2024 AMC 12A Problems/Problem 7

The following problem is from both the 2024 AMC 12A #7 and 2024 AMC 10A #9, so both problems redirect to this page.

Problem

Let $N$ be the least positive integer that is divisible by at least $3$ odd primes and at least $4$ perfect squares. What is the sum of the squares of the digits of $N$ ?

$\textbf{(A)}~ 41 \qquad \textbf{(B)}~ 65 \qquad \textbf{(C)}~ 80 \qquad \textbf{(D)}~ 89 \qquad \textbf{(E)}~ 100$

Solution

Consider the prime factorization of $N$ . Intuitively, the odd primes that divide $N$ should be as small as possible so we let $N$ be divisible by the three least odd primes: $3$ , $5$ , and $7$ . In order for $N$ to be divisible by at least four perfect squares, it must either be divisible by the sixth power of a prime, or the squares of at least two different primes. These can be obtained by multiplying $3 \cdot 5 \cdot 7$ by either $3^{5} = 243$ , $2^{6} = 64$ , $3\cdot 5 = 15$ , or $2^{2}\cdot 3 = 12$ ; any other combination of primes will clearly make $N$ larger. The least of these values is $12$ , giving $N = 2^{2} \cdot 3^{2} \cdot 5 \cdot 7 = 1260$ . The sum of the squares of the digits of $N$ is $1^{2}+2^{2}+6^{2}+0^{2} = \boxed{\textbf{(A)}~41}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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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2024 AMC 12A Problems/Problem 7

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