A) \[\frac{1}{4}\]
B) \[\frac{1}{7}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{49}\]
Correct Answer: C
Solution :
| \[{{7}^{m}}+{{7}^{n}}=[{{(5+2)}^{m}}+{{(5+2)}^{n}}]\]\[\equiv 5\times \operatorname{int}egar+{{2}^{m}}+{{2}^{n}}\] |
| \[\therefore \]\[{{7}^{m}}+{{7}^{n}}\]is divisible by 5 i ff \[{{2}^{m}}+{{2}^{n}}\] is divisible by 5 and so unit place of \[{{2}^{m}}+{{2}^{n}}\] must be 0 since it cannot be 5. |
| m possible n |
| 1 \[3,7,11,15,.......=\,\,25\] |
| 2 \[4,8,12,........=\,\,25\] |
| 3 \[1,5,9,........=\,\,25\] |
| 4 \[2,6,10,.......=\,\,25\] |
| Since \[{{2}^{1}}+{{2}^{3}}\equiv {{2}^{3}}+{{2}^{1}}\]so \[(1,3)\] and \[(3,1)\] are same as favourable cases. |
| \[\therefore \]Required probability \[=\frac{25\times 50}{100\times 100}=\frac{1}{8}\] |
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