multiples

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Uri: Master | Next Rank: 500 Posts; Posts: 229; Joined: Tue Jan 13, 2009 6:56 am; Thanked: 8 times; GMAT Score:700

by Uri » Mon Apr 27, 2009 10:28 pm

what could be the value of the product of the positive integers m, n and p, if 21m +28n = 24p?

I. 84
II. 168
III. 672

(A) I
(B) I & II
(C) I and III
(D) I, II and III
(E) None of the above

OA: [spoiler](C)[/spoiler]

Please explain your logic.

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Source: Beat The GMAT — Problem Solving | Mon Apr 27, 2009 10:28 pm

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: multiples

by Vemuri » Mon Apr 27, 2009 11:01 pm

This is an interesting question. Thanks for posting.

Given, 21m+28n=24p ==> 7*(3m+4n)=6*4p. We are asked to find out which of the products satisfy this condition.

Remember m, n & p are 3 different numbers

I. 84 = 2*2*7*3. If m=4, n=3 & p=7, the condition is satisfied. So, I is a possible product

II. 168 = 2*2*2*7*3. Now, based on I, we know that this condition is not possible, because a common number should multiply across all the numbers to make sure that the equation is satisfied. So, this condition is not valid.

III. 672 = 2*2*2*2*2*7*3. This is similar to I, with 2 as a common multiple across m,n & p. So, III is a possible product.

Hence C.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Mon Apr 27, 2009 11:03 pm

Wow; whats the source.....

21m +28n = 24p

7 (3m+4n) = 24 * p

p has to be alteast 7

3m+4n = 24

m=4 n=3 p=7

84 possible

672 = 84*8

Lets see if we can provide an extra 2 to each of m,n,and p and see

3m+4n = 24
3*4 + 4*3 = 24

3*4*2+4*3*2 = 24*2

24+24 = 48

m=8 n=6 p =14

672 possible

168 = 84*2

If either m,n or p has this extra 2 can we make 3m+4n = 24 work

3*4+4*3= 24

3*4*2 + 4*3 cannot equal 24

i.e m=8 n=3 p=7 not possible i.e 168 not possible

3*4+4*3*2 = 24 not possible

m=4 n=6 p=7 not possible i.e 168 not possible

3*4+4*3 = 24*2

m=4 n=3 p=14 not possible i.e 168 not possible

So rule out II

I am sure Ian/Stuart or others may have a easier way to rule out II

C

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sacx: Master | Next Rank: 500 Posts; Posts: 139; Joined: Wed Oct 22, 2008 4:36 am; Thanked: 17 times

by sacx » Tue Apr 28, 2009 2:25 am

21m +28n = 24p

Divide the equation by 28,

3/4*m + n = 6/7*p

The simple solution to this equation would be when m = 4, n = 3 and p = 7.

and every value of m,n and p will satisfy the equation as long as each of the variable is multiplied by the same number.

eg,
multiply each of the variable (m, n and p) by 2. m = 8, n = 6 and p = 14
multiply each of the variable by 3. m = 12, n = 9 and p = 21

1. 84 = 4*7*3. from this m = 4, p = 7 and n = 3. We know these value satisfy the equation

2. 168 = 4*7*3*2. Now there is an extra 2 in the factors hence it will not satisfy the equation

3. 672 = 4*7*3*2*2*2 OR 8*14*6 and that satisfies our equation

Choose C

SACX

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PAB2706: Master | Next Rank: 500 Posts; Posts: 260; Joined: Sun Oct 12, 2008 8:10 pm; Thanked: 4 times

by PAB2706 » Tue Apr 28, 2009 7:33 am

multiply each of the variable (m, n and p) by 2. m = 8, n = 6 and p = 14
multiply each of the variable by 3. m = 12, n = 9 and p = 21

I had a slightly different approach heading in the same direction..PLEASE CLARIFY WHETHER MY APPROACH IS RIGHT OR NOT.

the given equation becomes

3m+4n=24/7 p

tht means that m is a multiple of 3 shud be divisible by 3

n is multiple of 4 so shud be divisible by 4

and p shud be divisible by 7

thus the number mnp shud be the LCM of 3,4 and 7 and their equal multiples ie 6,8 and 14 or 9,12,21 etc.

this is satisfied by I and III

The problem gets solved within some seconds.

thus C

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Uri: Master | Next Rank: 500 Posts; Posts: 229; Joined: Tue Jan 13, 2009 6:56 am; Thanked: 8 times; GMAT Score:700

by Uri » Tue Apr 28, 2009 9:30 pm

PAB2706 wrote:the given equation becomes
3m+4n=24/7 p
tht means that m is a multiple of 3 shud be divisible by 3
n is multiple of 4 so shud be divisible by 4
and p shud be divisible by 7

Could you pleae explain the part in red font a little bit more? I don't think that m is a multiple of 3 and n is a multiple of 4 and others have already shown this above your post.

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Thu May 14, 2009 7:08 am

Uri wrote:
PAB2706 wrote:the given equation becomes
3m+4n=24/7 p
tht means that m is a multiple of 3 shud be divisible by 3
n is multiple of 4 so shud be divisible by 4
and p shud be divisible by 7
Could you pleae explain the part in red font a little bit more? I don't think that m is a multiple of 3 and n is a multiple of 4 and others have already shown this above your post.

Udi,
What is the source of the question here?

Charged up again to beat the beast

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Uri: Master | Next Rank: 500 Posts; Posts: 229; Joined: Tue Jan 13, 2009 6:56 am; Thanked: 8 times; GMAT Score:700

by Uri » Wed Aug 12, 2009 12:32 am

i got it from some other online forum....don't remember it now. sorry

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Wed Aug 12, 2009 1:10 am

IMO C because:
21m +28n = 24p
LCM of 21,28, and 24 =168
we can find the values of m,n and p where
21m=28n=24p
the values are m=8, n=6, p=7
for these values 21m+28n=2 *24p
so if we divide each m and n by 2, we get the values we want

m=4,n=3,p=7..product =84
for all values that are multiple of these numbers, our conditions will satisfy
m=8,n=6,p=14--product =672

you cannot get 168 because that means you are only multiply any one of the numbers with 2.

so C

The powers of two are bloody impolite!!

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