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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 $4$.

Some Important Links:

ISI Entrance Course

ISI Entrance Problems and Solutions

Inequality with Twist – Video

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 $4$.

Some Important Links:

ISI Entrance Course

ISI Entrance Problems and Solutions

Inequality with Twist – Video

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

  1. 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?

  2. 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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