Difference between revisions of "2023 AMC 10B Problems/Problem 15"
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== Solution 2 == | == Solution 2 == | ||
− | We can prime factorize the solutions: <math> | + | We can prime factorize the solutions: |
− | + | A = <math>2 \cdot 3 \cdot 5,</math> | |
− | B = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13, | + | B = <math>2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13, </math> |
− | C = 2 \cdot 5 \cdot 7, | + | C = <math>2 \cdot 5 \cdot 7, </math> |
− | D = 2 \cdot 5 \cdot 11 \cdot 13, | + | D = <math>2 \cdot 5 \cdot 11 \cdot 13, </math> |
− | E = 7 \cdot 11 \cdot 13, | + | E = <math>7 \cdot 11 \cdot 13, |
</math> | </math> | ||
Revision as of 17:46, 15 November 2023
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
...
So, 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)