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TOMATO Objective 21 | ISI Entrance Exam

This is a problem from TOMATO Objective 21 based on positive integers. This problem is helpful for ISI Entrance Exam. Try out the problem.

Problem: TOMATO Objective 21 

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

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Inequality with Twist – Video

By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

3 replies on “TOMATO Objective 21 | ISI Entrance Exam”

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?

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 !

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