IPMAT Indore 2025 · Quantitative Ability · Question 2

IPMAT Indore 2025 previous year question on Integral Solutions, with the answer and a step-by-step solution.

Integral SolutionsNumber SystemType in the answer (TITA)Medium
If m and n are two positive integers such that 7m+11n=200, then the minimum possible value of m+n is
Answer & solution

Answer: 20

Step 1 — why we want n large. Since 11>7, each unit of n “uses up” more of the 200 than a unit of m does. So to make m+n small we want as much of the 200 as possible carried by n.

Step 2 — find which n are allowed. We need 7m=200-11n to be a positive multiple of 7. Working modulo 7:

200\equiv 4 (\mod 7), 11n\equiv 4n (\mod 7)

4-4n\equiv 0 (\mod 7) \Rightarrow 4\left( 1-n \right)\equiv 0 \Rightarrow n\equiv 1 (\mod 7)

So n=1,8,15,22,\ldots

Step 3 — list the solutions (stopping when 11n exceeds 200):

n

200-11n

m

m+n

1

189

27

28

8

112

16

24

15

35

5

20

22

-42

—

not valid

Step 4 — read off the minimum. The largest usable n is 15, giving

m+n=5+15=20

Answer: 20

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