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 can reduce our original expression to
~Technodoggo ~minor edits by lucaswujc
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 ,
. So,
. 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) & ~Technodoggo (add more examples)
Solution 4 (Bashy method)
We know that a perfect square must be in the form where
are nonnegative integers, and
is the largest and
prime factor of our square number.
Let's assume . We need to prime factorize
and see which prime factors are raised to an odd power. Then, we can multiply one factor each of prime number with an odd number of factors to
. We can do this by finding the number of factors of
,
,
,
,
, and
.
Case 1: Factors of
We first count factors of in each of the factorials. We know there is one factor of
each in
and
, two in
and
, and so on until we have
factors of
in
. Adding them all up, we have
.
Now, we count factors of in each of the factorials. We know there is one factor of
each in
,
,
, and
, two in
,
,
, and
, and so on until we have
factors of
in
. Adding them all up, we have
.
Now we count factors of in each of the factorials. Using a similar method as above, we have a sum of
.
Now we count factors of in each of the factorials. Using a similar method as above, we have a factor of
in
, so there is
factor of
.
Adding all the factors of , we have
. Since
is odd,
has one factor of
.
Case 2: Factors of
We use a similar method as in case 1. We first count factors of . We obtain the sum
.
We count factors of . We obtain the sum
.
Adding all the factors of , we have
. Since
is even,
has
factors of
.
Case 3: Factors of
We count the factors of :
. Since
is odd,
has one factor of
.
Case 4: Factors of
We count the factors of :
. Since
is odd,
has one factor of
.
Case 5: Factors of
We count the factors of :
. Since
is even,
has
factors of
.
Case 6: Factors of
We count the factors of :
. Since
is even,
has
factors of
.
Multiplying out all our factors for , we obtain
.
~arjken
Video Solution 1 by OmegaLearn
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=sqVY5-h4vfo
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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