Difference between revisions of "2018 AMC 8 Problems/Problem 14"
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<math>\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math> | <math>\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20</math> | ||
| − | == Solution == | + | == Solution 1== |
| − | If we start off with the first digit, we know that it can't be <math>9</math> since <math>9</math> is not a factor of <math>120</math>. We go down to the digit <math>8</math>, which does work since it is a factor of <math>120</math>. Now, we have to know what digits will take up the remaining four spots. To find this result, just divide <math>\frac{120}{8}=15</math>. The next place can be <math>5</math>, as it is the largest factor, aside from <math>15</math>. Consequently, our next three values will be <math>3,1</math> and <math>1</math> if we use the same logic. Therefore, our five-digit number is <math>85311</math>, so the sum is <math>8+5+3+1+1=18\implies \boxed{\textbf{(D)}}</math>. | + | If we start off with the first digit, we know that it can't be <math>9</math> since <math>9</math> is not a factor of <math>120</math>. We go down to the digit <math>8</math>, which does work since it is a factor of <math>120</math>. Now, we have to know what digits will take up the remaining four spots. To find this result, just divide <math>\frac{120}{8}=15</math>. The next place can be <math>5</math>, as it is the largest factor, aside from <math>15</math>. Consequently, our next three values will be <math>3,1</math> and <math>1</math> if we use the same logic. Therefore, our five-digit number is <math>85311</math>, so the sum is <math>8+5+3+1+1=18\implies \boxed{\textbf{(D) }18}</math>. |
| − | ==Video Solution== | + | == Solution 2 (Factorial) == |
| + | |||
| + | <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> | ||
| + | |||
| + | Use this largest digit for the ten-thousands place: <math>8</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>.) | ||
| + | |||
| + | Fill the remaining places with 1: <math>85,311</math> | ||
| + | |||
| + | <math> = (5)(8)(3)(1)(1) =120</math> | ||
| + | |||
| + | <math>8+5+3+1+1=\boxed{\textbf{(D) }18}</math>. | ||
| + | |||
| + | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
| + | https://youtu.be/zxEO6amczPU | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solutions== | ||
https://youtu.be/IAKhC_A0kok | https://youtu.be/IAKhC_A0kok | ||
https://youtu.be/7an5wU9Q5hk?t=13 | https://youtu.be/7an5wU9Q5hk?t=13 | ||
| + | |||
| + | https://youtu.be/Q5YrDW62VDU | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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. 
