IPMAT Indore 2025 (SA) - If m and n are two positive integers such that 7m + 11n = 200, then the minimum possible value of m + n is | PYQs + Solutions

IPMAT Indore 2025

Number System

Integral Solutions

Medium

If $m$ and $n$ are two positive integers such that $7m + 11n = 200$ , then the minimum possible value of $m + n$ is

Entered answer:

✅ Correct Answer: 20

To find the minimum value of

m + n

where

7m + 11n = 200

and

m, n

are positive integers:

Rearranging the equation to isolate

m

7m + 11n = 200

7m = 200 - 11n

m = \dfrac{200 - 11n}{7}

Expressing

m + n

in terms of

n

m + n = \dfrac{200 - 11n}{7} + n

= \dfrac{200 - 11n + 7n}{7}

= \dfrac{200 - 4n}{7}

Since the coefficient of

n

is negative in our expression,

m + n

decreases as

n

increases. To minimize

m + n

, we need to maximize

n

For

m

to be a positive integer:

\dfrac{200 - 11n}{7}

must be a positive integer

200 - 11n

must be divisible by 7

n < \dfrac{200}{11} \approx 18.18

Testing values of

n

For

n = 15

200 - 11(15) = 200 - 165 = 35

m = \dfrac{35}{7} = 5

(an integer)

With

n = 15

and

m = 5

m + n = 5 + 15 = 20