing2
  ...a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>. ...er arrangement. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  80 KB (13,867 words) - 23:17, 22 December 2020
• User:Amalya
  ...st time, please add 1 to the following number. My goal is to someday reach 1000.
  1 KB (254 words) - 17:13, 20 February 2021
• Talk:2019 AMC 10B Problems/Problem 19
  For 1000, I did a pro gamer move and brute forced it so hard my pen ran out of ink a
  820 bytes (138 words) - 04:59, 21 July 2020
• 2021 AIME I Problems/Problem 3
  Find the number of positive integers less than <math>1000</math> that can be expressed as the difference of two integral powers of <m ...math> and <math>b</math>, the value of <math>2^a - 2^b</math> is less than 1000.
  7 KB (1,174 words) - 08:21, 13 May 2023
• User:Andrewc10
  size(1000);
  2 KB (288 words) - 14:26, 27 October 2020
• 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

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