Question

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 + 7ⁿ is divisible by 5, is -

• A. $\frac { 1 } { 5 }$
• B. $\frac { 1 } { 7 }$
• C. $\frac { 1 } { 4 }$
• D. $\frac { 1 } { 49 }$

Answer 1

Answer: (A)

$\frac { 1 } { 5 }$

Answer 2

We know 7^k, k Ī N, has 1, 3, 9, 7 at the units place for

k = 4p, 4p – 1, 4p – 2, 4p – 3 respectively, where p = 1, 2, 3,…..

Clearly, 7^m + 7ⁿ will be divisible by 5 if 7^m has 3 or 7 in the units place and 7ⁿ has 7 or 3 in the units place or 7^m has 1 or 9 in the units place and 7ⁿ has 9 or 1 in the units place \ for any choice of m, n the digit in the units place of 7^m + 7ⁿ is 2, 4, 6, 0, or 8. It is divisible by 5 only when this digits is 0.

\ The required probability = $\frac { 1 } { 5 }$ .

Hence (1) is the correct answer