Difference between revisions of "2003 AMC 10A Problems/Problem 14"
Mathgenius (talk | contribs) m (→Solution 1) |
Mathgenius (talk | contribs) m (→Solution 1) |
||
| Line 8: | Line 8: | ||
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>~~~</math> <math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>. | 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>~~~</math> <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_ | ||
| + | |||
| + | |||
| + | == Solution 2 == | ||
| + | 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>e</math> is a single digit prime number and is the units digit of the prime number <math>10d+e</math>, the set of possible values of <math>e</math> is <math>\{3,7\}</math>. | ||
| + | |||
| + | Using these values for <math>d</math> and <math>e</math>, the set of possible values of <math>10d+e</math> is <math>\{23,27,33,37,53,57,73,77\}</math> | ||
| + | |||
| + | Out of this set, the prime values are <math>\{23,37,53,73\}</math> | ||
| + | |||
| + | Therefore the possible values of <math>n</math> are: | ||
| + | |||
| + | <math>2\cdot3\cdot23=138</math> | ||
| + | |||
| + | <math>3\cdot7\cdot37=777</math> | ||
| + | |||
| + | <math>5\cdot3\cdot53=795</math> | ||
| + | |||
| + | <math>7\cdot3\cdot73=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> | ||
== See Also == | == See Also == | ||
Revision as of 11:44, 18 March 2020
Contents
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 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 
~ MathGenius_
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
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. 
