Question:

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

Updated On: Sep 16, 2024
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Correct Answer: 82

Approach Solution - 1

Given the provided information, we can calculate the values of A and B as follows: 

For the numbers of the form 3k + 4k + 5k

For k = 1: A = HCF(31 + 41 + 51) = HCF(12) = 12 

For k = 2: A = HCF(32 + 42 + 52) = HCF(50) = 2 

For k = 3: A = HCF(33 + 43 + 53) = HCF(216) = 2 

The highest common factor (HCF) of the values of A is 2. 

For the numbers of the form 4k + 3(4k) + 4(k+2)

For k = 1: B = 41 + 3(41) + 4(1+2) = 80 

For k = 2: B = 42 + 3(42) + 4(2+2) = 136 

For k = 3: B = 43 + 3(43) + 4(3+2) = 560 

The highest common factor (HCF) of the values of B is 80. 

Therefore, A = 2 and B = 80. 

Hence, a + B = 2 + 80 = 82.

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Approach Solution -2

Given, \(A\) is the HCF of \(3^k+4^k+5^k\) for different values of \(k\).
For \(k = 1\),
\(3^k+4^k+5^k=12\)
For \(k = 2\),
\(3^k+4^k+5^k=50\)
For \(k = 3\),
\(3^k+4^k+5^k=216\)
HCF \(= 2\)
Hence, \(A = 2\)
\(4^k+3(4^k)+4^{k+2}\)
\(= 4^k (1 +3) + 4^{k+2}\)
\(=4^{k+1}+4^{k+2}\)
\(=4^{k+1}(1+4)\)
\(=5.4^{k+1}\)
When, \(k = 1,\)
HCF of the values \(= 5\times16 = 80\)
Hence, \(B = 80\)
Now, 
\(A + B =80+2= 82\)

So, the answer is \(82\).

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