One of the factors of (8 2k + 5 2k ), where k is an odd number, is:
89
Rewrite the exponents with a common power: \(8^{2k}=(8^2)^k=64^k\) and \(5^{2k}=(5^2)^k=25^k\). The expression becomes \(64^k+25^k\).
For any odd positive integer \(n\), \(a^n+b^n\) is divisible by \(a+b\) (substituting \(a=-b\) makes the polynomial zero, so \((a+b)\) is a factor).
Since \(k\) is odd, \(64^k+25^k\) is divisible by \(64+25=89\). Therefore 89 is a factor.
Similar Questions
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Important Questions from Surds and Indices
The value of \(\frac{{{{\left( {251} \right)}^3} + {{\left( {249} \right)}^3}}}{{25.1 \times 25.1 - 624.99 + 24.9 \times 24.9}}\) is 5 × 10 k , where the value of k is :
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If √625 = 25; then√(.00000625/25)is:
A. 0.0025
B. 0.001
C. 0.0001
D. 0.0005Find the value of:
\(\sqrt{150}-\sqrt{54}-\sqrt{24}\)
If \(\sqrt{4624}=68\) , then the value of:
\(\sqrt{46.24}+\sqrt{0.4624}+\sqrt{0.004624}\)
