Number theorem

Question 1

Don’t divide by a potential negative number without flipping the inequality

Answer 1

x-1 > 6/x

square both sides or draw diagram, cant times by x as could be zero or negative

Question 2

f(-x) = -f(x) therefore

Answer 2

odd function

Question 3

3^4^x has last digit

Answer 3

3^4 = 81

therefore last digit 1

Question 4

a^b^c has last digit =

Answer 4

last digit of a^b

Question 5

a = 1 mod(n) therefore

Answer 5

a^k = 1^k mod(n) = 1 mod(n)

Question 6

Zeroes in n! =

Answer 6

n/5 + n/25…n/5^x

for 5^x <= n

Question 7

Divisible by 3

Answer 7

sum of digits is a multiple of 3

Question 8

divisible by 4

Answer 8

last two digits are multiples of 2

Question 9

divisible by 6

Answer 9

divisible by 3 and 2

Question 10

divisible by 7

Answer 10

double the last digit and subtract from rest of the digits. If 0 or multiple of 7 then divisible otherwise repeat to get smaller number

Question 11

divisible by 8

Answer 11

last three digits divisible by 8

Question 12

divisible by 11

Answer 12

alternating sum of digits is divisible by 11 or equals zero

Question 13

divisible by 12

Answer 13

number divisible by 3 and by 4

Question 14

A number is the square if

Answer 14

the powers in its prime factorisation are even

Question 15

A number is cube if

Answer 15

its powers are divisible by 3

Question 16

Number of factors a number has

Answer 16

by adding one to each power

Question 17

Square rooting gives

Answer 17

two possible solutions

Question 18

(2^7) * 5^4 =

Answer 18

(2*5)^4 * 8

Question 19

Approximate the value of an expression when variables become large.

Answer 19

The key here is that constant values become inconsequential when combined with a growing variable

Question 20

Number of factors =

Answer 20

prime factorization + 1 timesed together

Question 21

geometric mean <

Answer 21

arithmetic mean

Question 22

A number is a square if and only if

Answer 22

the powers in its factorization are even

Question 23

a number is a cube if and only if

Answer 23

its powers are divisible by 3

Question 24

To find the number of factors a number has

Answer 24

do prime factorisation, add one to each power, times all new powers together

Question 25

Arithmetic mean ___ geometric mean

Answer 25

>

Question 26

x^2…

Answer 26

> 0

Question 27

6^(x+y) x 12^(x-y) =

Answer 27

(32)^(x+y) x (22*3)^(x-y)

Question 28

1/a^x is an integer for

Answer 28

x<=0

Question 29

if 3(root3) > 4 then to prove

Answer 29

square both sides

27 > 16

Question 30

Two numbers, a and b, are coprime or relatively prime if

Answer 30

a and b share no common factors
LCM(a,b) = ab,
(a, b) = 1

Question 31

K and k+1 are coprime because

Answer 31

suppose k had some factor q, k+1 must have a remainder of 1 when divided by q so is not divisible by q

Question 32

k-1 and k+1 are coprime if k is

Answer 32

even

Question 33

Perfect square =

Answer 33

square number

Question 34

Number of zeros

Answer 34

via prime factorisation (2’s and 5’s and 10’s)

Question 35

Number of zeros on the end of n!

Answer 35

n/5^x

Question 36

Every two consecutive numbers are even,

Answer 36

every three are divisible by 3

Question 37

a/x + b/y = c –>

Answer 37

(cx-a)(cy-b)=ab

Question 38

Integer solutions of divisions tip

Answer 38

put variables in denominator only

Question 39

remeber negative

Answer 39

factors

Question 40

If you have two consecutive factors i.e. (n)(n+1) then we can say

Answer 40

they are coprime and thus share no factors

Question 41

moduluo arithmetic allows:

Answer 41

addition, multiplication, powers

Question 42

If x divided by y gives remainder z, then

Answer 42

x-z is divisible by y

Question 43

Show that if 2+2(1+12n^2)^0.5 is an integer then it is a perfect square

Answer 43

First note that the question says IF [..] is an integer, THEN it is a square. We need to start with the assumption, and reason towards the conclusion – don’t be tempted to prove the opposite. If it is an integer, then we can say that the bit in the root is a perfect square, since the square root has to be an integer, and that it is odd. We can therefore rewrite the root as (2k+1)^2. We can rearrange this to 3n^2 = k(k+1). The RHS are coprime, one is square the other is three times a square. If k+1 was a multiple of 3, then by modular arithmetic, k couldn’t be square. Therefore k+1 is a square, and k is three times a square

Question 44

Let n=ab, m = abc

Represent n and m in terms of single digits

Answer 44

n = 10a + b
m = 100a + 10b + c

Question 45

if p is prime and a is any integer such that a is not a multiple of p then fermats little theorem states:

Answer 45

a^p-1 = 1(modp)

Question 46

A square chessboard of sides 2^n is tiled with L-shapes, each of three square, such that tiles do not overlap. Show that you will always have one square on the chessboard left untiled

Answer 46

We are finding the remainder when we divide (2^n)^2 = 4^n by 3
4=1 (mod3)
So 4^n = 1^n = 1 (mod3)

Question 47

Show that x^2 - y^2 = 2002 has no integer solutions

Answer 47

x = 0, 1, 2, 3 (mod4)
x^2 = 0, 1 (mod4) as is y^2

So all possibilities of x^2 - y^2 = 0, 1, 3 (mod4)
However 2002 = 2 (mod4)
Thus there can be no integer solutions

Question 48

List all the prime numbers that are one less than a square number

Answer 48

Let p be a prime number, p+1 = n^2  p=n^2 – 1 = (n+1)(n-1) for p to be prime one bracket must be a one, the other must be p, as p>1 we can say n-1 = 1, n=2m therefore one solution when p=3

Question 49

List all primes that are one more than a cubed number

Answer 49

Let p be a prime number, p-1=n^3  p=n^3+1 = (n+1)(n^2-n+1) therefore n+1 = p or 1, but as n cant equal zero then n+1 =p, n^2-n+1 = 1 therefore n(n-1) = 0, therefore n=1 therefore p=2

Question 50

Can K^3 + 2k^2 +2k +1 ever be a cubed number for any positive integer k

Answer 50

K^3 + 2k^2 +2k +1 = n^3  we can say that n^3 > K^3  n^3 = (k+1)^3 + k^2 + k therefore n^3