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Zorluk: ZorPrime Numbers and Prime Factorization

Let n=2a3b5cn = 2^a \cdot 3^b \cdot 5^c, where aa, bb, and cc are positive integers. If the integer n6\frac{n}{6} has exactly 3232 positive divisors and the integer 10n10n has exactly 9090 positive divisors, what is the value of a+b+ca + b + c?

  1. A
    8
  2. 9Cevap
  3. C
    10
  4. D
    11
  5. E
    12

Cevap

The value of a+b+ca + b + c is 99.
Writing the prime factorizations gives n6=2a13b15c\frac{n}{6} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c with ab(c+1)=32a \cdot b \cdot (c+1) = 32, and 10n=2a+13b5c+110n = 2^{a+1} \cdot 3^b \cdot 5^{c+1} with (a+2)(b+1)(c+2)=90(a+2)(b+1)(c+2) = 90. Solving for positive integers aa, bb, and cc yields a=4a = 4, b=2b = 2, and c=3c = 3. Thus, the sum a+b+c=9a + b + c = 9.

Adım Adım Çözüm

1
Express the prime factorization of n6\frac{n}{6} and write its total divisor count equation.
n6=2a3b5c23=2a13b15c\frac{n}{6} = \frac{2^a \cdot 3^b \cdot 5^c}{2 \cdot 3} = 2^{a-1} \cdot 3^{b-1} \cdot 5^c. The number of positive divisors is ab(c+1)=32a \cdot b \cdot (c + 1) = 32.
Dividing nn by 6=236 = 2 \cdot 3 decreases the exponents of 22 and 33 by 11 each. The formula for the total number of divisors of a number p1xp2yp3zp_1^{x} p_2^{y} p_3^{z} is (x+1)(y+1)(z+1)(x+1)(y+1)(z+1).
2
Express the prime factorization of 10n10n and write its total divisor count equation.
10n=(25)(2a3b5c)=2a+13b5c+110n = (2 \cdot 5) \cdot (2^a \cdot 3^b \cdot 5^c) = 2^{a+1} \cdot 3^b \cdot 5^{c+1}. The number of positive divisors is (a+2)(b+1)(c+2)=90(a + 2) \cdot (b + 1) \cdot (c + 2) = 90.
Multiplying by 10=2510 = 2 \cdot 5 increases the exponents of 22 and 55 by 11 each.
3
Solve the system of equations for the positive integer exponents aa, bb, and cc.
From ab(c+1)=32a \cdot b \cdot (c + 1) = 32, test integer factors of 3232. Setting c+1=4    c=3c + 1 = 4 \implies c = 3, we get ab=8a \cdot b = 8. Substituting c=3c = 3 into (a+2)(b+1)(c+2)=90(a + 2)(b + 1)(c + 2) = 90 yields (a+2)(b+1)(5)=90    (a+2)(b+1)=18(a + 2)(b + 1)(5) = 90 \implies (a + 2)(b + 1) = 18. Testing pairs where ab=8a \cdot b = 8: if a=4a = 4 and b=2b = 2, then (4+2)(2+1)=63=18(4 + 2)(2 + 1) = 6 \cdot 3 = 18, which satisfies both equations.
Because a,b,ca, b, c are positive integers, factor analysis uniquely pinpoints a=4a = 4, b=2b = 2, and c=3c = 3.
4
Calculate a+b+ca + b + c.
a+b+c=4+2+3=9a + b + c = 4 + 2 + 3 = 9.
Summing the derived values of the positive integer exponents.

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Prime Factorization and Total Number of Divisors
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