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Zorluk: ZorProperties of Integers and Divisibility

Let N=2a3b5cN = 2^a \cdot 3^b \cdot 5^c, where aa, bb, and cc are positive integers. The integer NN is divisible by 2424, has exactly 3636 positive integer divisors, and N5\frac{N}{5} is not divisible by 2525. What is the least possible value of a+b+ca + b + c?

  1. A
    66
  2. 77Cevap
  3. C
    88
  4. D
    99
  5. E
    1010

Cevap

7
The correct answer is 77. Since 24=233124 = 2^3 \cdot 3^1 divides NN, a3a \ge 3 and b1b \ge 1. The condition that N5\frac{N}{5} is not divisible by 2525 restricts cc to either 11 or 22. For c=2c=2, the divisor equation becomes (a+1)(b+1)(3)=36(a+1)(b+1)(3) = 36, so (a+1)(b+1)=12(a+1)(b+1) = 12. Taking a+1=4a+1=4 (a=3a=3) and b+1=3b+1=3 (b=2b=2) satisfies a3a \ge 3 and gives a minimal sum of a+b+c=3+2+2=7a+b+c = 3+2+2 = 7.

Adım Adım Çözüm

1
Analyze divisibility conditions on exponents
Since NN is divisible by 24=233124 = 2^3 \cdot 3^1, we must have a3a \ge 3 and b1b \ge 1. Since N5=2a3b5c1\frac{N}{5} = 2^a \cdot 3^b \cdot 5^{c-1} is not divisible by 25=5225 = 5^2, c1<2    c2c-1 < 2 \implies c \le 2. Because cc is a positive integer, cc can be 11 or 22.
Establish structural constraints on prime exponents from the divisibility statements.
2
Set up the total divisor formula
The total number of positive integer divisors of NN is (a+1)(b+1)(c+1)=36(a+1)(b+1)(c+1) = 36.
Apply the standard formula for counting divisors using prime factorization.
3
Evaluate candidate values for c=1c = 1 and c=2c = 2
If c=1c = 1, then c+1=2c+1 = 2, so (a+1)(b+1)=18(a+1)(b+1) = 18. Since a3    a+14a \ge 3 \implies a+1 \ge 4, possible integer pairs for (a+1,b+1)(a+1, b+1) are (6,3)(6, 3) giving a=5,b=2    a+b+c=8a=5, b=2 \implies a+b+c = 8, or (9,2)(9, 2) giving a=8,b=1    a+b+c=10a=8, b=1 \implies a+b+c = 10.
If c=2c = 2, then c+1=3c+1 = 3, so (a+1)(b+1)=12(a+1)(b+1) = 12. With a+14a+1 \ge 4, possible pairs for (a+1,b+1)(a+1, b+1) are (4,3)(4, 3) giving a=3,b=2    a+b+c=3+2+2=7a=3, b=2 \implies a+b+c = 3+2+2 = 7, or (6,2)(6, 2) giving a=5,b=1    a+b+c=8a=5, b=1 \implies a+b+c = 8.
Test allowable values of cc to find all valid exponent triples (a,b,c)(a, b, c) and minimize their sum.
4
Identify the minimum sum
The minimum sum of a+b+ca + b + c is 77, achieved when a=3a = 3, b=2b = 2, and c=2c = 2.
Compare sums across all valid exponent combinations.

Anahtar Kavram

Prime Factorization, Divisibility Rules, and Number of Divisors Formula
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