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Zorluk: Çok zorDivisibility, Factors, and Multiples

A positive integer NN has no prime factors other than 22 and 33. If NN is a multiple of 1212 and has exactly 1818 positive divisors, what is the sum of all possible values of NN?

Cevap: 2028

Cevap

2028
The correct numerical answer is 2028, obtained by determining all exponent combinations of 2 and 3 that yield 18 total factors while guaranteeing divisibility by 12.

Adım Adım Çözüm

1
Set up the prime factorization of NN with constraints.
N=2a3bN = 2^a \cdot 3^b with a2a \ge 2 and b1b \ge 1.
Since the only prime factors are 22 and 33, and 12=223112 = 2^2 \cdot 3^1 divides NN, the exponents must satisfy a2a \ge 2 and b1b \ge 1.
2
Apply the total number of divisors formula.
(a+1)(b+1)=18(a+1)(b+1) = 18.
The number of positive divisors of p1e1p2e2p_1^{e_1} p_2^{e_2} is (e1+1)(e2+1)(e_1 + 1)(e_2 + 1).
3
Identify all valid integer solution pairs for (a+1,b+1)(a+1, b+1).
Valid pairs are (9,2)(9,2), (6,3)(6,3), and (3,6)(3,6), corresponding to (a,b)=(8,1),(5,2),(2,5)(a,b) = (8,1), (5,2), (2,5).
We require a+13a+1 \ge 3 and b+12b+1 \ge 2. Pairs (18,1)(18,1) and (2,9)(2,9) violate these lower bounds.
4
Compute the corresponding values of NN and find their sum.
768+288+972=2028768 + 288 + 972 = 2028.
Evaluating 2831=7682^8 \cdot 3^1 = 768, 2532=2882^5 \cdot 3^2 = 288, and 2235=9722^2 \cdot 3^5 = 972 yields a total sum of 20282028.

Anahtar Kavram

Divisor Count Formula & Multiplicativity Constraints
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