2018 AIME I Problems/Problem 11
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)