Difference between revisions of "2002 AMC 10A Problems/Problem 15"

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(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is <math>23</math>, <math>41</math>, <math>59</math>, and <math>67</math>.)
 
(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is <math>23</math>, <math>41</math>, <math>59</math>, and <math>67</math>.)
  
(you could also guess the numbers, if you have a lot of spare time)
+
(You could also guess the numbers, if you have a lot of spare time.)
  
 
==See Also==
 
==See Also==

Revision as of 23:18, 15 April 2017

Problem

Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?

$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$

Solution

Only odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite, hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{(E)= 190}$.

(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is $23$, $41$, $59$, and $67$.)

(You could also guess the numbers, if you have a lot of spare time.)

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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

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