Integer Properties

Question 1

Divisibility rule for 4

Answer 1

if the last two digits form a two-digit nb divisible by 4, then the entire number is divisible by 4

Question 2

Divisibility rule for 9

Answer 2

if the sum of the digits is divisible by 9, then the number is divisible by 9

Question 3

Divisibility rule for 6

Answer 3

in order to be divisible by 6, a number must be divisible by 2 AND divisible by 3

Question 4

is 1296 divisible by 6?

Answer 4

yes

Question 5

true or false

divisor = factor

Answer 5

true
7 is a factor of 91
7 is a divisor of 91
91 is divisible by 7
91 is a multiple of 7
If P is a multiple of r, r is a factor of P and a divisor of P
If P and Q are multiples of r, then P+Q and P-Q must also be multiples of r
If P is a multiple of r, then any multiple of P is also a multiple of R
If P and Q are multiples of r, then P*Q must also be a multiple of r

Question 6

If K, (K+200), (K+350) and 15*K are all multiples of P, then P could equal which of the following? 
A - 20
B - 25
C - 75
D - 100
E - 150

Answer 6

15*K is a useless piece of information: if K is a multiple of P, then 15K has to be a multiple of P.

differences amount K, K+200 and K+350 will also be multiples of P
K + 200 - K = 200
K + 350 - K = 350
350 - 200 = 150
200 - 150 = 50
P is a factor of 50, 150, 200 and 350

The answer can’t be anything bigger than 50 because anything bigger than 50 can’t be a factor of 50
can’t be 20 (others not divisible by it)

All the numbers are divisible by 25
Answer B

Question 7

List the prime numbers less than 20

Answer 7

2
3
5
7
11
13
17
19

Question 8

List the prime numbers between 20 and 60

Answer 8

23
29
31
37
41
43
47
53
59

Question 9

how to test whether a large number (less than 100) is prime?

Answer 9

check whether it is divisible by one of the prime nb less than 10 (2 - 3 - 5 - 7 )

Question 10

If 4680 = 2^3 * 3^2 * 5 * 13
Which of the following nb are factors of 4680?
25
45
65
85
120
180

Answer 10

25 = 5*5 NO
45 = 3*3*5 YES
65 = 5*13 YES
85 = 5*17 NO
120 = 2^3 * 3 * 5 YES
180 = 2^2 * 3^2 * 5 YES

Question 11

How to find the nb of factors of an integer?

Answer 11

prime factorization
list of exponents of the prime factors
add one to every nb on the list
multiply all those nb together

Question 12

How to find the nb of odd factors of an integer?

Answer 12

prime factorization
list of exponents of odd prime factors
add one to each
product of nb on new list = nb of odd factors

Question 13

How to find the nb of even factors of an integer?

Answer 13

no direct way

calculate the total nb of factors and the nb of odd factors, then subtract

Question 14

How many factors does 8400 have?

Answer 14

Prime factorization
8400 = 84 x 100 = 7 x 12 x 10 x 10
8400 = 7 x (2 x 2 x 3) x ( 2 x 5 ) x (2 x 5 )
8400 = 2^4 x 3 x 5^2 x 7

make a list of the exponents of the prime factors
{4, 1, 2, 1}

add one to every number on the list
{5, 2, 3, 2}

multiply all those numbers together
5 x 2 x 3 x 2 = 60

8400 has 60 factors

Question 15

How many even factors does 21600 have?

Answer 15

21 600 = 2^5 x 3^3 x 5^2

{5+1, 3+1, 2+1}
6x4x3 = 72 factors

{3+1, 2+1}
4x3 = 12 odd factors

72-12 = 60 even factors

Question 16

List the first fifteen perfect squares

Answer 16

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121
12² = 144
13² = 169
14² = 196
15² = 225

Question 17

The exponents of the prime factors of a square …

Answer 17

… all must be even

Question 18

If K = 2^6 x 3^4 x 5^2

Is K a perfect square?

Answer 18

Yes
all the exponents of the prime factors are even
K = 360²

Question 19

A perfect square always has …

Answer 19

… an odd nb of factors

Question 20

What is the GCF of 720 and 1200?

Answer 20

GCF = Greatest common factor / greatest common divisor

720 = 2^4 x 3^2 x 5
1200 = 2^4 x 3 x 5^2
Highest power of 2 in common : 4
Highest power of 3 in common : 1
Highest power of 5 in common : 1
GCF = 2^4 x 3 x 5 = 240

Question 21

What is the LCM of 24 and 32?

Answer 21

LCM = Least common multiple = LCD = least common denominator

24 = 2^3 x 3
32 = 2^5
GCF = 8

write each nb in the form : GCF x another factor
24 = 8 x 3
32 = 8 x 4

LCM = product of these three factors
LCM = 8 x 3 x 4 = 96

Question 22

1/10 - 1/35 = ?

Answer 22

GCF of 10 and 35 is 5
10 = 5 x 2
35 = 5 x 7
LCD = LCM = 5 x 2 x 7 = 70
1/10 - 1/35 = 7/70 - 2/70 = 5/70 = 1/14

Question 23

LCM / GCF formula

Answer 23

LCM = ( P x Q ) / GCF

Question 24

0 is… (x3)

Answer 24

an even integer
not positive
not negative

Question 25

Evens and Odds (adding & subtracting)
E + E =
E - E =

O + O =
O - O =

E + O =
E - O =

Answer 25

E + E = E
E - E = E

O + O = E
O - O = E

E + O = O
E - O = O

Add or subtract “likes”, we get EVEN
Add or subtract “unlikes”, we get ODD

Question 26

Evens and Odds (multiplying)
E x E =

O x O =

E x O =

Answer 26

E x E = E

O x O = O

E x O = E

as long as there is at least one even factor, the product will be even

Question 27

P, Q, R and S are integers. If P is even and (PxQ + RxS) is an odd integer, then which of the following must be true?
I - Q is odd
II - R is odd
III - S is odd

A : I only
B : I and II only
C : I and III only
D : II and III only
E ; I, II and II

Answer 27

D : II and III only

Question 28

P is an odd integer and (P² + QxR) is an even integer, then which of the following must be true?

A : either Q or R is an odd integer
B : either Q or R is an even integer
C : both Q and R are odd integers
D : both Q and R are even integers
E : nothing can be concluded

Answer 28

P² is odd so QxR must be odd
we have no guarantee that Q and R are integers !!!
E : nothing can be concluded

Question 29

If P and Q are integers, and if (P² + PQ) is an odd number, what do we know about P and Q?

Answer 29

P is odd and Q is even.

either plug 1 and 2 for odd and even nb and try the 4 cases.
or use logic: if P is even, it makes both terms even which doesn’t work (the result would be even) so P must be odd
if P is odd, the first term is odd. If the sum is odd, the second term must be even. if P is odd, the only way PQ can be even is if Q is even

Question 30

If P and Q are integers, and (4P + Q) is odd, what must be true?

Answer 30

Q is odd

4P must be even, so Q is odd (or use the 4 cases method)

Question 31

When positive integer N is divided by positive integer P, the quotient is 18, with a remainder of 7. When N is divided by (P + 2), the quotient is 15 and the remainder is 1. What is the value of N?

Answer 31

N/P = 18 + 7/P
N/(P+2) = 15 + 1/(P+2)
SO !!!
N = 18P + 7
N = 15(P+2) + 1 = 15P + 31

18P + 7 = 15P + 31
3P = 24
P = 8

N = 15 (8+2) + 1 = 151