Difference between revisions of "2018 AMC 8 Problems/Problem 14"
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− | 120 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>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> | <math>8=4 \cdot 2</math> | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
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+ | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 17:52, 6 June 2025
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.