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7⁶ⁿ- 6⁶ⁿ, where n is an integer >0, is divisible by

113
2127
3559
4All of these

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 7^{6 n} - 6^{6 n} = 7^{6} - 6^{6} \\ = {(7^{3})}^{2} - {(6^{3})}^{2} \\ = (7^{3} - 6^{3}) (7^{3} + 6^{3}) \\ = (343 - 216) \times (343 + 216) \\ = 127 \times 559 \\ = 127 \times 13 \times 43 \end{aligned}

$\eqalign{ & {7^{6n}} - {6^{6n}} = {7^6} - {6^6} \cr & = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr & = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr & = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr & = 127 \times 559 \cr & = 127 \times 13 \times 43 \cr}$
Clearly, it is divisible by 127, 13 as well as 559.

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Clearly, it is divisible by 127, 13 as well as 559. ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 7^{6 n} - 6^{6 n} = 7^{6} - 6^{6} \\ = {(7^{3})}^{2} - {(6^{3})}^{2} \\ = (7^{3} - 6^{3}) (7^{3} + 6^{3}) \\ = (343 - 216) \times (343 + 216) \\ = 127 \times 559 \\ = 127 \times 13 \times 43 \end{aligned}

$\eqalign{ & {7^{6n}} - {6^{6n}} = {7^6} - {6^6} \cr & = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr & = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr & = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr & = 127 \times 559 \cr & = 127 \times 13 \times 43 \cr}$
Clearly, it is divisible by 127, 13 as well as 559. ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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