If $8^{83} + 6^{83}$ is divided by $49$ , then the remainder is

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Solution

$8^{83} + 6^{83} = (1 + 7)^{83} - (1 - 7)^{83} = 2 [^{83} C_{1} 7 +^{83} C_{3} 7^{3} + \dots +^{83} C_{83} 7^{83}] = 2 \times^{83} C_{1} 7 + 49 k, k \in Z = (2 \times 83 \times 7) + 49 k = 166 \times 7 + 49 k = 161 \times 7 + 49 k + 5 \times 7 = 49 (k + 23) + 35$

Therefore, the remainder is $35$ .

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Similar questions

\frac{6^{83} + 8^{83}}{49}

22^{3} + 23^{3} + 24^{3} + . . . . . . . . . + 88^{3}

22^{3} + 23^{3} + 24^{3} + \dots + 87^{3} + 88^{3}

6^{83} + 8^{83}

49,

The remainder obtained when $49^{49} + 51^{24}$ is divided by 25 is

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