Transcript Year 9 Similarity - DrFrostMaths.com

KS3 Divisibility
Dr J Frost ([email protected])
Last modified: 30th November 2014
Divisibility Rules
How can we tell if a number is divisible by: !
2
3
4
5
6
7
?
Digits add up to multiple of 3. e.g: ?
1692: 1+6+9+2 = 18 
Last two digits are divisible by 4. e.g.
? 143328
Last digit is 0 or 5.
?
Number is divisible by 2 and 3 (so use
? tests for 2 and 3).
Last number is even.
There isn’t really any trick that would save time. You could double the last digit
and subtract it from the remaining digits, and see if the result is divisible by 7.
e.g: 2464 -> 246 – 8 = 238 -> 23 – 16 = 7. But you’re only removing a digit each time,
so you might as well long divide!
?
?
?
?
8
Last three digits divisible by 8.
9
Digits add up to multiple of 9.
10
Last digit 0.
11
When you sum odd-positioned digits and subtract even-positioned
digits, the result is divisible by 11. ?
e.g. 47949: (4 + 9 + 9) – (7 + 4) = 22 – 11 = 11, which is divisible by 11.
12
Number divisible by 3 and by 4.
?
Quickfire Divisibility
4
6 7
? ?
?
726
168
? ? ?
9196 ? ? ?
252
? ? ?
?
? ?
1001
?
? ?
91
216
? ? ?
87912 ? ? ?
9
11
? ?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
Breaking Down Divisibility Rules
Are these statements true or false?
If we want to show that a number
is divisible by 15, we can show it
is divisible by 3 and 5.
False 
True 
If we want to show that a number
is divisible by 24, we can show it
is divisible by 6 and 4.
False 
True 
The problem is that 12 is divisible by 6 and 4, but it is not
divisible by 24!
We need to pick two numbers which are coprime, i.e. do
not share any factors.
How can we therefore test if a number is divisible by 24?
Quickfire
What divisibility rules would we use if we wanted to test divisibility by:
18
45
36
40
2 and 9 rules
5 and 9
4 and 9
5 and 8
An easy Year 10 Maths Olympiad problem:
Find the smallest positive integer
which consists only of 0s and 1s,
and which is divisible by 12.
• Since in must be divisible by 4, the only
possibility for the last two digits is 00.
• It must have at least?three 1s to be divisible
by 3 (as we can’t have zero 0s).
• Therefore 11100 is the answer.
Exercises
Problem sheet of Junior and Intermediate Olympiad problems.
Work in pairs/groups if you wish.
Answers on next slides.
(File Ref: KS3_DivisibilityQuestions.docx)
Question 1
[J31] Every digit of a given positive integer is either a 3 or a 4 with each occurring at least
once. The integer is divisible by both 3 and 4. What is the smallest such integer?
?
Question 2
[J50] The eight-digit number “ppppqqqq”, where p and q are digits, is a multiple of 45.
What are the possible values of p?
?
Question 3
[M07] (a) A positive integer N is written using only the digits 2 and 3, with each appearing
at least once. If N is divisible by 2 and by 3, what is the smallest possible integer N?
(b) A positive integer M is written using only the digits 8 and 9, with each appearing at
least once. If M is divisible by 8 and by 9, what is the smallest possible integer M?
?
?
Question 4
[M55] A palindromic number is one which reads the same when its digits are reversed, for
example 23832. What is the largest six-digit palindromic number which is exactly divisible
by 15?
?
Question 5
[J16] Find a rule which predicts exactly when five consecutive integers have sum divisible
by 15.
?
Question 6
[M96] Find the possible values of the digits p and q, given that the five-digit number
‘p543q’ is a multiple of 36.
?
Question 7
[M127] The five-digit number ‘𝑎679𝑏’, where 𝑎 and 𝑏 are digits, is divisible by 36. Find all
possible such five-digit numbers.
?
Question 8
[M31] Find the smallest positive multiple of 35 whose digits are all the same as each
other.
?
Question 9
Show that:
𝑛3 − 𝑛 is divisible by 6 for all integers 𝑛.
𝑛4 + 2𝑛3 − 𝑛2 − 2𝑛 is divisible by 24 for all integers 𝑛.
a) 𝑛3 − 𝑛 = 𝑛 𝑛 − 1 𝑛 + 1
This is the product of three consecutive numbers. One of the three
numbers must be divisible by 3, so the product is divisible by 3.
Similarly, at least one of the three numbers is divisible by 2, so the
product is divisible by 2. Therefore, the product is divisible by 6.
b) = 𝑛3 𝑛 + 2 − 𝑛 𝑛 + 2
= 𝑛3 − 𝑛 𝑛 + 2
?
= 𝑛−1 𝑛 𝑛+1 𝑛+2
This is the product of 4 consecutive numbers.
At least one of the four is divisible by 3.
Exactly two of the numbers will be divisible by 2.
However, one of the four numbers will be divisible by 4, giving an
extra factor of 2. Overall, this means the product is divisible by
2 × 3 × 4 = 24.
Question 10
[Based on NRich] If the digits 5, 6, 7 and 8 are inserted at random in 3_1_4_0_92 (one in
each space), what is the probability that the number created will be a multiple of 396 if:
a) Each of 5, 6, 7, 8 is used exactly once in each of the four gaps.
b) Each of 5, 6, 7, 8 can be used multiple times.
[JAF solution] If number is divisible by 396, it is divisible by 4 × 9 × 11.
If we used one each of 5, 6, 7, 8 to fill the gaps, all three divisibility rules, for 4, 9 and
11, would be satisfied regardless of order, since the last two digits are fixed and all
inserted digits are in even positions. Thus the probability is 1.
b) If we used 5, 6, 7, 8, there are 4! = 24 possible orderings.
However, we could also use 5, 5, 8, 8 to fill the gaps, as this will not affect the digit sums
involved in either the divisibility by 9 or 11 rules (since all digits are inserted in even
?
positions), and the last two digits are fixed for the purposes of the 4 rule. There are 6
possible orderings of these digits.
6, 6, 7, 7 is also possible, which by the same reasoning, gives 6 possible orderings.
5, 7, 7, 7 is also possible, which gives 4 orderings.
6, 6, 6, 8 is also possible, which gives 4 orderings.
There are therefore 4 + 4 + 6 + 6 + 24 = 44 possible numbers divisible by 396.
However, there are 44 total possible ways of using 5, 6, 7 or 8 for each of the four gaps.
44
11
The probability is therefore 44 = 64