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Zorluk: OrtaDivisibility, Factors, and Multiples

A positive integer nn has a prime factorization of the form 2a⋅3b⋅5c2^a \cdot 3^b \cdot 5^c, where aa, bb, and cc are non-negative integers. If nn is a multiple of 2020 but not a multiple of 4040, is not divisible by 99, and has exactly 1212 positive integer divisors, what is the sum of all possible values of nn?

Cevap: 560

Cevap

The sum of all possible values of nn is 560.
To determine the sum of all possible values of nn, first analyze the given conditions on the prime exponents of n=2a⋅3b⋅5cn = 2^a \cdot 3^b \cdot 5^c. Divisibility by 20=22⋅5120 = 2^2 \cdot 5^1 requires a≥2a \ge 2 and c≥1c \ge 1. Since nn is not divisible by 40=23⋅5140 = 2^3 \cdot 5^1, the exponent aa must equal 2. Since nn is not divisible by 9=329 = 3^2, the exponent bb must be strictly less than 2, meaning bb can be 0 or 1. The total number of positive integer divisors is given by (a+1)(b+1)(c+1)=12(a+1)(b+1)(c+1) = 12. Substituting a=2a = 2 gives 3(b+1)(c+1)=123(b+1)(c+1) = 12, or (b+1)(c+1)=4(b+1)(c+1) = 4. If b=0b = 0, then c+1=4  ⟹  c=3c+1 = 4 \implies c = 3, which gives n=22⋅30⋅53=500n = 2^2 \cdot 3^0 \cdot 5^3 = 500. If b=1b = 1, then c+1=2  ⟹  c=1c+1 = 2 \implies c = 1, which gives n=22⋅31⋅51=60n = 2^2 \cdot 3^1 \cdot 5^1 = 60. The sum of these two valid values is 500+60=560500 + 60 = 560.

Adım Adım Çözüm

1
Analyze the divisibility constraints to find values/ranges for exponents aa, bb, and cc.
a=2a = 2, c≥1c \ge 1, and b∈{0,1}b \in \{0, 1\}.
Divisibility by 20 (22⋅512^2 \cdot 5^1) requires a≥2a \ge 2 and c≥1c \ge 1. Non-divisibility by 40 (23⋅512^3 \cdot 5^1) restricts a<3a < 3, so a=2a = 2. Non-divisibility by 9 (323^2) restricts b<2b < 2.
2
Apply the divisor count formula (a+1)(b+1)(c+1)=12(a+1)(b+1)(c+1) = 12.
(b+1)(c+1)=4(b+1)(c+1) = 4.
Plugging a=2a = 2 into (a+1)(b+1)(c+1)=12(a+1)(b+1)(c+1) = 12 yields 3(b+1)(c+1)=123(b+1)(c+1) = 12, which simplifies to (b+1)(c+1)=4(b+1)(c+1) = 4.
3
Evaluate the cases for b=0b = 0 and b=1b = 1.
The valid values for nn are 500500 and 6060.
When b=0b = 0, c+1=4  ⟹  c=3c+1 = 4 \implies c = 3, yielding n=22⋅30⋅53=500n = 2^2 \cdot 3^0 \cdot 5^3 = 500. When b=1b = 1, c+1=2  ⟹  c=1c+1 = 2 \implies c = 1, yielding n=22⋅31⋅51=60n = 2^2 \cdot 3^1 \cdot 5^1 = 60.
4
Calculate the sum of all valid integers nn.
560
500+60=560500 + 60 = 560.

Anahtar Kavram

Divisor Count Formula and Divisibility Constraints
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Divisibility, Factors, and Multiples

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