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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

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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

by M7MBA » Sun Jun 14, 2020 8:30 am

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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

[spoiler]OA=E[/spoiler]

Source: Princeton Review
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Re: If \(7(a - 1) = 17(b - 1),\) and \(a\) and \(b\) are both positive integers the product of which is greater than 1,

by Jay@ManhattanReview » Mon Jun 15, 2020 9:17 pm
M7MBA wrote: ↑
Sun Jun 14, 2020 8:30 am
 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

[spoiler]OA=E[/spoiler]

Source: Princeton Review
Observing \(7(a - 1) = 17(b - 1),\) we find that 7 and 17 are co-prime to each other; thus, we have \((a - 1) = 17=> a = 18\) and \(7=(b - 1)=b=8\)

So, the least possible sum of \(a\) and \(b\) = 18 + 8 = 26

Correct answer: E

Hope this helps!

-Jay
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Re: If \(7(a - 1) = 17(b - 1),\) and \(a\) and \(b\) are both positive integers the product of which is greater than 1,

by swerve » Tue Jun 16, 2020 4:26 am
M7MBA wrote: ↑
Sun Jun 14, 2020 8:30 am
 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

[spoiler]OA=E[/spoiler]

Source: Princeton Review
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, we have
\(a-1=17\)
\(b-1=7\)

Hence, \(a=18\) and \(b=8\) and \(a+b=26\)
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Re: If \(7(a - 1) = 17(b - 1),\) and \(a\) and \(b\) are both positive integers the product of which is greater than 1,

by gentvenus » Tue Jun 16, 2020 8:51 am
E.
7 and 17 are co-prime to each other
Least possible sum = 18 + 8 = 26
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Re: If \(7(a - 1) = 17(b - 1),\) and \(a\) and \(b\) are both positive integers the product of which is greater than 1,

by Scott@TargetTestPrep » Sat Jun 20, 2020 1:56 pm
M7MBA wrote: ↑
Sun Jun 14, 2020 8:30 am
 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

[spoiler]OA=E[/spoiler]

Solution:

Since a and b are positive integers and ab > 1, then a and b are both greater than 1.
Since 7 and 17 do not have any common factors other than 1, we see that a - 1 must be divisible by 17 and b -1 must be divisible by 7. So a and b are both greater than 1, we need a - 1 = 17 and b - 1 = 7 in order to have the least possible sum of a and b. In that case, a = 18 and b = 8, and their sum is 26.

Answer: E

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