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If the integers \(m\) and \(n\) are chosen at random from \(1\) to \(100\), then find the probability that the number of the form \(7^n+7^m\) is divisible by \(5\). $$ .$$
Number of numbers of the form \(7^n+7^m\) is 100×100= \(100^2\) , since m,n ∈{1 ,…………,100}.
Now \(7^1=7,7^2=49,7^3=343\) and \(7^4=2401\) , the digits at the unit places are 7,9,3 and 1.
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For n ≥5 the digit at the unit place for the number \(7^n\) will be one of among the numbers 1,3,7 and 9 $$ .$$
A number is divisible by 5 iif the digit at the unit’s place is either 0 or 5.
Therefore a number of the form \(7^n+7^m\) will be divisible by 5 iff the digits at the unit place of them add up to 10. The possible cases being {1,9} and {3,7} in any order.
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Since there are 4 distinct digits at the unit place for the number \(7^n\) and they are repeated modulo 4, we have 100/4=25 distinct number of the form \(7^n\) having the digit 1 or 3 or 7 or 9 at the unit place.
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When \(7^n\) has the digit 1 at the unit place, \(7^m\) need to have the digit 9 at the unit place giving 25 such choices. In total 2 x 25 x 25 choices ( n and m can be interchanged!)
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Similarly for the pair {3,7} we have \( 2 \times 25 \times 25 \) choices, in total giving \( 4 \times 25 \times 25 \) choices!!
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Required probability $$ = {4 \times 25 \times 25 \over 100 \times 100} = {1 \over 4} .$$