2023 AMC 10B Problems/Problem 15
Contents
Problem
What is the least positive integer such that
is a perfect square?
Solution 1
Consider 2,
there are odd number of 2's in (We're not counting 3 2's in 8, 2 3's in 9, etc).
There are even number of 3's in
...etc,
So, we original expression reduce to
Solution 2
We can prime factorize the solutions:
A =
B =
C =
D =
E =
We can immediately eliminate B, D, and E since 13 only appears in , so
is a perfect square.
Next, we can test if 7 is possible (and if it is not we can use process of elimination)
7 appears in
to
and
14 appears in
to
.
So, there is an odd amount of 7's since there are 10 7's from
to
and 3 7's from
to
, and
which is odd. So we need to multiply by 7 to get a perfect square. Since 30 is not a divisor of 7, our answer is 70 which is
.
~aleyang
Solution 3
First, we note . Simplifying the whole sequence and cancelling out the squares, we get
. Prime factoring
and cancelling out the squares, the only numbers that remain are
and
. Since we need to make this a perfect square,
. Multiplying this out, we get
.
~yourmomisalosinggame (a.k.a. Aaron)