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divisibilty

by sgr21 » Mon Oct 21, 2013 10:55 am
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
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by Brent@GMATPrepNow » Mon Oct 21, 2013 11:07 am
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
ASIDE: 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

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
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by theCodeToGMAT » Mon Oct 21, 2013 11:10 am
(x)*4 is divisible be 24, 36, 42 and 60

24 = 2x2x2x3
36 = 2x2x3x3
42 = 2x3x7
60 = 2x2x3x5

Least X = 2x3x3x7x5 = 630
Last edited by theCodeToGMAT on Mon Oct 21, 2013 11:31 am, edited 1 time in total.
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by theCodeToGMAT » Mon Oct 21, 2013 11:13 am
Brent, what does quadrupled mean? is it x^4?
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by Brent@GMATPrepNow » Mon Oct 21, 2013 11:26 am
theCodeToGMAT wrote:Brent, what does quadrupled mean? is it x^4?
It means "multiplied by 4"

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by [email protected] » Mon Oct 21, 2013 11:45 am
Brent@GMATPrepNow wrote:
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
ASIDE: 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

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
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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by Brent@GMATPrepNow » Mon Oct 21, 2013 2:38 pm
[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
No problem.

Let's look at each number on its own.

To begin, we want the number, N, to be divisible by 24
24 = (2)(2)(2)(3)
So, we need three 2's and one 3 in the prime factorization of N.
In other words, N = (2)(2)(2)(3) [so far]

We also we want N to be divisible by 36
36 = (2)(2)(3)(3)
So, we need two 2's and two 3's in the prime factorization of N.
We already have more than two 2's in the prime factorization of N, so we're good there. We also have one 3 so far, so we need to add one more 3.
So, now we have N = (2)(2)(2)(3)(3) [so far]

We also we want N to be divisible by 42
42 = (2)(3)(7)
So, we need one 2, one 3 and one 7 in the prime factorization of N.
We already have one 2 and one 3, so we're good there. But we do need to add a 7
So, now we have N = (2)(2)(2)(3)(3)(7)

Finally, we want N to be divisible by 60
60 = (2)(2)(3)(5)
So, we need to add a 5
So, now we have N = (2)(2)(2)(3)(3)(7)(5)

So, when N = (2)(2)(2)(3)(3)(5)(7), N is divisible by 24, 36, 42 and 60
Since we're multiplying some number by 4 to get N, we'll take (2)(2)(2)(3)(3)(5)(7) and divide by 4 to get 630

Cheers,
Brent
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