hard numbers problem from Princeton CAT

Yes. This is an EXCEPT question, so you're looking for the choice that cannot be the value of j – k. Since we know j – 2 and k – 5 are both divisible by 4, we can come up with values of j = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50... and k = 5, 9, 13, 17, 21, 25, 29, 33... j – k = 14 – 9 = 5, or 26 – 13 = 13, or 34 – 13 = 21 or 46 – 13 = 33. The only answer that you can't eliminate is A, so the answer is A.

"Shortcut" version....

j = 4*(integer)+2
k = 4*(integert) +5

j - k = 4i + 2 - 4i -5

= 4i - 4i -3

= (multiple of 4 - multiple of 4) - 3

= (multiple of 4) - 3

so, if an answer choice plus 3 is a multiple of 4 it is a possible solution.


***Also note that instead of k = 4i +5, we could have used k = 4i +1 since 5 is larger than 4. Then we would have had j-k = multiple of 4 + 1 and we would have subtracted 1 from each choice and checked if it was a multiple of 4***

Thanks very much.
khurram

the best way to aswer these questions is to draw them out and map out hte possibilities.