A probability problem by Prabhav Bansal
How many 7-digit numbers are there formed by using only the digits 5 and 7 and divisible by both 5 and 7?
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1 solution
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Clearly, the last digit must be 5 and we have to determine the remaining 6 digits. For divisibility by 7, it is sufficient to consider the number obtained by replacing 7 by 0; eg, 5775755 is divisible by 7, if and only if 5005055 is divisible by 7. Each such number is obtained by adding some of the numbers from the set{50, 500, 5000 ,50000, 500000, 5000000} along with 5.
We look at the remainders of these when divided by 7; they are {1, 3, 2, 6, 4, 5}. Thus it is sufficient to check for those combinations of remainders which add up to a number of the form 2+7k, since the last digit is already 5. There are {2 ; 3,6 ; 4,5 ; 2,3,4 ; 1,3,5 ; 1,2,6 ; 2,3,5,6 ; 1,4,5,6 ; 1,2,3,4,6}.
These correspond to the numbers 7775775, 7757575, 5577775, 7575575, 5777555, 7755755, 5755575, 5557755, 7555555. Hence the answer comes out to be 9.