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>. | ||
| − | == See | + | == 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== | ||
| + | {{2008 iTest box|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 16:43, 24 November 2018
Contents
Problem
The difference between two prime numbers is 
. Find their sum.
Solution
We know that any prime number, excluding 
, is congruent to 
. Thus, if both of the primes are not , their difference would be congruent to 
. Because 
, one of the primes must be . It follows that the other prime must then be 
. Therefore, the sum of the two is 
.
Solution 2
Since the difference is , one number must be even for the difference to be even. is the only even prime number, and therefore is one of the two numbers. The other number is 
, and their sum: 
.
See Also
| 2008 iTest (Problems) | ||
| Preceded by: Problem 3 |
Followed by: Problem 5 | |
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