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Zorluk: OrtaPrime Numbers and Prime Factorization

The positive integer nn has a prime factorization of the form 2a×3b×5c2^a \times 3^b \times 5^c, where aa, bb, and cc are positive integers. If nn is divisible by both 12 and 15, and nn has exactly 24 positive integer divisors, what is the least possible value of nn?

  1. A
    180
  2. 360Cevap
  3. C
    480
  4. D
    540
  5. E
    600

Cevap

The least possible value of nn is 360.
To find the least value of n=2a×3b×5cn = 2^a \times 3^b \times 5^c divisible by 12=22×312 = 2^2 \times 3 and 15=3×515 = 3 \times 5, we require a≥2a \ge 2, b≥1b \ge 1, and c≥1c \ge 1. The number of positive divisors is (a+1)(b+1)(c+1)=24(a+1)(b+1)(c+1) = 24. Testing valid factor triples for 24 with a+1≥3a+1 \ge 3, b+1≥2b+1 \ge 2, and c+1≥2c+1 \ge 2 gives possible values n=480n = 480 (from a=5,b=1,c=1a=5, b=1, c=1), n=360n = 360 (from a=3,b=2,c=1a=3, b=2, c=1), n=540n = 540 (from a=2,b=3,c=1a=2, b=3, c=1), and n=600n = 600 (from a=3,b=1,c=2a=3, b=1, c=2). The minimum among these valid integers is 360.

Adım Adım Çözüm

1
Determine the minimum exponent constraints from divisibility conditions.
Since nn is divisible by 12=22×3112 = 2^2 \times 3^1, we must have a≥2a \ge 2 and b≥1b \ge 1. Since nn is divisible by 15=31×5115 = 3^1 \times 5^1, we must have b≥1b \ge 1 and c≥1c \ge 1. Combining these, a≥2a \ge 2, b≥1b \ge 1, and c≥1c \ge 1.
Divisibility requires that the prime factorization of nn contains at least the prime powers present in the prime factorizations of 12 and 15.
2
Set up the total divisor count equation.
The total number of positive integer divisors of n=2a×3b×5cn = 2^a \times 3^b \times 5^c is given by (a+1)(b+1)(c+1)=24(a+1)(b+1)(c+1) = 24.
The divisor counting formula states that for a number with prime factorization p1e1p2e2…p_1^{e_1} p_2^{e_2} \dots, the total number of positive divisors is (e1+1)(e2+1)…(e_1+1)(e_2+1)\dots
3
Find all integer factor triples (a+1,b+1,c+1)(a+1, b+1, c+1) that multiply to 24 subject to a+1≥3a+1 \ge 3, b+1≥2b+1 \ge 2, and c+1≥2c+1 \ge 2.
The valid factorizations of 24 into three factors meeting the bounds are:
- (6,2,2)  ⟹  (a,b,c)=(5,1,1)(6, 2, 2) \implies (a, b, c) = (5, 1, 1), giving n=25×31×51=480n = 2^5 \times 3^1 \times 5^1 = 480
- (4,3,2)  ⟹  (a,b,c)=(3,2,1)(4, 3, 2) \implies (a, b, c) = (3, 2, 1), giving n=23×32×51=360n = 2^3 \times 3^2 \times 5^1 = 360
- (3,4,2)  ⟹  (a,b,c)=(2,3,1)(3, 4, 2) \implies (a, b, c) = (2, 3, 1), giving n=22×33×51=540n = 2^2 \times 3^3 \times 5^1 = 540
- (4,2,3)  ⟹  (a,b,c)=(3,1,2)(4, 2, 3) \implies (a, b, c) = (3, 1, 2), giving n=23×31×52=600n = 2^3 \times 3^1 \times 5^2 = 600
To minimize n=2a×3b×5cn = 2^a \times 3^b \times 5^c, we must evaluate all valid permutations of exponents consistent with the divisor product constraint.
4
Compare the resulting values of nn to find the minimum.
Comparing 480, 360, 540, and 600, the smallest value is 360.
Assigning the larger exponent 3 to the smallest prime base 2 and exponent 2 to prime base 3 minimizes the overall product.

Anahtar Kavram

Divisor Counting Formula and Prime Factorization Constraints
Tahmini Süre:2m 0s
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Konular

Prime Numbers and Prime Factorization

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