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Sets and Relations question

2020 · 6 Sep · Shift 1 · Q25
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Sets and Relations question

2020 · 6 Sep · Shift 1 · Q25

JEE MainMathematicsSets and RelationsNumerical+4 / −1
Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m.n is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 28

  1. The number of subsets of a set with kkk elements is 2k2^k2k

  2. Therefore:

    • Number of subsets of set AAA =2m= 2^m=2m
    • Number of subsets of set BBB =2n= 2^n=2n

  3. Given that the total number of subsets of AAA is 112112112 more than that of BBB: 2m−2n=1122^m - 2^n = 1122m−2n=112

  4. Factor out 2n2^n2n: 2n(2m−n−1)=1122^n(2^{m-n} - 1) = 1122n(2m−n−1)=112

  5. Now factorize 112112112: 112=24⋅7112 = 2^4 \cdot 7112=24⋅7

  6. Since 2m−n−12^{m-n} - 12m−n−1 is odd, it must divide the odd part of 112112112, i.e. 777. So, 2m−n−1=72^{m-n} - 1 = 72m−n−1=7

  7. Hence, 2m−n=8=232^{m-n} = 8 = 2^32m−n=8=23 so m−n=3m-n = 3m−n=3

  8. Also, 2n=1127=16=242^n = \frac{112}{7} = 16 = 2^42n=7112​=16=24 so n=4n = 4n=4

  9. Therefore, m=n+3=7m = n+3 = 7m=n+3=7

  10. Now compute m⋅nm\cdot nm⋅n: m⋅n=7×4=28m\cdot n = 7 \times 4 = 28m⋅n=7×4=28

So the required integer is 28\boxed{28}28​

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