Difference between revisions of "2018 AMC 8 Problems/Problem 14"
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5+8+3+1+1 = 18 | 5+8+3+1+1 = 18 | ||
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+ | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
+ | https://youtu.be/zxEO6amczPU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solutions== | ==Video Solutions== |
Revision as of 14:18, 30 March 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 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) =120
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 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.