Dr. Cordero
- Define a set of “triplet primes” to be three consecutive odd integers (p, p+2, p+4) which are all prime. A quick scan of our list of primes less than 100 shows that there is exactly one set of triplet primes in that range: (3, 5, 7). Are there any other sets of triplet primes (ever)? If so, give an example. If not, prove that there can’t be.
- (a). What is the smallest positive integer which is congruent to 1 modulo 2, 1 modulo 3, 3 modulo 4, and 3 modulo 5?
(b). Make a conjecture of the form (fill this in) is congruent to 1 modulo 2, 1 modulo 3, 3 modulo 4, and 3 modulo 5. Furthermore, every positive integer which is congruent to 1 modulo 2, 1 modulo 3, 3 modulo 4, and 3 modulo 5 has this form. Write a sentence or two explaining why you believe your conjecture.
- Here is a test for divisibility by 7:
Take the last digit of your number, double it, and subtract what you get from the number formed by the remaining digits of your number. If what you get is divisible by 7, then so is your original number. Otherwise, your original number is not divisible by 7. For example, to see if the number 231 is divisible by 7, you look at 23-2(1) =21. Since 21 is divisible by 7, 231 must be also. (Indeed, 231= (7) (33).) On the other hand to see if 426 is divisible by 7, you look at 42-2(6) =30 and since 30 is not divisible by 7, neither is 426. (Indeed, 426= (7) (60) +6.) Prove that this test always work. Hint: You will want to start by expressing your original number in terms of its digits.
- Some computations. Be sure to show your work.
- Find the last two digits of
. - Is
divisible by 77?
Hint: Remember that a number is divisible by m if and only if it is congruent to 0 modulo m. Also notice that a number is divisible by 77 if and only if it is divisible by 7 and by 11. You’ll want to use congruences and deal with 7 and 11 separately.
- Find the last two digits of
- A perfect number is equal to the sum of its divisors (other than itself). For example, the divisors of 6 (other that 6) are 1, 2, and 3, and since 1+2+3=6, we say that 6 is a perfect number. On the other hand, the divisors of 15 (other than 15) are 1, 3, and 5 but 1+3+5=9, which means that 15 is not a perfect number.
If we look at the product instead of the sum, we could say that a number is product perfect if the product of all its divisors (other than itself) is equal to the original number. For example 6 and 15 are product perfect because 6=1.2.3 and 15=1.3.5.
- List all the product perfect numbers between 2 and 50.
Based on the data you generated in (a), find a characterization of all product perfect numbers. Your characterization should be precise enough to easily answer such questions as “Is 47, 920 product perfect?” and “Find a product perfect number larger that 100, 000”. (Demonstrate that your characterization actually works by providing answers to these questions.)