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Zorluk: ZorDivisibility, Factors, and Multiples

Let nn be a positive integer whose only prime factors are 22, 33, and 55. If n2\dfrac{n}{2} has 3636 positive divisors, n3\dfrac{n}{3} has 3636 positive divisors, and n5\dfrac{n}{5} has 3232 positive divisors, how many positive divisors does n2n^2 have?

  1. A
    96
  2. B
    180
  3. C
    216
  4. 245Cevap
  5. E
    343

Cevap

245
Writing nn as 2a⋅3b⋅5c2^a \cdot 3^b \cdot 5^c, the number of divisors of n2\frac{n}{2}, n3\frac{n}{3}, and n5\frac{n}{5} leads to the system a(b+1)(c+1)=36a(b+1)(c+1) = 36, (a+1)b(c+1)=36(a+1)b(c+1) = 36, and (a+1)(b+1)c=32(a+1)(b+1)c = 32. Equating the first two yields a=ba = b. Substituting b=ab = a into the third gives (a+1)2c=32(a+1)^2 c = 32. Since aa is a positive integer, (a+1)2(a+1)^2 must be a square dividing 3232, so a+1=4  ⟹  a=3a+1 = 4 \implies a = 3, giving b=3b = 3 and c=2c = 2. Therefore, n2=26⋅36⋅54n^2 = 2^6 \cdot 3^6 \cdot 5^4, and the total number of positive divisors of n2n^2 is (6+1)(6+1)(4+1)=245(6+1)(6+1)(4+1) = 245.

Adım Adım Çözüm

1
Express nn in terms of its prime factorization
n=2a⋅3b⋅5cn = 2^a \cdot 3^b \cdot 5^c where a,b,c≥1a, b, c \ge 1
The problem states that 22, 33, and 55 are the only prime factors of nn.
2
Set up equations for the number of divisors of n2\dfrac{n}{2}, n3\dfrac{n}{3}, and n5\dfrac{n}{5}
a(b+1)(c+1)=36a(b+1)(c+1) = 36, (a+1)b(c+1)=36(a+1)b(c+1) = 36, and (a+1)(b+1)c=32(a+1)(b+1)c = 32
Dividing nn by 22, 33, or 55 decreases the corresponding prime exponent by 11. The formula for total positive divisors of 2p3q5r2^p 3^q 5^r is (p+1)(q+1)(r+1)(p+1)(q+1)(r+1).
3
Solve the system of equations for exponents aa, bb, and cc
a=3a = 3, b=3b = 3, c=2c = 2
Since a(b+1)(c+1)=(a+1)b(c+1)=36a(b+1)(c+1) = (a+1)b(c+1) = 36, we get a(b+1)=b(a+1)  ⟹  a=ba(b+1) = b(a+1) \implies a = b. Substituting b=ab = a into (a+1)2c=32(a+1)^2 c = 32 gives (a+1)2(a+1)^2 as a factor of 3232. Testing perfect squares yields a+1=4  ⟹  a=3a+1 = 4 \implies a = 3, so c=2c = 2. Checking in the first equation gives 3(4)(3)=363(4)(3) = 36, which is consistent.
4
Determine the prime factorization and number of positive divisors of n2n^2
n2=26⋅36⋅54n^2 = 2^6 \cdot 3^6 \cdot 5^4, total divisors =(6+1)(6+1)(4+1)=7⋅7⋅5=245= (6+1)(6+1)(4+1) = 7 \cdot 7 \cdot 5 = 245
Squaring nn doubles each exponent in its prime factorization.

Anahtar Kavram

Prime Factorization and Number of Divisors Formula
Tahmini Süre:2m 0s
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Divisibility, Factors, and Multiples

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