for sure the answer is B, but to me this problem is too time consuming,
it will be great if somebody shares any shortcut

17a+19b=?, find b-?
(1) a+b=8 not suff
(2) 17a+19b=140, we must prove the only integers for a and b exist
here b=2,a=6 (17*6+19*2=140)
and there are no others integers for a and b that give proper result

I agree!

clock60 wrote:for sure the answer is B, but to me this problem is too time consuming,
it will be great if somebody shares any shortcut

17a+19b=?, find b-?
(1) a+b=8 not suff
(2) 17a+19b=140, we must prove the only integers for a and b exist
here b=2,a=6 (17*6+19*2=140)
and there are no others integers for a and b that give proper result

If you wanted to SOLVE the problem (ie a PS problem) first I would subtract 19b from both sides, then divide by 17

(17b/17) = (140/17) - (19a/17)

You know that the left side is an integer so the right side must also be an integer

integer = (136/17) + (4/17) - (17/17) * a - (2/17) * a
integer = integer + 4/17 - integer - (2/17) * a

therefore 4/17 - (2/17) * a must be an integer to remove fractions. a = 2 is the only positive solution that could also satisfy the stem equation (19, etc would work but couldn't possibly satisfy the original equation)

hi Idoolitt
thanks for sharing, but can you elaborate i little bit more
why first I would subtract 19b from both sides, then divide by 17
it sounds great but why in that order
what if substract 17a from both parts and then to divide by 19?
or it does not matter?

clock60 wrote:hi Idoolitt
thanks for sharing, but can you elaborate i little bit more
why first I would subtract 19b from both sides, then divide by 17
it sounds great but why in that order
what if substract 17a from both parts and then to divide by 19?
or it does not matter?

Does not matter

(19a/19) = (140/19) - (17b/19)
integer = (133/19) + (7/19) - (19/19) * b + (2/19) * b
integer = integer + 7/19 - integer + (2/19) * b
7/19 + (2/19)*b = integer

b=6 works there

great reply!!!
thanks ldoolitt