Probability Ques 4
- If the integers $m$ and $n$ are chosen at random between $1$ and $100$ , then the probability that a number of the form $7^m+7^n$ is divisible by $5$ , equals
(1999, 2M)
(a) $\frac{1}{4}$
(b) $\frac{1}{7}$
(c) $\frac{1}{8}$
(d) $\frac{1}{49}$
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Answer:
Correct Answer: 4.(a)
Solution: (a) $7^1=7,7^2=49,7^3=343,7^4=2401, \ldots$
Therefore, for $7^r, r \in N$ the number ends at unit place $7,9,3,1,7, \ldots$
$\therefore \quad 7^m+7^n$ will be divisible by $5$ if it end at $5$ or $0$ .
But it cannot end at $5$ .
Also for end at $0$ .
For this $m$ and $n$ should be as follows
| $m$ | $n$ | |
|---|---|---|
| 1 | $4 r$ | $4 r-2$ |
| 2 | $4 r-1$ | $4 r-3$ |
| 3 | $4 r-2$ | $4 r$ |
| 4 | $4 r-3$ | $4 r-1$ |
For any given value of $m$, there will be $25$ values of $n$.
Hence, the probability of the required event is
$ \frac{100 \times 25}{100 \times 100}=\frac{1}{4} $
NOTE: Power of prime numbers have cyclic numbers in their unit place.
BETA
