November 13, 2017
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Today is 11/13/17, which makes it a “prime time” to solve the following problems involving prime numbers!

For each of the prime numbers less than 30, Wallace writes down all the number’s positive factors. What is the sum of all the numbers Wallace writes down?

There are ten prime numbers less than 30. They are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Each of these numbers has two positive factors, 1 and the number itself. The total of all these factors, then, is 10(1) + 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 10 + 129 = 139.

 

If the sum of three prime numbers is 31 and their product is 1001, what is the largest of the three prime numbers?

There are two sets of three prime numbers that have a sum of 31: 3 + 11 + 17 and 7 + 11 + 13. Since 3 × 11 × 17 = 561 and 7 × 11 × 13 = 1001, the three primes we are looking for are 7, 11 and 13. The largest of these is 13.

 

What is the greatest prime factor common to both 12! + 10! and 12! – 10!?              

We can rewrite 12! + 10! as 12 × 11 × 10! + 10! = 10!(12 × 11 + 1) = 10!(133), and 12! − 10! can be rewritten as 12 × 11 × 10! − 10! = 10!(12 × 11 − 1) = 10!(131). The greatest common factor of 12! + 10! and 12! – 10! is 10!. Since 10! = 10 × 9 × 8 × 7 × ⋯ × 2 × 1, we know that its greatest prime factor is 7.

 

If n = 15! + 35!, how many of the nine prime numbers less than 25 are divisors of n?

We can rewrite 15! + 35! as 15! + 15! × 16 × 17 × ⋯ × 34 × 35 = 15!(1 + 16 × 17 × ⋯ × 34 × 35). Looking at the factors of 15! = 15 × 14 × 13 × ⋯ × 2 × 1, we see that its prime factors, all of which are less than 25, are 2, 3, 5, 7, 11 and 13. There are 6 of them.

 

Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2
contains ONLY PROBLEMS.