Question

There are two sets A and B with \( |A| = m \) and \( |B| = n \). If \( |P(A)| - |P(B)| = 112 \) then choose the wrong option (where \( |A| \) denotes the cardinality of A, and P(A) denotes the power set of A)

• A. \( m+n = 11 \)
• B. \( 2n-m = 1 \)
• C. \( 2m-n = 1 \)
• D. \( 3n-m = 5 \)

Hint:

For equations of the form \( 2^a - 2^b = C \), factor out the smaller power \( 2^b(2^{a-b}-1) \). The term \( 2^{a-b}-1 \) will be odd. Find the prime factorization of \( C \) into a power of 2 and an odd number, and then equate the corresponding parts to solve for the exponents.

Answer 1

The Correct Option is C

The cardinality of the power set of a set with \( k \) elements is \( 2^k \). So, \( |P(A)| = 2^m \) and \( |P(B)| = 2^n \). We are given the equation: \[ 2^m - 2^n = 112 \] Since the difference is positive, we must have \( m > n \). Let's factor out the smaller power, \( 2^n \). \[ 2^n(2^{m-n} - 1) = 112 \] Now, we find the prime factorization of 112. \[ 112 = 2 \times 56 = 2 \times 8 \times 7 = 2 \times 2^3 \times 7 = 2^4 \times 7 \] So, our equation is: \[ 2^n(2^{m-n} - 1) = 2^4 \times 7 \] By comparing the terms, the power of 2 on the left must match the power of 2 on the right. \[ 2^n = 2^4 \implies n = 4 \] The odd factor on the left must match the odd factor on the right. \[ 2^{m-n} - 1 = 7 \] \[ 2^{m-n} = 8 = 2^3 \] \[ m-n = 3 \] Substitute \( n=4 \) to find m: \[ m - 4 = 3 \implies m = 7 \] So we have \( m=7 \) and \( n=4 \). Now we check which of the options is wrong. (A) \( m+n = 7+4 = 11 \). (Correct) (B) \( 2n-m = 2(4) - 7 = 8 - 7 = 1 \). (Correct) (C) \( 2m-n = 2(7) - 4 = 14 - 4 = 10 \). The option says it is 1. This is wrong. (D) \( 3n-m = 3(4) - 7 = 12 - 7 = 5 \). (Correct) The wrong option is (C).