Difference between revisions of "2003 AMC 10A Problems/Problem 14"
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<math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24 </math> | <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24 </math> | ||
− | == Solution == | + | == Solution 1 == |
Since <math>d</math> is a single digit prime number, the set of possible values of <math>d</math> is <math>\{2,3,5,7\}</math>. | Since <math>d</math> is a single digit prime number, the set of possible values of <math>d</math> is <math>\{2,3,5,7\}</math>. | ||
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The largest possible value of <math>n</math> is <math>1533</math>. | The largest possible value of <math>n</math> is <math>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> | ||
+ | |||
+ | == Solution 2 == | ||
+ | Since you want <math>n</math> to be the largest number possible, you will want <math>d</math> in <math>10d+e</math> to be as large as possible. So <math>d = 7</math>.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>.<math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 7s = 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> | So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> | ||
Revision as of 09:25, 14 January 2020
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 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
Solution 2
Since you want to be the largest number possible, you will want in to be as large as possible. So .Then, cannot be because which is not prime. So .. So, the sum of the digits of is
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.