Difference between revisions of "2005 iTest Problems/Problem 17"
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On the <math>2004</math> iTest, we defined an optimus prime to be any prime number whose digits sum to a prime number. (For example, <math>83</math> is an optimus prime, because it is a prime number and its digits sum to <math>11</math>, which is also a prime number.) Given that you select a prime number under <math>100</math>, find the probability that is it not an optimus prime. | On the <math>2004</math> iTest, we defined an optimus prime to be any prime number whose digits sum to a prime number. (For example, <math>83</math> is an optimus prime, because it is a prime number and its digits sum to <math>11</math>, which is also a prime number.) Given that you select a prime number under <math>100</math>, find the probability that is it not an optimus prime. | ||
==Solution 1== | ==Solution 1== | ||
| − | Consider the set of primes less than <math>100</math>: <math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97</math>. There are <math>25</math> primes in this set. Since the last digit of a multi-digit prime number is odd, and the sum of two odd numbers is even, we can immediately disregard all primes with an odd tens digit. In addition, all the one-digit primes have a prime digit sum. We can consider the set of primes with a nonzero even tens digit and find their sums: <math>5, 11, 5, 7, 11, 7, 13, 11, 17</math>. All of these sums are prime, so this case contributes <math>9</math> primes. We can now find the desired probability: <math>1-\frac{9+4\text{for the one digit primes}}{25}=\boxed{\frac{12}{25}}</math>. | + | Consider the set of primes less than <math>100</math>: <math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97</math>. There are <math>25</math> primes in this set. Since the last digit of a multi-digit prime number is odd, and the sum of two odd numbers is even, we can immediately disregard all primes with an odd tens digit. In addition, all the one-digit primes have a prime digit sum. We can consider the set of primes with a nonzero even tens digit and find their sums: <math>5, 11, 5, 7, 11, 7, 13, 11, 17</math>. All of these sums are prime, so this case contributes <math>9</math> primes. We can now find the desired probability: <math>1-\frac{9+4\text{(for the one digit primes)}}{25}=\boxed{\frac{12}{25}}</math>. |
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==See Also== | ==See Also== | ||
{{iTest box|year=2005|num-b=16|num-a=18}} | {{iTest box|year=2005|num-b=16|num-a=18}} | ||
[[Category: Introductory Number Theory Problems]] | [[Category: Introductory Number Theory Problems]] | ||
Revision as of 13:43, 19 October 2025
Problem
On the 
iTest, we defined an optimus prime to be any prime number whose digits sum to a prime number. (For example, 
is an optimus prime, because it is a prime number and its digits sum to 
, which is also a prime number.) Given that you select a prime number under 
, find the probability that is it not an optimus prime.
Solution 1
Consider the set of primes less than : 
. There are 
primes in this set. Since the last digit of a multi-digit prime number is odd, and the sum of two odd numbers is even, we can immediately disregard all primes with an odd tens digit. In addition, all the one-digit primes have a prime digit sum. We can consider the set of primes with a nonzero even tens digit and find their sums: 
. All of these sums are prime, so this case contributes 
primes. We can now find the desired probability: 
.
See Also
| 2005 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 16 |
Followed by: Problem 18 | |
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