Let $N = 55^{3} + 17^{3} - 72^{3}$ then N is divisible by

both 7 and 13

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both 3 and 13

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both 17 and 7

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both 3 and 17

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Solution

The correct option is D both 3 and 17
N can be written either $(54 + 1)^{3} + (18 - 1)^{3} - 72^{3}$
or $(51 + 4)^{3} + 17^{3} - (68 + 4)^{3}$
Hence, the first form is divisible by $3$ and second form is divisible by $17$

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Similar questions

N = 55^{3} + 17^{3} - 72^{3} .

N = 55^{3} + 17^{3} - 72^{3} .

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P = \frac{3}{17} + \frac{33}{17^{2}} + \frac{333}{17^{3}} + \dots \infty

P

\to

\in

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