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

Revision as of 09:59, 10 April 2024

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 1

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 2 (Factorial)

120 is 5!, so we have $(5)(4)(3)(2)(1) = 120$ . Now look for the largest digit you can create by combining these factors.

$8=4 \cdot 2$

Use this largest digit for the ten-thousands place: $8$ _ , _ _ _

Next you use the $5$ and the $3$ for the next places: $85,3$ _ _ (You can't use $3 \cdot 2=6$ because the $2$ was used to make $8$ .)

Fill the remaining places with 1: $85,311$

$= (5)(8)(3)(1)(1) =120$

$8+5+3+1+1=\boxed{\textbf{(D) }18}$ .

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/zxEO6amczPU

~Education, the Study of Everything

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.

@@ Line 9: / Line 9: @@
 == Solution 2 (Factorial) ==
-is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
+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>(5)(4)(3)(2)(1) = 120</math>
+<math>8=4 \cdot 2</math>
-Making the greatest integer,
-<math>(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)</math>
+Use this largest digit for the ten-thousands place: <math>8</math>_ , _ _ _
-<math> = (5)(8)(3)(1)(1) =120</math>
+Next you use the <math>5</math> and the <math>3</math> for the next places: <math>85,3</math> _ _ (You can't use <math>3 \cdot 2=6</math> because the <math>2</math> was used to make <math>8</math>.)
-is the largest value and will go in the front.
+Fill the remaining places with 1: <math>85,311</math>
-We can express the number as <math>85311</math>.
+<math> = (5)(8)(3)(1)(1) =120</math>
 <math>8+5+3+1+1=\boxed{\textbf{(D) }18}</math>.

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