Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | + | How many positive integer divisors of <math>201^9</math> are perfect squares or perfect cubes (or both)? | |
<math>\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41</math> | <math>\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41</math> | ||
Revision as of 13:05, 26 November 2023
Contents
Problem
How many positive integer divisors of are perfect squares or perfect cubes (or both)?
Solution 1 (PIE)
Prime factorizing , we get
.
A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are
for both
and
. This yields
perfect squares.
Perfect cubes must have multiples of for each of their prime factors' exponents, so we have either
, or
for both
and
, which yields
perfect cubes, for a total of
.
Subtracting the overcounted powers of (
,
,
, and
), we get
.
Solution 2
Observe that . Now divide into cases:
Case 1: The factor is . Then we can have
,
,
,
,
, or
.
Case 2: The factor is . This is the same as Case 1.
Case 3: The factor is some combination of s and
s.
This would be easy if we could just have any combination, as that would simply give . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for
.
is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of
.
is a "square".
is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.
is a "square"
is a "cube".
Now let's consider subcases:
Subcase 1: The squares are with each other.
Since we have square terms, and they would pair with
other square terms, we get
possibilities.
Subcase 2: The cubes are with each other.
Since we have cube terms, and they would pair with
other cube terms, we get
possibilities.
Subcase 3: A number pairs with .
Since any number can pair with (as it gives both a square and a cube), there would be
possibilities. Remember however that there can be two different bases (
and
), and they would produce different results. Thus, there are in fact
possibilities.
Finally, summing the cases gives .
Solution 3 (Quick)
We first prime factorize . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of
in the exponent. The largest square for
can be
, so their must be
ways. The largest cube is
, so there must be
. Minus one
due to overlapping and we get
ways for
to be a cube/square. We can see that this same thing happens for
due to the same exponent. Adding
as a case, we have our answer;
~ Wiselion
Solution 4 (A Little Long)
Notice that . We factorize
to get
. We then list perfect squares and cubes.
,
,
,
.
,
,
.
,
,
,
.
,
,
. Notice that the powers of
overlap. We must not forget
though. Of course, all of these factors already work. This gives us
. Next, we count the perfect squares. Since there are
options we have
. We do the same for the perfect cubes except with 3 options this time, and we have
. However, we accidentally overcounted
. We add our answers and subtract
to get
~ PerseverePlayer
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=2402
~ pi_is_3.14
Video Solution
Education, the Study of Everything
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.