High School Mathematics Extensions/Primes/Problem Set/Solutions

HSME
Content
100% developed Primes
100% developed Modular Arithmetic
Problems & Projects
100% developed Problem Set
100% developed Project
Solutions
100% developed Exercise Solutions
50% developed Problem Set Solutions
Misc.
100% developed Definition Sheet
100% developed Full Version
25% developed PDF Version

At the moment, the main focus is on authoring the main content of each chapter. Therefore this exercise solutions section may be out of date and appear disorganised.

If you have a question please leave a comment in the "discussion section" or contact the author or any of the major contributors.


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
2nd category
 
 
so p + 2 is not prime
3rd category
 
 
therefore p + 4 is not prime

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