What will be the least number which when quadrupled will be exactly divisible by 24, 36, 42 and 60?
(A)630
(B) 1260
(C) 315
(D)420
(A)630
(B) 1260
(C) 315
(D)420
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ASIDE: A lot of integer property questions can be solved using prime factorization.sgr21 wrote:What will be the least number which when quadrupled will be exactly divisible by 24, 36, 42 and 60?
(A)630
(B) 1260
(C) 315
(D)420
It means "multiplied by 4"theCodeToGMAT wrote:Brent, what does quadrupled mean? is it x^4?
Hey Brent. I'm still a bit confused as why you chose these prime factors to multiply (2x2x2x3x3x5x7), and why you excluded others. Can you shed some light on that?Brent@GMATPrepNow wrote:ASIDE: A lot of integer property questions can be solved using prime factorization.sgr21 wrote:What will be the least number which when quadrupled will be exactly divisible by 24, 36, 42 and 60?
(A)630
(B) 1260
(C) 315
(D)420
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
Examples:
24 is divisible by 3 <--> 24 = (2)(2)(2)(3)
70 is divisible by 5 <--> 70 = (2)(5)(7)
330 is divisible by 6 <--> 330 = (2)(3)(5)(11)
56 is divisible by 8 <--> 56 = (2)(2)(2)(7)
Okay, now to the question.
24 = (2)(2)(2)(3)
36 = (2)(2)(3)(3)
42 = (2)(3)(7)
60 = (2)(2)(3)(5)
Let's first find a number that it is divisible by 24, 36, 42 and 60
That number will = (2)(2)(2)(3)(3)(5)(7) [notice that 24, 36, 42 and 60 are all "hiding in this prodcut]
Of course, we're quadrupling the number first. So, we can take (2)(2)(2)(3)(3)(5)(7) and divide by 4 to get (2)(3)(3)(5)(7) = 630
Answer: A
Cheers,
Brent
No problem.[email protected] wrote: Hey Brent. I'm still a bit confused as why you chose these prime factors to multiply (2x2x2x3x3x5x7), and why you excluded others. Can you shed some light on that?
Thanks,
Tom
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