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  4. The number 1399-1993 must be divisible by

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Q. The number $13^{99}-19^{93}$ must be divisible by

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

Let, $M=13^{99}-19^{93}$
$= \left(\right. 1 + 4 \times 3 \left.\right)^{99} - \left(\right. 1 + 2 \times 3^{2} \left.\right)^{93}$
$= \left\{\right. 1 + \,{}^{99} C_{1} \times 4 \times 3 + \,{}^{99} C_{2} \times \left(\right. 4 \times 3 \left.\right)^{2} + ... \left.\right\} - \left\{\right. 1 + \,{}^{93} C_{1} \times \left(\right. 2 \times 3^{2} \left.\right) + \,{}^{93} C_{2} \times \left(\right. 2 \times 3^{2} \left.\right)^{2} + ... \left.\right\}$
$= - 2 \times 3^{5} + 2 \times 3^{4} \times $ integer
$= 2 \times 3^{4} \times $ an integer
Hence, the number is divisible by $162$