vatsavayi Posted July 26, 2015 Share Posted July 26, 2015 if 252^6 is divisible by 6^n, then what is the maximum possible value of n A) 1 B) 2 C) 3 D) 6 E) 12 Quote Link to comment Share on other sites More sharing options...
Brent Hanneson Posted July 27, 2015 Share Posted July 27, 2015 if 252^6 is divisible by 6^n, then what is the maximum possible value of n A) 1 B) 2 C) 3 D) 6 E) 12 A lot of integer property questions can be solved using prime factorization. For questions involving divisibility, divisors, factors and multiples, we can say: If N is divisible by k, then k is "hiding" within the prime factorization of N Consider these examples: 24 is divisible by 3 because 24 = (2)(2)(2)(3) Likewise, 70 is divisible by 5 because 70 = (2)(5)(7) And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7) And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7) -------------------------------- Okay, onto the question: We want to know how many 6's are hiding in the prime factorization of 252^6 252 = (2)(2)(3)(3)(7) Notice that there are already two 6's hiding in the prime factorization of 252 Since 252^6 = (252)(252)(252)(252)(252)(252), we know that there must be a total of twelve 6's hiding in the prime factorization of 252^6 Answer: E Cheers, Brent Quote Link to comment Share on other sites More sharing options...
smuhamov Posted November 28, 2020 Share Posted November 28, 2020 You can break 262 into prime factors as below: 262^6 = (2^2*3^2*7)^6 If you expand the above you get below: 2^12 * 3^12 * 7^6 6^n = (2 * 3) ^ n If you look at prime factorization of 262 ^ 6, you have 2 ^ 12 * 3 ^ 12. As such answer should be 12. Quote Link to comment Share on other sites More sharing options...
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