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Math Expert V
Joined: 02 Sep 2009
Posts: 57290
Integer d is the product of the integers a, b, and c and 1 < a < b < c  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 74% (02:25) correct 26% (02:53) wrong based on 85 sessions

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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

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Math Expert V
Joined: 02 Aug 2009
Posts: 7754
Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c  [#permalink]

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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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GMAT 1: 730 Q50 V38 Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c  [#permalink]

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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
Intern  B
Joined: 19 Sep 2016
Posts: 35
Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c  [#permalink]

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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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Senior PS Moderator V
Joined: 26 Feb 2016
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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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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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Re: Integer d is the product of the integers a, b, and c and 1 < a < b < c  [#permalink]

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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

We see that 233 = dq + 79 where q is a positive integer. Simplifying, we have dq = 154. Since 154 = 14 x 11 = 2 x 7 x 11 and d is a product of 3 distinct positive integers greater than 1, we see that q must be 1 and d must be 154, and hence a = 2, b = 7 and c = 11. So a + c = 2 + 11 = 13.

Answer: C
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