Find the Sum of the Digits
This problem is from Hai He, Hunter College of CUNY.
Let A be the sum of the digits of 1999! (that is 1999*1998*1997*..*2*1). Let B be the sum of the digits of A. Let C be the sum of the digits of B. What is the value of C?
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Solution to Find the Sum of the Digits
Correct solutions were submitted by Wyatt Witemeyer.
| 1999! = | 1.2…999.1000.1001…1998.1999 |
| = | (1.1999)(2.1998)…(999.1001).1000 |
| < | (1000.1000)(1000.1000)…(1000.1000).1000 |
| = | 10001999 |
| = | 105997 |
Therefore 1999! has at most 5997 digits. This implies that A (the sum of the digits of 1999!) is at most 5997.9 = 53, 973.
So B (the sum of the digits of A) is at most 5 + 4.9 = 41.
Hence C (the sum of the digits of B) is at most 4 + 9 = 13.
But the sum of the digits of a positive integer N has the same remainder upon division by 9 as does N itself. So 1999! and A and B and C all have the same remainder upon division by 9. But 9 divides 1999!, so 9 must divide C. Since C is at most 13 and C is not 0, then
C = 9.
