Year 9 Proof
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Objectives: Understand what is meant by a proof, and examples of
what does and what doesn’t constitute a proof.
Be able to solve problems involving oddness/evenness, consecutive
numbers and involving digits.
Last modified: 20th March 2016
How Many Examples Needed?
In 1772 Euler noticed that the following equation
gives prime numbers for many positive integers n:
𝒏𝟐 – 𝒏 + 𝟒𝟏
We might wonder if the statement “For all positive
integers n, n2 – n + 41 is prime” is true.
Try this for a few 𝒏. How many examples of n
would we need to show this statement is:
a) True
One
A Few
Infinitely Many
b) False
One
A Few
Infinitely Many
The smallest value of 𝑛 for which this statement is false is: 41?
An Oxford entrance test question (yes, really!)
Source: MAT
A

B

C
D

The statement “If 𝐴, then 𝐵” is only falsified when 𝐴 is true but 𝐵 is false.
If the condition of the statement is not true, then the statement is not
applicable, and we can neither say it is true or false.
Counterexamples
! A counterexample is an example used to disprove a statement.
Discussing in pairs, find counterexamples for the following statements:
1
Statement
Possible counterexample
Prime numbers are always
odd.
2 is prime, but not odd.
The square root of a number is
2 always smaller than the
number itself.
3
If 𝑝 is prime then 𝑝 + 2 is
prime.
2𝑛2 + 11 is prime for all
4 integer values of 𝑛
?
0.25 = 0.5, but 0.5 > 0.25.
?
7 is prime but 9 is not.
?
(2 × 112 ) + 11 will be
divisible by 11 ?
because 2 ×
2
11 and 11 both are.
Types of Proofs
Geometry
Sequence Proofs
“Prove that the
exterior angle
of a triangle is
the sum of the
two other
interior angles.”
Consecutive Numbers
“Prove that the sum of any three
consecutive integers is a multiple of 3.”
“The first two terms of a sequence
are the numbers 1, 2. From then on,
each term is obtained by adding 1 to
the previous term and then dividing
by the term before that. Prove that
the sequence repeats.”
We already covered these in the
Sequences II topic.
Game Proofs
Even/Odd
If I have a ‘perfect Connect 4’ strategy
for winning, how can I show it’s perfect?
e.g. “Show that for all integers 𝑛,
𝒏𝟐 + 𝒏 is always even.”
Even/Odd Proofs
You will do this kind of proof at GCSE…
“Show that for any integer n,
𝑛2 + 𝑛 is always even.”
The key here is that
all integers are either
odd or even.
We can therefore do
a case analysis and
consider these two
different cases.
1. If n is odd:
𝑜𝑑𝑑 × 𝑜𝑑𝑑 + 𝑜𝑑𝑑 = 𝑜𝑑𝑑 + 𝑜𝑑𝑑 = 𝑒𝑣𝑒𝑛
2. If n is even:
𝑒𝑣𝑒𝑛 × 𝑒𝑣𝑒𝑛 + 𝑒𝑣𝑒𝑛 = ?𝑒𝑣𝑒𝑛 + 𝑒𝑣𝑒𝑛 = 𝑒𝑣𝑒𝑛
Therefore 𝑛2 + 𝑛 is even for all integers.
Test Your Understanding
A
B
Prove that 𝑛 𝑛 − 1 + 1 is odd for all integers 𝑛.
If 𝒏 is odd, 𝒐𝒅𝒅 × 𝒆𝒗𝒆𝒏 + 𝒐𝒅𝒅 = 𝒐𝒅𝒅
If 𝒏 is even, 𝒆𝒗𝒆𝒏 × 𝒐𝒅𝒅 +
? 𝒐𝒅𝒅 = 𝒐𝒅𝒅
Thus 𝒏 𝒏 − 𝟏 + 𝟏 is odd for all integers 𝒏.
[IMC 2009 Q16] How many different positive integers 𝑛 are there for which 𝑛
and 𝑛2 + 3 are both prime numbers?
A 0
B 1
C 2
D 3
E infinitely many
(Hint: Think odd/even)
Solution: B
All prime numbers except 2 are odd. But 𝒐𝒅𝒅𝟐 + 𝒐𝒅𝒅 = 𝒆𝒗𝒆𝒏, which can’t be
prime (and note 𝒏𝟐 + 𝟑 can’t be 2). This?the only possibility is if 𝒏 = 𝟐.
Indeed 𝟐𝟐 + 𝟑 = 𝟕 is prime. So there is one solution only.
Exercise 1
1
Prove that 2𝑛 + 1 is odd for all integers 𝑛.
If 𝒏 is odd, 𝒆𝒗𝒆𝒏 × 𝒐𝒅𝒅 + 𝒐𝒅𝒅 = 𝒐𝒅𝒅
If 𝒏 is even, 𝒆𝒗𝒆𝒏 × 𝒆𝒗𝒆𝒏 + 𝒐𝒅𝒅 = 𝒐𝒅𝒅
Therefore 𝟐𝒏 + 𝟏 is odd for all integers 𝒏.
6
?
2
Prove that 𝑛3 − 𝑛 is always even for all integers 𝑛.
If 𝒏 is odd, 𝒐𝒅𝒅 × 𝒐𝒅𝒅 × 𝒐𝒅𝒅 − 𝒐𝒅𝒅 = 𝒆𝒗𝒆𝒏
If 𝒏 is even, 𝒆𝒗𝒆𝒏 × 𝒆𝒗𝒆𝒏 × 𝒆𝒗𝒆𝒏 − 𝒆𝒗𝒆𝒏 = 𝒆𝒗𝒆𝒏
Thus 𝒏𝟑 − 𝒏 is always even for all integers 𝒏.
?
3
?
7
Find the smallest positive counterexample for the
statement “If 𝑝 is prime, then 𝑝2 + 𝑝 + 1 is prime.”
Smallest is 𝒑 = 𝟕 as 𝟕𝟐 + 𝟕 + 𝟏 = 𝟓𝟕 = 𝟏𝟗 × 𝟑
?
4
[Kangaroo Pink 2015 Q16] Which of the following
values of 𝑛 is a counterexample to the statement, ‘If 𝑛
is a prime number, then exactly one of 𝑛 − 2 and 𝑛 + 2
is prime’?
A 11
B 19
C 21
D 29
E 37
Solution: E
?
5
[SMC 2013 Q15] For how many positive integers 𝑛 is
4𝑛 − 1 a prime number?
A 0
B 1
C 2
D 3
E infinitely many
Solution: B
?
[SMC 2001 Q9] Which of the following numbers 𝑛
gives a counter-example for the statement: “If 𝑛 is a
prime number then 𝑛2 + 2 is also a prime number’?
A 3
B 5
C 6
D 9
E none of them
Solution: B
[SMC 2006 Q12] The factorial of 𝑛, written 𝑛! Is
defined by 𝑛! = 1 × 2 × 3 × ⋯ × 𝑛.
Which of the following values of 𝑛 provides a
counterexample to the statement:
“If 𝑛 is a prime number, then 𝑛! + 1 is also a prime
number”?
A 1
B 2
C 3
D 4
E 5
Solution: E
?
8
[SMC 2013 Q16] Andrew states that every composite
number of the form 8𝑛 + 3, where 𝑛 is an integer,
has a prime factor of the same form. Which of these
numbers is an example showing that Andrew’s
statement is false?
A 19
B 33
C 85
D 91
E 99
Solution: D
?
Divisibility Proofs
Show that the sum of any three consecutive integers is a multiple of 3.
Kyle’s proof:
“5 + 6 + 7 = 18, which is divisible by 3”.
We could represent 3 consecutive integers as:
𝒏
𝒏+𝟏 𝒏+𝟐
?
Then:
𝒏+ 𝒏+𝟏 + 𝒏+𝟐
= 𝟑𝒏 + 𝟑
= 𝟑(𝒏 +?𝟏)
which is divisible by 3.
Que Problemo?
He hasn’t shown it’s
true for all?
possible
integers.
Divisibility Proofs
How could I algebraically represent:
!
An odd number
2𝑘?+ 1
An even number
2𝑘
?
Two consecutive odd numbers
2𝑘 + 1,
?
2𝑘 + 3
(but 2𝑘 − 1 and 2𝑘 + 1 might make
your subsequent algebra easier)
The sum of the squares of two
consecutive numbers.
A power of 2.
One less than a multiple of 5.
𝑘2 + 𝑘 + 1
?
2?𝑘
5𝑘?− 1
2
Check Your Understanding
Prove that the sum of three consecutive even numbers is a multiple
of 6.
2𝑛 + 2𝑛 + 2 + 2𝑛 + 4
= 6𝑛 + 6
= 6(𝑛 + 1) ?
which is a multiple of 6.
Prove that the product of two consecutive odd numbers is always
one less than a multiple of 4.
2𝑛 + 3 2𝑛 + 1
= 4𝑛2 + 8𝑛 + 3
= 4 𝑛2 + 2𝑛 + 3
Which is 3 more than a multiple of 4, i.e. 1 less than a multiple of 4.
Alternatively:
?
2𝑛 + 1 2𝑛 − 1
= 4𝑛2 − 1
Which is clearly one less than a multiple of 4.
Exercise 2
Q1
Prove algebraically that the sum of two consecutive odd
numbers is divisible by 4.
𝟐𝒙 + 𝟏 + 𝟐𝒙 + 𝟑
= 𝟒𝒙 + 𝟒 = 𝟒(𝒙 + 𝟏)
N1
?
Q2
[GCSE] I think of two consecutive integers. Prove that the
difference of the squares of these integers is equal to the
sum of the two integers.
Two numbers are: 𝒙 and 𝒙 + 𝟏
Difference of squares: 𝒙 + 𝟏 𝟐 − 𝒙𝟐 = 𝟐𝒙 + 𝟏
Sum of numbers: 𝒙 + 𝒙 + 𝟏 = 𝟐𝒙 + 𝟏
These are equal.
?
Q3
𝒏 + 𝟏 𝟑 − 𝒏𝟑
= 𝒏𝟑 + 𝟑𝒏𝟐 + 𝟑𝒏 + 𝟏 − 𝒏𝟑
= 𝟑𝒏𝟐 + 𝟑𝒏 + 𝟏
= 𝟑𝒏 𝒏 + 𝟏 + 𝟏
The product of two consecutive integers is even,
thus 𝟑𝒏(𝒏 + 𝟏) is divisible by 6.
?
N2
Prove that the difference between the squares of two
consecutive odd numbers is a multiple of 8.
𝟐𝒏 + 𝟏 𝟐 − 𝟐𝒏 − 𝟏 𝟐
= 𝟒𝒏𝟐 + 𝟒𝒏 + 𝟏 − 𝟒𝒏𝟐 − 𝟒𝒏 + 𝟏
= 𝟒𝒏𝟐 + 𝟒𝒏 + 𝟏 − 𝟒𝒏𝟐 + 𝟒𝒏 − 𝟏
= 𝟖𝒏
which is divisible by 8.
[JMO] Find a rule which predicts exactly when five
consecutive integers have sum divisible by 15.
𝒏+ 𝒏+𝟏 + 𝒏+𝟐 + 𝒏+𝟑 + 𝒏+𝟒
= 𝟓𝒏 + 𝟏𝟎
= 𝟓(𝒏 + 𝟐)
Thus to be divisible by 15, 𝒏 + 𝟐 must be divisible by 3,
i.e. the middle number of the five numbers must be
divisible by 3.
?
[IMO] Prove that there is exactly one
sequence of five consecutive positive
integers in which the sum of the squares of
the first three integers is equal to the sum
of the squares of the other two integers.
𝒙 − 𝟏 𝟐 + 𝒙𝟐 + 𝒙 + 𝟏 𝟐
= 𝒙+𝟐 𝟐+ 𝒙+𝟑 𝟐
𝟑𝒙𝟐 + 𝟐 = 𝟐𝒙𝟐 + 𝟏𝟎𝒙 + 𝟏𝟑
𝒙𝟐 − 𝟏𝟎𝒙 − 𝟏𝟏 = 𝟎
𝒙 + 𝟏 𝒙 − 𝟏𝟎 = 𝟎
𝒙 = −𝟏 𝒐𝒓 𝒙 = 𝟏𝟎
So 𝟗𝟐 + 𝟏𝟎𝟐 + 𝟏𝟏𝟐 = 𝟏𝟐𝟐 + 𝟏𝟑𝟐 is only
solution.
?
?
Q4
Prove that the difference between two
consecutive cubes is one more than a
multiple of 6.
N3
Prove that the product of four consecutive
numbers is one less than a square number.
𝒂 𝒂+𝟏 𝒂+𝟐 𝒂+𝟑 +𝟏
= 𝒂𝟐 + 𝒂 𝒂𝟐 + 𝟓𝒂 + 𝟔 + 𝟏
= 𝒂𝟒 + 𝟔𝒂𝟑 + 𝟏𝟏𝒂𝟐 + 𝟔𝒂 + 𝟏
?
= 𝒂𝟐 + 𝟑𝒂 + 𝟏
𝟐
Digit Proofs
“I think of a two-digit number. I then reverse the
digits. Prove that the difference between the two
numbers is a multiple of 9”
e.g. 71 – 17 = 54
Let “ab” be a two digit number. Then its reverse is “ba”.
The value of each
number
is 10𝑎Algebraically
+ 𝑏 and 10𝑏 +?𝑎
Model
Problem
respectively.
Thus their difference is 10𝑎 + 𝑏 − 10𝑏 + 𝑎 = 9𝑎 − 9𝑏
− 𝑏)conclusion
(Note how factoring
Manipulate =to9(𝑎
reach
? out the 9 is a
good explicit way to show the
which is a multiple of 9.
expression is a multiple of 9)
Test Your Understanding
a
What would be the value of the 4-digit number with digits "𝑎𝑏𝑐𝑑”?
𝟏𝟎𝟎𝟎𝒂 + 𝟏𝟎𝟎𝒃 + 𝟏𝟎𝒄 + 𝒅
b
What would be the value if its digits were reversed?
𝟏𝟎𝟎𝟎𝒅 + 𝟏𝟎𝟎𝒄
? + 𝟏𝟎𝒃 + 𝒂
?
c
What is the sum of these two numbers?
𝟏𝟎𝟎𝟏𝒂 + 𝟏𝟏𝟎𝒃?+ 𝟏𝟏𝟎𝒄 + 𝟏𝟎𝟎𝟏𝒅
d
Hence prove that the sum of a four digit number and its reverse is a multiple of 11.
= 𝟏𝟏(𝟗𝟏𝒂 + 𝟏𝟎𝒃
? + 𝟏𝟎𝒄 + 𝟗𝟏𝒅)
Rubbish Proofs
Darth Kitty’s solution:
“Pick the number 31 for example. The
reverse of this is 13. And
31 – 13 = 18, which is divisible by 9.
Hence the statement is true.”
What is wrong with their proof?
Showing something is true for one example
is not sufficient. We need to show the
? POSSIBLE starting
statement is true for ANY
number we choose.
Rubbish Proofs
Photoshop Kitty’s solution:
“Here’s all possible two-digit
numbers:
11 – 11 = 0, 12 – 12 = 0, ...
21 – 12 = 9, 22 – 22 = 0, 32 – 23 = 9, 42 – 24 = 18, ...
98 – 89 = 9, 99 – 99 = 0
These are all divisible by 9.
What is wrong with their proof?
Technically it’s a valid proof, because we’ve
shown the statement is true for every
possible two-digit number. But it’s bad maths
having to list out every?case, and would
become difficult if for example we tried to
generalise the statement to 3-digit numbers.
Proof of 9 divisibility rule
Prove for 2-digit numbers that if the sum of the digits of a number is a
multiple of 9, then the number itself is divisible by 9.
How to represent a two digit number 𝑛 with digits "𝑎𝑏":
𝑛 = 𝟏𝟎𝒂 + 𝒃
?
How do we represent property that digits add up to a multiple of 9?
𝒂 + 𝒃 = 𝟗𝒌
?
How do we combine these equations to therefore show that the original number 𝑛
is a multiple of 9?
𝒏 = 𝟏𝟎𝒂 + 𝒃
= 𝟗𝒂 + 𝒂 + 𝒃
= 𝟗𝒂 +?𝟗𝒌
=𝟗 𝒂+𝒌
Hence 𝒏 is divisible by 9.
Exercise 3 – Digit Problems
[JMO] Observe that 49 = 4 x 9 + 4 + 9
I think of a 3-digit number and then reverse its digits.
Q1 Prove that the difference between these two numbers 4 Use algebra to find all other two-digit numbers
which are equal to the product of their digits plus
the sum of their digits.
is a multiple of 11.
If a number has the digits “𝒂𝒃𝒄” then its reverse is "𝒄𝒃𝒂“.
Each number has the value 𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 + 𝒄 and 𝟏𝟎𝟎𝒄 +
𝟏𝟎𝒃 + 𝒂.
𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 + 𝒄 − 𝟏𝟎𝟎𝒄 + 𝟏𝟎𝒃 + 𝒂
= 𝟗𝟗𝒂 − 𝟗𝟗𝒄 = 𝟏𝟏 𝟗𝒂 − 𝟗𝒄
𝟏𝟎𝒂 + 𝒃 = 𝒂𝒃 + 𝒂 + 𝒃
𝟗𝒂 = 𝒂𝒃
Given 𝒂 is not 0:
𝟗=𝒃
Thus numbers are 𝟏𝟗, 𝟐𝟗, … , 𝟗𝟗
?
[IMO] All the digits of a certain positive three-digit
Q2 number are non-zero. When the digits are taken in
?
Prove that there are no three-digit numbers
b which are equal to the product of their digits plus
the sum of their digits.
reverse order a different number is formed. The
difference between the two numbers is divisible by
eight. Given that the original number is a square
number, find its possible values.
Difference of “𝒂𝒃𝒄” and "𝒄𝒃𝒂“ is 𝟗𝟗(𝒂 − 𝒄)
Since 99 is not divisible by 8, 𝒂 − 𝒄 is. Only possible if 𝒂 =
𝟏, 𝒄 = 𝟗 or vice versa. Only square in former case is 169.
Only square in latter case is 961.
?
[IMO] An ‘unfortunate’ number is a positive integer
Q3 which is equal to 13 times the sum of its digits. Find all
‘unfortunate’ numbers. (Hint: start with 2 digit first)
For two-digit 𝟏𝟎𝒂 + 𝒃 = 𝟏𝟑 𝒂 + 𝒃 → 𝟑𝒂 + 𝟏𝟐𝒃 =
𝟎 which is impossible. Using same approach for 3digit: 117, 156, 195
Need to also show not possible for more than 3 digits.
?
𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 + 𝒄 = 𝒂𝒃𝒄 + 𝒂 + 𝒃 + 𝒄
𝟗𝟗𝒂 + 𝟗𝒃 = 𝒂𝒃𝒄
But since 𝒃𝒄 is at most 81, 𝟗𝟗 > 𝒃𝒄 and thus
𝟗𝟗𝒂 > 𝒂𝒃𝒄 and hence the LHS is always greater.
?
5
N
Can you prove the 9 divisibility rule for 3 digit numbers?
A divisibility test for 7 is to subtract twice the last digit
from the remaining numbers, and see if the result is
divisible by 7. e.g. 392 -> 39 – 4 = 35, which is clearly
divisible by 7, so 392 is. Prove this works for all threedigit numbers.
Let our number be n with the digits “abc”.
𝒏 = 𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 + 𝒄. If we subtract twice the last digit
from the remaining numbers, we get 10a + b – 2c. Now if
this is divisible by 7, 10a + b – 2c = 7k for some integer k.
Thus 𝟏𝟎𝟎𝒂 + 𝟏𝟎𝒃 − 𝟐𝟎𝒄 = 𝟕𝟎𝒌
Then n = 100a + 10b + c = 70k + 21c = 7(10k + 3c) which is
divisible by 7.
?
Another Example of Good/Bad Proofs
[IMO] “An arithmetic sequence is one in which the
difference between successive terms remains constant
(for example, 4, 7, 10, 13, …). Suppose that a rightangled triangle has the property that the lengths of its
sides form an arithmetic sequence. Prove that the
sides of the triangle are in the ratio 3:4:5.”
Example of a bad proof:
“If the sides of the triangle are in the ratio
3:4:5, we could have any multiple of this.
e.g. 6, 8, 10. These are still in an
arithmetic sequence and satisfy
Pythagoras Theorem.”
Why is it bad?
We’re supposed to be showing that IF the sides
form an arithmetic sequence THEN the sides have
ratio 3:4:5. They’ve just shown the reverse!
?
What would a good proof look like?
Use our general structure for an algebraic
proof!
• “Assume that the sides form an arithmetic
sequence.
• Model this assumption,
? using algebraic
expressions for side lengths that represent
all possible arithmetic sequences, e.g. a-b,
a, a+b.
• Do some manipulation using Pythagoras...
• Therefore, the sides have ratio 3:4:5.”
QED
QED is short for the latin “Quad erat demonstrandum”.
?
It translates roughly to “Which had to be
? demonstrated”.
Its symbol in maths is □ ?
I think of a two-digit number. I then reverse the digits. Prove that
the difference between the two numbers is a multiple of 9.
“Let the number have the digits ‘ab’. Then the number is 10a + b.
If we reverse the digits to get ‘ba’, which has the value 10b + a.
Subtracting to find the difference:
10a + b – (10b + a) = 9a – 9b = 9(a-b).
□
9(a-b) is divisible by 9.”
‘Without loss of generality’
I think of a two-digit number. I then reverse the digits. Prove that
the difference between the two numbers is a multiple of 9.
“Let the number have the digits ‘ab’. Then the number is 10a + b.
If we reverse the digits to get ‘ba’, which has the value 10b + a.
Subtracting to find the difference:
10a + b – (10b + a) = 9a – 9b = 9(a-b).
□
9(a-b) is divisible by 9.”
When we find the difference between two numbers we tend to do the bigger number
subtracting the smaller number. Here in our proof we did “𝑎𝑏”−”𝑏𝑎”. If this difference is
positive what have we assumed?
That 𝒂 ≥ 𝒃
?
If we stated this assumption that 𝑎 ≥ 𝑏, have we made our proof ‘less general’, in the
sense that we’ve only now proved the property for restricted two-digit numbers where
the first digit is greater than (or equal to) the second?
No, it doesn’t matter, because if we had say 𝟓𝟕, then after we’d reversed the digits to
get 75, we could have subtracted the other way round. The proof still implicitly covers
?
all two-digit numbers because the original number
and the reversed number are
interchangeable when we come to subtract.
‘Without loss of generality’
I think of a two-digit number. I then reverse the digits. Prove that
the difference between the two numbers is a multiple of 9.
“Let the number have the digits ‘𝑎𝑏’. Without loss of generality, let
𝒂 ≥ 𝒃. Then the number is 10𝑎 + 𝑏.
If we reverse the digits to get ‘ba’, which has the value 10b + a.
Subtracting to find the difference:
10a + b – (10b + a) = 9a – 9b = 9(a-b).
□
9(a-b) is divisible by 9.”
! ‘Without loss of generality’ (or w.l.o.g. for short)
means that we have assumed a specific case, but this
doesn’t make the proof any less ‘general’.
Other Types of Proof
[Source: JMO] “X and Y play a game in which X
starts by choosing a number, which must be either
1 or 2.
Y then adds either 1 or 2 and states the total of the
two numbers chosen so far. X does likewise, adding
either 1 or 2 and stating the total, and so on. The
winner is first player to make the total reach (or
exceed) 20.
i) Explain how X can always win.
…
For (i), what do we actually have to show?
That for any way in which
? Y plays, we can always
find a resulting move that guarantees us to win.
Games
Summary
So what makes a good (and valid!) proof?
1. Use of algebra:
a) When appropriate, use variables to represent something that could take any value
(e.g. the two digits!) This keeps your argument as general as possible.
b) Then use algebra to manipulate these variables based on the question.
2. Keep things general!
Avoid making assertions that would ignore certain possibilities.
3. Don’t make leaps of logic
Try not to ‘assume’ too much. You need to clearly show your steps without making
assumptions, which may turn out to be wrong!
4. Avoid circular arguments
If the statement to prove is something like “If X, then show that Y”, then your proof needs
to be something like: “Assume that X, then [manipulation involving X]... And thus Y is true.”
You can’t assume Y is true at any point, because that’s what you’re trying to prove!
5. Consider edge cases
Consider unusual cases. e.g. What if the first digit is 0? What if the digits are the
same? Does our proof still work?
6. Avoid listing out lots of cases
If you’re having to list out lots and lots of cases, then you’re possibly missing
ways to narrow down your search. Could you use algebra?