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Three company of soldiers containing 120, 192, and 144 soldiers are to

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Bunuel
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Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink] New post   24 Feb 2020, 05:39
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Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

A. 8
B. 12
C. 19
D. 24
E. 38


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C
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nick1816
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Re: Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink] New post   24 Feb 2020, 13:35
HCF (120,192,144)= 24

Least number of such groups= (120/24) + (192/24)+ (144/24)= 5+8+6= 19


Bunuel wrote:
Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

A. 8
B. 12
C. 19
D. 24
E. 38


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Re: Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink] New post   25 Feb 2020, 08:34
for 120,192 & 144 l HCF ; 2^3*3 ; 24
120/24+192/24+144/24 = 5+8+6 ; 19
IMO C


Bunuel wrote:
Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

A. 8
B. 12
C. 19
D. 24
E. 38


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Re: Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink] New post   25 Feb 2020, 09:46
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Bunuel wrote:
Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

A. 8
B. 12
C. 19
D. 24
E. 38


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HCF (120,192,144) = 24

Number of groups (min) = 120/24 + 192/24 + 144/24 = 5 + 8 + 6 = 19

IMO C
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Re: Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink] New post   02 Mar 2020, 07:00
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Bunuel wrote:
Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

A. 8
B. 12
C. 19
D. 24
E. 38




We need to find the greatest common factor (GCF) of 120, 192, and 144. First we can prime factorize them:

120 = 12 x 10 = 2^3 x 3^1 x 5^1

192 = 2 x 96 = 2 x 2 x 48 = 2^6 x 3^1

144 = 12 x 12 = 2^4 x 3^2

So, the GCF of 120, 192, and 144 is 2^3 x 3^1 = 24, so there will be 24 soldiers in each group.

Thus, the smallest number of groups is 120/24 + 192/24 + 144/24 = 5 + 8 + 6 = 19.

Answer: C
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Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink] New post   02 Mar 2020, 08:07
We need to find the biggest number that divides all three. Find prime factors and find numbers that are common i.e....

120 --> (2^3) * 3 * 5. (5 not common so ignore)
144 --> (2^4) * 3^2
192 --> (2^5) * 3

What's common amongst all three? 2^3 * 3 = 24.

Divide 120, 144, & 192 by 24.
5 + 6 + 8 = 19.
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Three company of soldiers containing 120, 192, and 144 soldiers are to [#permalink]
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