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Difference between revisions of "2008 iTest Problems/Problem 4"

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We know that any prime number, excluding <math>2</math>, is congruent to <math>1 \pmod 2</math>. Thus, if both of the primes are not <math>2</math>, their difference would be congruent to <math>0 \pmod 2</math>. Because <math>11 \equiv 1 \pmod 2</math>, one of the primes must be <math>2</math>. It follows that the other prime must then be <math>13</math>. Therefore, the sum of the two is <math>13+2=\boxed{15}</math>.
 
We know that any prime number, excluding <math>2</math>, is congruent to <math>1 \pmod 2</math>. Thus, if both of the primes are not <math>2</math>, their difference would be congruent to <math>0 \pmod 2</math>. Because <math>11 \equiv 1 \pmod 2</math>, one of the primes must be <math>2</math>. It follows that the other prime must then be <math>13</math>. Therefore, the sum of the two is <math>13+2=\boxed{15}</math>.
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== Solution 2 ==
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Since the difference is <math>11</math>, one number must be even for the difference to be even. <math>2</math> is the only even prime number, and therefore is one of the two numbers. The other number is <math>2 + 11 = 13</math>, and their sum: <math>13 + 2 = \boxed{15}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 16:43, 24 November 2018

Problem

The difference between two prime numbers is $11$. Find their sum.

Solution

We know that any prime number, excluding $2$, is congruent to $1 \pmod 2$. Thus, if both of the primes are not $2$, their difference would be congruent to $0 \pmod 2$. Because $11 \equiv 1 \pmod 2$, one of the primes must be $2$. It follows that the other prime must then be $13$. Therefore, the sum of the two is $13+2=\boxed{15}$.

Solution 2

Since the difference is $11$, one number must be even for the difference to be even. $2$ is the only even prime number, and therefore is one of the two numbers. The other number is $2 + 11 = 13$, and their sum: $13 + 2 = \boxed{15}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 3 		Followed by:
Problem 5
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