Difference between revisions of "2018 AMC 8 Problems/Problem 14"
(→Solution(factorial)) |
(→Solution(factorial)) |
||
| Line 16: | Line 16: | ||
Making the greatest integer, | Making the greatest integer, | ||
| − | <math>(5)(4 \cdot 2)(3)(2 | + | <math>(5)(4 \cdot 2)(3)(\frac{2}{2})(1)</math> |
<math> = (5)(8)(3)(1)(1) =120</math> | <math> = (5)(8)(3)(1)(1) =120</math> | ||
Revision as of 20:06, 24 November 2023
Contents
Problem
Let 
be the greatest five-digit number whose digits have a product of 
. What is the sum of the digits of ?
Solution
If we start off with the first digit, we know that it can't be 
since is not a factor of . We go down to the digit 
, which does work since it is a factor of . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide 
. The next place can be 
, as it is the largest factor, aside from 
. Consequently, our next three values will be 
and 
if we use the same logic. Therefore, our five-digit number is 
, so the sum is 
.
Solution(factorial)
120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
Making the greatest integer,
8 is the largest value and will go in the front so we can express it as 8,5,3,1,1
We don't even need the number just add
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solutions
https://youtu.be/7an5wU9Q5hk?t=13
~savannahsolver
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
