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integers medium level gmat prob

Expert replies
by quantskillsgmat » Tue Dec 27, 2011 9:10 pm
If 4x-17y=1, where x and y are positive integers also x is less than or equals to 1000.find number of pairs (x,y) that satisfy above.
a)59
b)57
c)55
d)58
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Source: — Problem Solving |

by arpan20690 » Tue Dec 27, 2011 11:13 pm
Thanx, quantskillsgmat for posting this question. But i think this question is little harder w.r.t
GMAT standard. I took 5 mins to solve it. and this question is also somewhat tricky. So the anser is this
.............
4x-17y=1 .......(1)
and x<= 1000 ......(2)
so we can get inequality for y also.
from (1) x = (1+17y)/4 ....(3)
so, from (2) (1+17y)/4 <= 1000
y<= 235.23
as y and x both are integer y <= 235

now, as x is integer from (3) we can write (1+17y) will be divisible by 4 i.e
remainder will be 0.
or we can write 17y will have a remainder 3 when it is divided by 4. (understand it)
17 gives remainder 1 when it is divided by 4. so y gives a remainder of 3 when it
is divided by 4. so y can be 3, 7 , 11, 15, 19........this entire series. .....(a)
but y must be less than equal to 235.
so series (a) will have 235 as last term.
we can get no. of terms in that series using AP formula.
3 + (n-1)*4 = 235
so, n = 59.
so no of solution will be 59 for y. hence, x also.
so, (x,y) pair will have 59 values. (answer)
so A IS THE ANSWER. IT NOT UNDERSTOOD PLEASE REPLY.
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by quantskillsgmat » Tue Dec 27, 2011 11:43 pm
thanx arpan i know this approach but this is little time taking approach. i am looking for fast approach
thanx.
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by arpan20690 » Wed Dec 28, 2011 12:19 am
It will not that much time if one is good at equation and remainder calculation.
Because in any approach ultimately u have to apply the AP term formula.
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