2021 AIME II Problems/Problem 13
Contents
Problem
Find the least positive integer for which
is a multiple of
.
Solution 1
1000 divides this expression iff 8, 125 both divide it. It should be fairly obvious that ; so we may break up the initial condition into two sub-conditions.
(1) . Notice that the square of any odd integer is 1 modulo 8 (proof by plugging in
,
,
,
into modulo 8), so the LHS of this expression goes "5,1,5,1, ...", while the RHS goes "1,2,3,4,5,6,7,8,1, ...". The cycle length of the LHS is 2, RHS is 8, so the cycle length of the solution is
. Indeed, the n that solve this equation are exactly those such that
.
(2) . This is extremely computationally intensive if you try to calculate all
, so instead, we take a divide-and-conquer approach. In order for this expression to be true
is necessary; it shouldn't take too long for you to go through the 20 possible LHS-RHS combinations and convince yourself that
or
.
With this in mind we consider modulo 25. By the generalized Fermat's little theorem,
, but we already have n modulo 20! Our calculation is greatly simplified. The LHS cycle length is 20, RHS cycle length is 25, the lcm is 100, in this step we need to test all the numbers between 1 to 100 that
or
. In the case that
,
, and RHS goes "3,23,43,63,83"; clearly
. In the case that
, by a similar argument,
.
In the final step, we need to calculate modulo 125. The former is simply
; because
modulo 125,
.
is
.
. This time, LHS cycle is 100, RHS cycle is 125, so we need to figure out what is n modulo 500. It should be
.
Put everything together. By the second subcondition, the only candidates < 100 are . Apply the first subcondition, n=797 is the desired answer.
-Ross Gao
Solution 2
We have that , or
and
by CRT. It is easy to check
don't work, so we have that
. Then,
and
, so we just have
and
. Let us consider both of these congruences separately.
First, we look at . By Euler's Totient Theorem (ETT), we have
, so
. On the RHS of the congruence, the possible values of
are all nonnegative integers less than
and on the RHS the only possible values are
and
. However, for
to be
we must have
, a contradiction. So, the only possible values of
are when
.
Now we look at . Plugging in
, we get
. Note, for
to be satisfied, we must have
and
. Since
as
, we have
. Then,
. Now, we get
. Using the fact that
, we get
. The inverse of
modulo
is obviously
, so
, so
. Plugging in
, we get
.
Now, we are finally ready to plug into the congruence modulo
. Plugging in, we get
. By ETT, we get
, so
. Then,
. Plugging this in, we get
, implying the smallest value of
is simply
. ~rocketsri
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.