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Bunuel
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 20 Nov 2018, 00:17
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E
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54% (02:29) correct 46% (02:35) wrong based on 142 sessions

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If 7(a – 1) = 17(b – 1), and a and b are both positive integers the product of which is greater than 1, then what is the least possible sum of a and b?

A. 2
B. 7
C. 17
D. 24
E. 26

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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr Updated on: 01 Dec 2018, 17:17
Originally posted by KSBGC on 20 Nov 2018, 01:47.
Last edited by KSBGC on 01 Dec 2018, 17:17, edited 1 time in total.
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Bunuel
If 7(a – 1) = 17(b – 1), and a and b are both positive integers the product of which is greater than 1, then what is the least possible sum of a and b?

A. 2
B. 7
C. 17
D. 24
E. 26
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7(a-1) =17(b-1)

17 is not divisible by 7. thus b-1 must be divisible by 7. So, b - 1 = 7k . least value of b= 8.

on the other hand, 7 is not divisible by 17. (a -1 ) is must be the multiple of 17. so least value of a = 18.


least value of a + b = 18 + 8 = 26.

E is our answer.
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 20 Nov 2018, 00:42
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\(7a + 10 = 17b\)

Least possible value apart from (1,1) is (18,8)

\(7(18) + 10 = 17(8)\)

So a+b = 26.

OPTION: E
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 20 Nov 2018, 05:55
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eswarchethu135
\(7a + 10 = 17b\)

Least possible value apart from (1,1) is (18,8)

\(7(18) + 10 = 17(8)\)

So a+b = 26.

OPTION: E
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Hi eswarchethu135
How did you get 18 , 8 in one step. Is that any shortcut ? Please explain.
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eswarchethu135
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 20 Nov 2018, 05:58
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Haha! I just plugged in numbers. I kept on going from 1 to 8 multiplying with 17 and checking for divisibility by 7 after subtracting 10 from the product. It took me 45 seconds to verify from 1 to 8. I think this time is not a big deal in GMAT.

Posted from my mobile device
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 27 Nov 2018, 01:15
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Bunuel
If 7(a – 1) = 17(b – 1), and a and b are both positive integers the product of which is greater than 1, then what is the least possible sum of a and b?

A. 2
B. 7
C. 17
D. 24
E. 26
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Since both 7 and 17 are prime numbers, take LCM of 7 and 17,i.e. 119.
For both RHS and LHS to be minimum 119, (a-1) must take the value of 17. Thus, a can be a minimum of 18.
By this logic,(b-1) must take minimum value of 7. So b has to be 8.
Thus, a+b=18+8=26.

Hope this helps.
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 27 Nov 2018, 01:24
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Bunuel
If 7(a – 1) = 17(b – 1), and a and b are both positive integers the product of which is greater than 1, then what is the least possible sum of a and b?

A. 2
B. 7
C. 17
D. 24
E. 26
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\(a - 1= \frac{17(b – 1)}{7}\)

As a is an Integer, b =8. Hence, a = 18.

a+b = 26.
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If 7(a 1) = 17(b 1), and a and b are both positive integers the pr 30 Nov 2018, 08:33
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Two numbers are equal if the expression on each side is equal to the given number on opposite side.
For e.g:

7(a-1)=17(b-1)

so,
a-1=17
b-1=7

Hence, a=18
b=8

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