Answer exercise 1.3 #3 ONLY #3 Definition 1.1.2 The natural number n is called p

Question

Answer exercise 1.3 #3 ONLY #3

Definition 1.1.2 The natural number n is called prime if it has only two divisors. It is called composite if it is bigger than 1 and is not prime The first several prime numbers are: 2,3,5,7,11, 13, 17, 19,.... Only one of the prime numbers is even, of course. Some people say that the even prime number is the oddest prime number! There is a reason for that saying, but we will not be able to appreciate it in our (quite elementary) course on Number Theory Exercise 1.3 1) Is the number 999991 prime? 2) Is it true that 3 never divides n 1, and that 5 never divides n2 2? Give arguments for your answer RAA: suppose 3 n2 1; then n2 1 3m for some natural number m. From n2 3 1 we get that 3tn2, and so, 31 n. Therefore n 3k +1 or n 3k+2. We have that n2 3A+1 where A-3k2 + 2k in the former case, and n2-3B + 1 where B = 3k2 + 4k + 1 in the latter case. Since neither 3A +1-3m 1 nor 3B+1-3m -1 is possible, we get a contradiction. The second claim of the Exercise is treated similarly. 3) Let N be a two digit natural number, and let M be the number obtained by reordering the digits of N. Show that 9 | |M - N|, and find all N such that |N M 18 4) Show that the product of three consecutive integers is divisible by 6, and that the product of four consecutive integers is divisible by 24 5) Prove that, for n E N, we have that 6 n(n 1) (2n 1) [ Notice that n(n-1)(2n _ 1) = n(n-1)(2n-4 +3) = n(n-1 )(2n-1) + 3n(n-1)- 2n(n -1) (n 2) 3n(n 1), and that the summands in the last ezpressions are divisible by 6. 6) Prove that if neither 2 n nor 3|n, then 24 (n 23) Prove first that 3 and 8 divide n2 -1 for n non-divisible by 2 and 3.] 7) Find all natural numbers n such that n1 n2 1 8) How many natural numbers, less than 100, are there such that neither 2, nor 3, nor 5 divides them? 9) Find ten consecutive composite natural numbers. For every n > 1, find n consecutive composite natural numbers

Explanation / Answer

Let the digits of N be a(unit's digit) and b(ten's digit)

Then, N = 10b+a

Now, M = 10a+b

So, |M-N| = 9|a-b| , hence, 9 divides |M-N|

Now, |N-M| = 9|b-a| = 18 implies, |b-a| = 2

That means a and b differ by 2

again, a and b are digits of a number, so each of them range from 0 to 9.

So, we have the pair (a,b) as (0,2) , (1,3) , (2,4) , (3,5) , (4,6) , (5,7) , (6,8) , (7,9) and also, (2,0) , (3,1) , (4,2) , (5,3) , (6,4) , (7,5) , (8,6) , (9,7)

So, the number N can have the following choices :

02, 13, 24, 35, 46, 57, 68, 79, 20, 31, 42, 53, 64, 75, 86, 97

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