2018 AMC 8 Problems/Problem 14
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!!!)
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Video Solutions
https://youtu.be/7an5wU9Q5hk?t=13
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See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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