Question

Difficulty: MediumPrime Numbers and Prime Factorization

Let KK be a positive integer with prime factorization K=2a×5b×11cK = 2^a \times 5^b \times 11^c, where aa, bb, and cc are positive integers. If 10K10K has 36 more positive divisors than KK, and 11K11K has 12 more positive divisors than KK, what is the value of a+b+ca + b + c?

Answer: 9

Answer

The value of a+b+ca + b + c is 9.
For K=2a5b11cK = 2^a 5^b 11^c, the number of divisors is given by (a+1)(b+1)(c+1)(a+1)(b+1)(c+1). Multiplying by 11 increases the exponent of 11 by 1, yielding (a+1)(b+1)(a+1)(b+1) additional divisors. Since this difference is 12, (a+1)(b+1)=12(a+1)(b+1) = 12. Multiplying by 10 increases the exponents of both 2 and 5 by 1, yielding an additional (c+1)(a+b+3)(c+1)(a+b+3) divisors. Since this difference is 36, (c+1)(a+b+3)=36(c+1)(a+b+3) = 36. For positive integers aa and bb, the only factor pair of 12 that allows 36/(a+b+3)36 / (a+b+3) to be an integer is when a+b=6a+b = 6. Consequently, c+1=4c+1 = 4, so c=3c = 3. Therefore, a+b+c=6+3=9a+b+c = 6+3 = 9.

Step-by-Step Solution

1
Write the divisor count formulas for KK, 10K10K, and 11K11K.
d(K)=(a+1)(b+1)(c+1)d(K) = (a+1)(b+1)(c+1), d(11K)=(a+1)(b+1)(c+2)d(11K) = (a+1)(b+1)(c+2), and d(10K)=(a+2)(b+2)(c+1)d(10K) = (a+2)(b+2)(c+1).
The number of positive divisors of a number 2x5y11z2^x 5^y 11^z is (x+1)(y+1)(z+1)(x+1)(y+1)(z+1).
2
Use d(11K)d(K)=12d(11K) - d(K) = 12 to solve for (a+1)(b+1)(a+1)(b+1).
(a+1)(b+1)(c+2)(a+1)(b+1)(c+1)=12    (a+1)(b+1)=12(a+1)(b+1)(c+2) - (a+1)(b+1)(c+1) = 12 \implies (a+1)(b+1) = 12.
Factoring out (a+1)(b+1)(a+1)(b+1) simplifies the equation directly.
3
Use d(10K)d(K)=36d(10K) - d(K) = 36 to find the relationship between a,b,a, b, and cc.
(c+1)[(a+2)(b+2)(a+1)(b+1)]=36    (c+1)(a+b+3)=36(c+1)[(a+2)(b+2) - (a+1)(b+1)] = 36 \implies (c+1)(a+b+3) = 36.
Expanding (a+2)(b+2)(a+1)(b+1)=ab+2a+2b+4(ab+a+b+1)=a+b+3(a+2)(b+2) - (a+1)(b+1) = ab + 2a + 2b + 4 - (ab + a + b + 1) = a + b + 3.
4
Determine a+ba+b and cc using positive integer constraints.
a+b=6a+b = 6 and c=3c = 3.
Since a,b1a, b \ge 1, the pairs for (a+1,b+1)(a+1, b+1) yielding 12 are (2,6)(2,6) or (3,4)(3,4). If (a+1,b+1)=(2,6)(a+1,b+1)=(2,6), a+b=6a+b=6, making a+b+3=9a+b+3=9 and c+1=36/9=4    c=3c+1=36/9=4 \implies c=3. If (3,4)(3,4), a+b=5a+b=5, so a+b+3=8a+b+3=8, but 36/836/8 is not an integer.
5
Sum aa, bb, and cc.
a+b+c=6+3=9a + b + c = 6 + 3 = 9.
Adding the derived sum a+b=6a+b=6 and c=3c=3 gives 9.

Key Concept

Divisor Count Function from Prime Factorization
Estimated Time:2m 0s
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