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Integer d is the product of the integers a, b, and c and 1 < a < b < c

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Integer d is the product of the integers a, b, and c and 1 < a < b < c [#permalink]

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Integer d is the product of the integers a, b, and c and 1 < a < b < c. If the remainder when 233 is divided by d is 79, what is the value of a+c?

A. 7
B. 9
C. 13
D. 14
E. 18
[Reveal] Spoiler: OA

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Kudos [?]: 128912 [0], given: 12183

Expert Post
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Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c [#permalink]

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New post 13 Sep 2017, 03:52
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Bunuel wrote:
Integer d is the product of the integers a, b, and c and 1 < a < b < c. If the remainder when 233 is divided by d is 79, what is the value of a+c?

A. 7
B. 9
C. 13
D. 14
E. 18


d=a*b*c.....
If 233 lives a Remainder 79 when div by d, 233-79 or 154 should be div by d...

154=1*2*7*11..
Now 1<a<b<c and a*b*c=2*7*11, a+c=2+11=13
C
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Kudos [?]: 5476 [0], given: 112

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Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c [#permalink]

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New post 13 Sep 2017, 03:59
given d= a*b*c
1<a<b<c
find: a+c

by equation 233 = d*x +79
d*x = 233-79 = 154.
if x=1 => d= 154

so a*b*c = 154 = 2*7*11
=> a=2 and c=11 => a+c = 2+11 =13

Answer: C

Kudos [?]: 59 [0], given: 59

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Joined: 19 Sep 2016
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Kudos [?]: 2 [0], given: 3

Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c [#permalink]

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New post 13 Sep 2017, 18:38
233=dx+79 which is the equation of the remainder. Simplifying the equation becomes - dx=154 and it is given that d=a*b*c so substituting the value of d in above equation we get (abc)*x=154. Now prime factorization of 154 is 2*7*11 and assuming the quotient i.e. x=1 and considering the condition given ie. 1<a<b<c. This states that a=2, b=7 & c=11. So a+c=2+11=13, hence the answer is C.


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Kudos [?]: 2 [0], given: 3

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Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c [#permalink]

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New post 13 Sep 2017, 20:47
Bunuel wrote:
Integer d is the product of the integers a, b, and c and 1 < a < b < c. If the remainder when 233 is divided by d is 79, what is the value of a+c?

A. 7
B. 9
C. 13
D. 14
E. 18


233 can be divided by 79 or 158 to give us a remainder of 79

When divided by 158, the difference is 75(3*5^2) but a < b < c
But, when divided by 79, the difference is 154(2*7*11).
Here 1 < a=2 < b=7 < c=11!

Therefore, a+c = 2+11 = 13(Option C)
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Kudos [?]: 594 [0], given: 16

Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c   [#permalink] 13 Sep 2017, 20:47
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