Bunuel wrote:
The expression x#y denotes the product of the consecutive multiples of 3 between x and y, inclusive. What is the sum of the exponents in the prime factorization of 21#42?
A. 23
B. 24
C. 25
D. 26
E. 27
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:First, translate the expression 21#42, using the definition given:
21#42 = 21×24×27×30×33×36×39×42
You need the prime factorization of this enormous product.
Since these are consecutive multiples of 3, a good move is to factor out that 3 from each multiple. You have 8 multiples, all multiplied together, so you get 3^8:
21#42 = 3^8(7×8×9×10×11×12×13×14)
Now replace each consecutive integer with its prime factorization:
21#42 = 3^8(7×2^3×3^2×(2×5)×11×(2^2×3)×13×(2×7))
Group up the prime bases:
21#42 = 2^7×3^11×5×7^2×11×13
Don’t forget to put in an understood 1 for the primes lacking explicit exponents:
21#42 = 2^7×3^11×5^1×7^2×11^1×13^1
Finally, add up the exponents:
7 + 11 + 1 + 2 + 1 + 1 = 23
The correct answer is A.
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