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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
Since, m + 2n is a multiple of k
∴ 3 × (m + 2n) will also be a multiple of k
It is given that (3m + 4n) is also a multiple of k
Now if we subtract (3m + 6n) – (3m + 4n)
We will get 2n which is also a multiple of k
Similarly 2 × (m + 2n) is a multiple of k
And (3m + 4n) – (2m + 4n) = m, which is also a multiple of k.
Hence, k must be a common divisor of m and 2n.