Question 21.

Let A be the largest positive integer that divides all the numbers of the form 3k+4k+5k3^k+4^k+5^k, and B be the largest positive integer that divides all the numbers of the form 4k+3(4k)+4k+24^k+3\left(4^k\right)+4^{k+2}, where k is any positive integer. Then (A+B) equals

A
B
C
D

Question Explanation

Text Explanation

A is the HCF of 3k+4k+5k3^k + 4^k + 5^k for different values of k.

For k = 1, value is 12

For k = 2, value is 50

For k = 3, value is 216

HCF is 2. Therefore, A = 2

4k+3(4k)+4k+2=4k(1+3)+4k+2=4k+1+4k+2=4k+1(1+4)=54k+14^k + 3(4^k) + 4^{k+2} = 4^k(1 + 3) + 4^{k+2} = 4^{k+1} + 4^{k+2} = 4^{k+1}(1 + 4) = 5 \cdot 4^{k+1}

HCF of the values is when k = 1, i.e. 516=805 \ast 16 = 80

Therefore, B = 80

A + B = 82

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