# High School Mathematics Extensions/Primes/Problem Set/Solutions

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### Question 1[edit]

Is there a rule to determine whether a 3-digit number is divisible by 11? If yes, derive that rule.

**Solution**

Let *x* be a 3-digit number We have

now

We can conclude a 3-digit number is divisible by 11 if and only if the sum of first and last digit minus the second is divisible by 11.

### Question 2[edit]

Show that *p*, *p* + 2 and *p* + 4 cannot all be primes. (*p* a positive integer and is great than 3)

**Solution**

We look at the arithmetic mod 3, then *p* slotted into one of three categories

- 1st category
- we deduce
*p*is not prime, as it's a multiple of 3

- we deduce
- 2nd category
- so p + 2 is not prime

- 3rd category
- therefore
*p*+ 4 is not prime

- therefore

Therefore *p*, *p* + 2 and *p* + 4 cannot all be primes.

### Question 3[edit]

Find *x*

**Solution**

Notice that

- .

Then

- .

Likewise,

and

- .

Then

### Question 4[edit]

9. Show that there are no integers *x* and *y* such that

**Solution**

Look at the equation mod 5, we have

but

therefore there does not exist *a* x such that

### Question 5[edit]

Let *p* be a prime number. Show that

**(a)**

where

E.g. 3! = 1×2×3 = 6

**(b)** Hence, show that

for *p* ≡ 1 (mod 4)

**Solution**

**a)** If *p* = 2, then it's obvious. So we suppose *p* is an odd prime. Since *p* is prime, some deep thought will reveal that every distinct element multiplied by some other element will give 1. Since

we can pair up the inverses (two numbers that multiply to give one), and (p - 1) has itself as an inverse, therefore it's the only element not "eliminated"

as required.

**b)** From part a)

since *p* = 4*k* + 1 for some positive integer *k*, (p - 1)! has 4*k* terms

there are an even number of minuses on the right hand side, so

it follows

and finally we note that p = 4k + 1, we can conclude