Soru

Zorluk: ZorDivisibility, Factors, and Multiples

Let kk be a positive integer with the prime factorization k=2x3y5zk = 2^x \cdot 3^y \cdot 5^z, where xx, yy, and zz are positive integers. If kk is divisible by both 1818 and 7575, and kk has exactly 3636 positive integer divisors, what is the maximum possible value of x+y+zx + y + z?

  1. A
    7
  2. 8Cevap
  3. C
    9
  4. D
    11
  5. E
    12

Cevap

The maximum possible value of x+y+zx + y + z is 8.
The correct answer is 8. The prime factorization of kk requires x1x \ge 1, y2y \ge 2, and z2z \ge 2 because kk is a multiple of 18=213218 = 2^1 \cdot 3^2 and 75=315275 = 3^1 \cdot 5^2. The number of positive divisors is given by (x+1)(y+1)(z+1)=36(x+1)(y+1)(z+1) = 36. Under the constraints x+12x+1 \ge 2, y+13y+1 \ge 3, and z+13z+1 \ge 3, the factorizations of 36 into three factors yield the sums x+y+z=8x+y+z = 8 (from factors 2,3,62, 3, 6) and x+y+z=7x+y+z = 7 (from factors 4,3,34, 3, 3). Therefore, the maximum possible value is 8.

Adım Adım Çözüm

1
Determine the lower bounds for the exponents xx, yy, and zz based on divisibility conditions.
Since kk is divisible by 18=213218 = 2^1 \cdot 3^2, we must have x1x \ge 1 and y2y \ge 2. Since kk is divisible by 75=315275 = 3^1 \cdot 5^2, we must have y2y \ge 2 and z2z \ge 2.
For a prime factorization to be divisible by another number, each prime factor's exponent in kk must be at least as large as its corresponding exponent in the divisor.
2
Set up the equation for the total number of positive integer divisors of kk.
(x+1)(y+1)(z+1)=36(x + 1)(y + 1)(z + 1) = 36, with constraints x+12x + 1 \ge 2, y+13y + 1 \ge 3, and z+13z + 1 \ge 3.
The total number of positive divisors of 2x3y5z2^x \cdot 3^y \cdot 5^z is given by (x+1)(y+1)(z+1)(x+1)(y+1)(z+1).
3
Find all valid integer factor triples (A,B,C)=(x+1,y+1,z+1)(A, B, C) = (x+1, y+1, z+1) multiplying to 36 under the given constraints.
The valid triples (A,B,C)(A, B, C) with A2,B3,C3A \ge 2, B \ge 3, C \ge 3 are (2,3,6)(2, 3, 6), (2,6,3)(2, 6, 3), and (4,3,3)(4, 3, 3).
Testing factorizations of 3636: 236=362 \cdot 3 \cdot 6 = 36, 263=362 \cdot 6 \cdot 3 = 36, and 433=364 \cdot 3 \cdot 3 = 36 all meet the inequality bounds for each term.
4
Calculate the sum x+y+z=(A+B+C)3x + y + z = (A + B + C) - 3 for each valid triple and identify the maximum.
For (2,3,6)(2, 3, 6): x+y+z=(21)+(31)+(61)=1+2+5=8x+y+z = (2-1) + (3-1) + (6-1) = 1 + 2 + 5 = 8.
For (2,6,3)(2, 6, 3): x+y+z=(21)+(61)+(31)=1+5+2=8x+y+z = (2-1) + (6-1) + (3-1) = 1 + 5 + 2 = 8.
For (4,3,3)(4, 3, 3): x+y+z=(41)+(31)+(31)=3+2+2=7x+y+z = (4-1) + (3-1) + (3-1) = 3 + 2 + 2 = 7.
The maximum possible value is 88.
Comparing all valid scenarios yields 88 as the maximum sum of exponents.

Anahtar Kavram

Calculating total positive integer divisors from prime factorizations and analyzing exponent constraints derived from divisibility conditions.
Tahmini Süre:2m 0s
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