Difference between revisions of "2016 AMC 8 Problems/Problem 15"
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Denote <math>v_p(k)</math> as the largest exponent of <math>p</math> in the prime factorization of <math>n</math>. In this problem, we have <math>p=2</math>. | Denote <math>v_p(k)</math> as the largest exponent of <math>p</math> in the prime factorization of <math>n</math>. In this problem, we have <math>p=2</math>. | ||
− | Lifting the Exponent on <math>n</math>, | + | By the Lifting the Exponent Lemma on <math>n</math>, |
<cmath>v_2(13^4-11^4)=v_2(13-11)+v_2(2)+v_2(13+11)</cmath> | <cmath>v_2(13^4-11^4)=v_2(13-11)+v_2(2)+v_2(13+11)</cmath> |
Revision as of 15:43, 7 March 2023
Contents
Problem
What is the largest power of that is a divisor of
?
Solution 1
First, we use difference of squares on to get
. Using difference of squares again and simplifying, we get
. Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of
that is a divisor
is
.
Solution 2 (a variant of Solution 1)
Just like in the above solution, we use the difference-of-squares factorization, but only once to get We can then compute that this is equal to
Note that
(we don't need to factorize any further as
is already odd) thus the largest power of
that divides
is only
while
so the largest power of
that divides
is
Hence, the largest power of
that is a divisor of
is
Aops-g5-gethsemanea2 (talk) 05:16, 4 January 2023 (EST)
Solution 3 (Lifting the exponent)
Let We wish to find the largest power of
that divides
.
Denote as the largest exponent of
in the prime factorization of
. In this problem, we have
.
By the Lifting the Exponent Lemma on ,
Therefore, exponent of the largest power of that divids
is
so the largest power of
that divides this number is
.
-Benedict T (countmath1)
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=3705
~ pi_is_3.14
Video Solution
~savannahsolver
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.