Difference between revisions of "2018 AIME I Problems/Problem 11"
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Quick inspection yields <math>3^5 \equiv 1(\mod 121)</math> and <math>3^3 \equiv 1(\mod 13)</math>. Now we must find the smallest <math>k</math> such that <math>3^3k \equiv 1(\mod 13)</math>. Euler's gives <math>3^156 \equiv 1(\mod 169)</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=2, 4, 13, 26, 52</math>. Some more inspection yields <math>k=13</math> is the smallest valid <math>k</math>. So <math>3^5 \equiv 1(\mod 121)</math> and <math>3^39 \equiv 1(\mod 169)</math>. The least <math>n</math> satisfying both is <math>lcm[5,39]=\boxed{195}</math>. (RegularHexagon) | Quick inspection yields <math>3^5 \equiv 1(\mod 121)</math> and <math>3^3 \equiv 1(\mod 13)</math>. Now we must find the smallest <math>k</math> such that <math>3^3k \equiv 1(\mod 13)</math>. Euler's gives <math>3^156 \equiv 1(\mod 169)</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=2, 4, 13, 26, 52</math>. Some more inspection yields <math>k=13</math> is the smallest valid <math>k</math>. So <math>3^5 \equiv 1(\mod 121)</math> and <math>3^39 \equiv 1(\mod 169)</math>. The least <math>n</math> satisfying both is <math>lcm[5,39]=\boxed{195}</math>. (RegularHexagon) | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2018|n=I|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Revision as of 00:58, 8 March 2018
Find the least positive integer such that when
is written in base
, its two right-most digits in base
are
.
Solutions
Modular Arithmetic Solution- Strange (MASS)
Note that . And
. Because
,
and
.
If , one can see the sequence
so
.
Now if , it is harder. But we do observe that
, therefore
for some integer
. So our goal is to find the first number
such that
. In other words, the
coefficient must be
. It is not difficult to see that this first
, so ultimately
. Therefore,
.
The first satisfying both criteria is
.
-expiLnCalc
Solution
Note that Euler's Totient Theorem would not necessarily lead to the smallest and that in this case that
is greater than
.
We wish to find the least such that
. This factors as
. Because
, we can simply find the least
such that
and
.
Quick inspection yields and
. Now we must find the smallest
such that
. Euler's gives
. So
is a factor of
. This gives
. Some more inspection yields
is the smallest valid
. So
and
. The least
satisfying both is
. (RegularHexagon)
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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