Suppose x & y are positive integers,x>y, and 3x+4y & 2x+3y when divided by 5,leave remainders 2&3 respectively. It follows that when (x-y) is divided by 5,the remainder necessarily equals

(a) 2 (b)1 (c) 4 (d) none of these

According to the problem,

\((3x+2y)\equiv 2 \bmod{5} \)

\((2x+3y)\equiv 3 \bmod{5} \)

subtracting the above 2 relations we get,

\((x-y)\equiv (-1) \bmod{5} \)

i.e. \((x-y)\equiv 4 \bmod{5} \)

Hence the remainder is 4

The question says 3x+4y and you have solved the problem with 3x+2y (as on 6 may 2015). According to the question shouldn’t the answer be 1?

I checked the question in my book. As per to book, in question there should be 3x+2y instead of 3x+4y. So answer is 4 and in question there is typing mistake!

3x + 2y = 5p + 2, where p is an integer. ———- (1)

[Since given 3x + 2y leaves remainder 2 when divided by 5]

Similarly, 2x + 3y = 5q + 3 ——– (2)

(1) – (2): x – y = 5(p – q) – 1

==> The remainder is -1, = -1 + 5 = 4

Thus when (x – y) is divided by 5, remainder is 4.

I think this is the solution !