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(m + 2n)(m - 6n)
(m - 6n)(m + 2n)
Suppose m and n are integers. Then 2m+1 is an odd number and 2n is an even number.(2m + 1) * 2n = 4mn + 2n = 2*(2mn + 1). Since m and n are integers, the closure of the set of integers under multiplication and addition implies that 2mn + 1 is an integer. Thus the product is a multiple of 2: that is, it is even.
Suppose m and n are integers. Then 2m + 1 and 2n +1 are odd integers.(2m + 1)*(2n + 1) = 4mn + 2m + 2n + 1 = 2*(2mn + m + n) + 1 Since m and n are integers, the closure of the set of integers under multiplication and addition implies that 2mn + m + n is an integer - say k. Then the product is 2k + 1 where k is an integer. That is, the product is an odd number.
It's clear that the first set has 2m subsets and second one has 2n subests. so we have to solve 2m - 2n = 56 (m,n are positive integers) Its also clear that m>n let m = k+n so 2k+n- 2n= 56 2n(2k- 1)= 23*7 clearly 2k-1 is odd so 2n= 23, and n = 3 and 2k-1= 7 so k =3 so m = 3+3 = 6 and n = 3