Multiplied by 9
Let's find a number, which satisfies the next condition:
Every digit, except the first, is larger than the previous one.
Now multiply it by . Let be the amount of the digits of the result . Find the maximum of .
For example: The number is . . The sum of the digits is . So is a possible value of .
The answer is 9.
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1 solution
The only possible value of S is 9 .
Let the number be a b c d e . We know that a b c d e ∗ 9 = 1 0 ∗ a b c d e − a b c d e = a b c d e 0 − a b c d e . The last digit (from the left) of the result is 1 0 − e . We have 1 remainder. The next digit is e − d − 1 . Now we don't have remainder, because e > d . The next digit is d − c , then the next is c − b , the second digit is b − a , and the first is a . (We don't have remainder, because 0 < a < b < c < d < e .) Now we can see that the amount of S digits is a + ( b − a ) + ( c − b ) + ( d − c ) + ( e − d − 1 ) + ( 1 0 − e ) = − 1 + 1 0 = 9 . .
When the number is not a five-digit number, then the way of thinking is the same as here. So the maximum value of S is 9 .