| What is this number? |
n is a positive three-digit integer in base ten with the following properties:
1) If the digits of n are reversed, the new three- digit number is 51 more than a third of n.
2) There are between 128 and 130 zeros, inclusive, following the final non-zero digit in n!.
Find n. |
Problem Moderated by: Douglas |
| Problem Solution |
Look at the second property first. For a minimum value of n, ten will divide n! 128 times, so five must divide n! 128 times. Therefore, 128 = [n/5]+[n/25]+[n/125], where [x] is the greatest integer function on x. [n/25] will be just about five times [n/125], as [n/5] will be just about 25 times [n/125]. So, we should start looking around numbers where 31(n/125)=128, or n=516.
From here, n! is followed by 127 zeros. n, then, is between 520 and 529, inclusive.
Let n = 520 + C, and the reversal of n = 100C + 25. From the first property:
(520+C)/3+51=100C+25
(520+C)/3=100C-26
520+C=300C-78
299C=598
C=2
Therefore, n is 522.
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