Difference between revisions of "2018 AMC 8 Problems/Problem 14"
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| − | ==Problem | + | ==Problem== |
Let <math>N</math> be the greatest five-digit number whose digits have a product of <math>120</math>. What is the sum of the digits of <math>N</math>? | Let <math>N</math> be the greatest five-digit number whose digits have a product of <math>120</math>. What is the sum of the digits of <math>N</math>? | ||
<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 | + | 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>. |
| + | |||
| + | == Solution 2 (Factorial) == | ||
| + | |||
| + | 120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers. | ||
| + | |||
| + | <math>(5)(4)(3)(2)(1) = 120</math> | ||
| + | Making the greatest integer, | ||
| + | |||
| + | <math>(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)</math> | ||
| + | <math> = (5)(8)(3)(1)(1) =120</math> | ||
| + | |||
| + | 8 is the largest value and will go in the front. | ||
| + | |||
| + | We can express the number as <math>85311</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/7an5wU9Q5hk?t=13 | ||
| + | |||
| + | https://youtu.be/Q5YrDW62VDU | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Revision as of 21:10, 24 November 2023
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)
120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
Making the greatest integer,
8 is the largest value and will go in the front.
We can express the number as .
.
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. 
