Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
 
Line 6: Line 6:
  
 
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>.
 +
 +
== Solution 2 ==
 +
 +
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 17:43, 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
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_4&oldid=99036"