Difference between revisions of "2018 AMC 8 Problems/Problem 14"

Revision as of 10:51, 11 March 2023

Problem

Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$ ?

$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

Solution

If we start off with the first digit, we know that it can't be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$ . The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18\implies \boxed{\textbf{(D) }18}$ .

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/IAKhC_A0kok

https://youtu.be/7an5wU9Q5hk?t=13

https://youtu.be/Q5YrDW62VDU

~savannahsolver

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.

Page

Toolbox

Search