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Question

# Make the correct alternative in the following question: $\mathrm{For}\mathrm{all}n\in \mathbf{N},3×{5}^{2n+1}+{2}^{3n+1}\mathrm{is}\mathrm{divisible}\mathrm{by}$ (a) 19 (b) 17 (c) 23 (d) 25

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Solution

## $\mathrm{Let}\mathrm{P}\left(n\right)=3×{5}^{2n+1}+{2}^{3n+1},\mathrm{for}\mathrm{all}n\in \mathbf{N}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}n=1,\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(1\right)=3×{5}^{2+1}+{2}^{3+1}\phantom{\rule{0ex}{0ex}}=3×{5}^{3}+{2}^{4}\phantom{\rule{0ex}{0ex}}=375+16\phantom{\rule{0ex}{0ex}}=391\phantom{\rule{0ex}{0ex}}=17×23\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}n=2,\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(2\right)=3×{5}^{4+1}+{2}^{6+1}\phantom{\rule{0ex}{0ex}}=3×{5}^{5}+{2}^{7}\phantom{\rule{0ex}{0ex}}=9375+128\phantom{\rule{0ex}{0ex}}=9503\phantom{\rule{0ex}{0ex}}=17×13×43\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As},\mathrm{HCF}\left(391,9503\right)=17\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{P}\left(n\right)\mathrm{is}\mathrm{divisible}\mathrm{by}17.$ Hence, the correct alternative is option (b).

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