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• Resources for mathematics competitions
  ...ving.com/resources/articles/lifting-the-exponent.pdf Lifting the Exponent (LTE)] by Amir Hossein Parvardi
  16 KB (2,152 words) - 21:46, 6 May 2024
• 2009 AMC 12A Problems/Problem 18
  ==Solution 5 (LTE)== ...6</math>, then <math>v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)</math>. By LTE, <math>v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3</math>. Since <math>
  6 KB (1,012 words) - 19:16, 14 September 2022
• 2011 AIME I Problems/Problem 15
  By LTE, <math>v_{2011}(a^3-b^3)=v_{2011}(a-b)</math> if <math>a-b</math> is divisi
  8 KB (1,302 words) - 04:07, 24 July 2023
• Olympiad books
  ...blemsolving.com/Forum/viewtopic.php?t=401494 ''Lifting the Exponent Lemma (LTE)''] - '''Amir Hossein Parvardi'''.
  17 KB (2,261 words) - 00:30, 22 April 2024
• 2018 AIME I Problems/Problem 11
  ==Solution 6 (LTE)== ==Solution 9 (Motivation and LTE)==
  10 KB (1,448 words) - 06:30, 21 April 2024
• Lifting the Exponent Lemma
  Let <math>p</math> refer to an odd prime. We can split up LTE into six identities (where <math>\nu_p(Z)</math> represents the largest fac
  1 KB (217 words) - 00:30, 1 May 2024
• 2019 AIME I Problems/Problem 7
  Remark: You can obtain the contradiction by using LTE. If <math>\nu_2{(x)}\geq{\nu_2{(y)}}, \nu_2{(y^2x)}=60</math>. However, <ma
  8 KB (1,351 words) - 16:07, 3 January 2024
• 2020 AIME I Problems/Problem 12
  ...-2^{4c})=v_5(149^4-16)+v_5(c)=1+v_5(c)</math> meaning that we have that by LTE, <math>5^4 | c</math> and <math>4 \cdot 5^4</math> divides <math>n</math>. We were able to use LTE with 3 and 7 but not 5 because in order to use LTE, we need \( p \mid x-y \).
  10 KB (1,650 words) - 03:50, 21 January 2024
• 2019 Mock AMC 10B Problems/Problem 19
  By LTE Lemma, we know that <math>v_2(x^n - y^n) = v_2(x-y) + v_2(n) + v_2(x+y)-1</
  1 KB (158 words) - 23:04, 3 October 2023
• 2021 AIME I Problems/Problem 14
  ...{a(a^n - 1^n)}{a-1}\right)</math> must be divisible by <math>43,</math> by LTE, we have <math>v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,</m
  13 KB (2,185 words) - 02:28, 13 November 2023
• 2021 GMC 10B Problems/Problem 18
  ...of <math>k</math>. By [//en.wikipedia.org/wiki/Lifting-the-exponent_lemma LTE],
  1 KB (214 words) - 19:40, 7 March 2022
• 2024 AIME I Problems/Problem 13
  ...th>2</math> from before. Note <math>17^2 \mid 2^{4 \cdot 17} + 1</math> by LTE. Hence the possible <math>n</math> are, <math>2^{17}, 2^{51}, 2^{85}, 2^{11
  4 KB (616 words) - 19:30, 23 May 2024
• 2024 USAJMO Problems/Problem 3
  Then, by LTE, <math>v_p(a(m))=v_p((a(m-1))^m-1)=v_p(a(m-1)-1)+v_p(m)>v_p(m)</math>. Sinc
  2 KB (469 words) - 11:39, 28 March 2024