Problem#15
I take a random three-digit number.I double the number.What is the probability that the sum of the digits of the doubled number is double that of the original number? The probability can be expressed in the form where a and b are co-prime.What is the sum of ?
This problem is a part of the series <One minute problems>
The answer is 10.
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1 solution
Any number comprised of the digits 0 − 4 (inclusive) clearly satisfies this property. A number with a digit j greater than 4 will, when doubled, "overflow" the neighboring digit by 1. Then, 2 j − 1 0 is that respective digit in the doubled number, and due to the "overflow", another 1 is added to the digit sum of the doubled number. Since 2 j > 2 j − 1 0 + 1 , the digit sum of the original number is thus always greater than half of that of the doubled.
So we're looking for the probability a random 3 digit number is comprised of the digits 0 − 4 . There are 4 ∗ 5 ∗ 5 = 1 0 0 such numbers ( 0 can't be first digit). Thus the desired probability is 1 0 0 / 9 0 0 = 1 / 9 , making the answer 1 + 9 = 1 0 .