Difference between revisions of "2018 AMC 8 Problems/Problem 14"
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== Solution 2 (Factorial) == | == Solution 2 (Factorial) == | ||
| − | 120 is 5!, so we have <math>(5)(4)(3)(2)(1) = 120</math>. Now look for the largest digit you can create by combining these factors. | + | <math>120</math> is <math>5!</math>, so we have <math>(5)(4)(3)(2)(1) = 120</math>. (Alternatively, you could identify the prime factors <math>(5)(3)(2)(2)(2) = 120</math>.) Now look for the largest digit you can create by combining these factors. |
<math>8=4 \cdot 2</math> | <math>8=4 \cdot 2</math> | ||
Latest revision as of 09:01, 10 April 2024
Contents
Problem
Let 
be the greatest five-digit number whose digits have a product of 
. What is the sum of the digits of ?
Solution 1
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 2 (Factorial)
is 
, so we have 
. (Alternatively, you could identify the prime factors 
.) Now look for the largest digit you can create by combining these factors.
Use this largest digit for the ten-thousands place: _ , _ _ _
Next you use the and the 
for the next places: 
_ _ (You can't use 
because the 
was used to make .)
Fill the remaining places with 1: 
.
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. 
