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.
multiples
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- Vemuri
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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.
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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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
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
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
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
I had a slightly different approach heading in the same direction..PLEASE CLARIFY WHETHER MY APPROACH IS RIGHT OR NOT.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
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
- Uri
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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.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
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Udi,Uri wrote: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.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
What is the source of the question here?
Charged up again to beat the beast
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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
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!!