This morning, Zirui sent me a question:
A number has a digit sum of 104. After multiplying the number by 2, the digit sum becomes 100. It is known that this number contains no 9s, but has four 8s, three 7s, and two 6s. How many 5s does this number have?
Of course, this problem is not difficult; a little analysis yields the answer. However, it must be said that this is an interesting problem, more like a numerical game.
Four 8s, three 7s, and two 6s sum to 65, which is 39 short of 104. The remaining part consists of x fives and some number of 0s, 1s, 2s, 3s, and 4s. Let the remaining 39 be composed of x fives and a set of digits \{0, 1, 2, 3, 4\} whose sum is y. That is: 39 = 5x + y
The key to the problem is the multiplication by 2. How does the sum of the digits change after multiplying by 2? Obviously, when 0, 1, 2, 3, or 4 is multiplied by 2, they become 0, 2, 4, 6, or 8, respectively. That is, their contribution to the digit sum simply doubles. However, when 5, 6, 7, 8, or 9 is multiplied by 2, they become 10, 12, 14, 16, or 18. When calculating the digit sum, the tens and units places are each counted as a single digit. Thus, after multiplying by 2, the digits 5, 6, 7, 8, and 9 “correspondingly” become 1, 3, 5, 7, and 9.
Therefore, after multiplying the number by 2, the four 8s, three 7s, and two 6s equivalently become four 7s, three 5s, and two 3s. Their sum is 49, which is 51 short of 100. This gives us: 51 = x + 2y
Solving the system of equations, we get x = 3, which means there are three 5s.
Clearly, this number can be arbitrarily large because it can contain infinitely many zeros. If there are no zeros, the remaining digits could all be 1s. Therefore, it is obvious that there is no way to determine exactly what this number is.
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