Difference between revisions of "2018 AMC 8 Problems/Problem 14"
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120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multple numbers. | 120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multple numbers. | ||
| − | (5)(4)(3)(2)(1)=120 | + | (5)(4)(3)(2)(1) = 120 |
make the greatest integer | make the greatest integer | ||
(5)(4 times 2)(3)(2 divided by 2)(1) | (5)(4 times 2)(3)(2 divided by 2)(1) | ||
| + | = | ||
(5)(8)(3)(1)(1) | (5)(8)(3)(1)(1) | ||
Revision as of 10:51, 11 March 2023
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 multple numbers.
(5)(4)(3)(2)(1) = 120 make the greatest integer
(5)(4 times 2)(3)(2 divided by 2)(1)
=
(5)(8)(3)(1)(1)
8 is the largest value and will go in the front so we can express it as 5,8,3,1,1
We don't even need the number just add
5+8+3+1+1 = 18
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. 
