Difference between revisions of "2003 AMC 10A Problems/Problem 14"
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== Solution 1 == | == Solution 1 == | ||
− | Since | + | Since we want <math>n</math> to be as large as possible, we would like <math>d</math> in <math>10d+e</math> to be as large as possible. So, <math>d=7,</math> the greatest single-digit prime. Then, <math>e</math> cannot be <math>5</math> because <math>10(7)+5 = 75,</math> which is not prime. So <math>e = 3</math>. Then, <math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>. |
− | So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_ | + | So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_ (Edited by Sophia866) |
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== Solution 2 == | == Solution 2 == |
Revision as of 14:47, 19 August 2023
Contents
[hide]Problem
Let be the largest integer that is the product of exactly 3 distinct prime numbers , , and , where and are single digits. What is the sum of the digits of ?
Solution 1
Since we want to be as large as possible, we would like in to be as large as possible. So, the greatest single-digit prime. Then, cannot be because which is not prime. So . Then, . So, the sum of the digits of is ~ MathGenius_ (Edited by Sophia866)
Solution 2
Since is a single digit prime number, the set of possible values of is .
Since is a single digit prime number and is the units digit of the prime number , the set of possible values of is .
Using these values for and , the set of possible values of is
Out of this set, the prime values are
Therefore the possible values of are:
The largest possible value of is .
So, the sum of the digits of is
Video Solution by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=yApq-Vny_A0 ~David
See Also
2003 AMC 10A (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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.