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Special Numbers

Stage: 3 Challenge Level: Challenge Level:1

We received a mass of good solutions to this problem.

Aileen, Becca, and Paulina from Walter J. Paton School described how they arrived at the solution:

The problem "Special Numbers" was really fun to figure out. At first, we didn't understand what it meant. But after a little figuring we noticed a pattern. Here is what we did:
 
We just tried random numbers and came across the number 18. It was only 1 off, so we tried 19.
The sum of 1+9 was 10 and the product of 1x9 was 9.
After that, we added 10+9 and it equaled 19!
That was our original number!
We found a special number!

But, there was still another question: is there more than one special numbers?
As we tried to figure out more we tried the number 29.
That worked too!

After a little observation, we noticed 9 was in the ones (units) place both times!

We tried more numbers such as 39 and 49 and so on.
Those all worked! We had found a pattern!
That was how we figured out that problem.
 
Joseph, from the same school, and May, from Heartlands High School, showed some examples that worked:
 
29:
2 + 9 = 11
2 x 9 = 18
11 + 18 = 29

39:
3 + 9 = 12
3 x 9 = 27 
12 + 27 = 39

49:
4 + 9 = 13
4 x 9 = 36
13 + 36 = 49

59:
5 + 9 = 14
5 x 9 = 45 
14 + 45 = 59

69:
6 + 9 = 15
6 x 9 = 54 
15 + 54 = 69


Joanna, Sophie and Gaby from Kings Sutton Primary School discovered these numbers and then moved on to other sets of special two-digit numbers:

On the first problem we divided up the numbers between the 3 of us and worked on different sections. Joanna started with 0 to 33, Sophie did 34 to 66 and Gaby did the rest.
 
Gaby found 69 worked and at about the same time Joanna found that 19 worked. So then we decided to try every 2 digit number ending in a 9. They all worked.

The algebra was a bit tricky so we looked at the second part and this is what we found:

We found that if you double the tens then the units have to be 8, and if you triple the tens it has to end in 7.

We then noticed a pattern, as the number we multiplied by increases the units value decreases.
We tested out our hypothesis and we found it worked.
 
Eddie, from Wilson's School, did use some algebra to explain his results:
 
Firstly, to make this problem simpler to solve, we can use algebra. This means substituting the first digit of the special number (the tens digit) for 'a', and substituting the second digit (the units digit) for 'b'.

So, the product of the digits (ab) + the sum of the digits (a + b) must equal the original number (10a + b).

As an equation, this is written as:
ab + a + b = 10a + b.

If you take away 'b' from both sides this equates to:
ab + a = 10a

Then, if you take away one 'a' from both sides you get:
ab = 9a
This is because 10 lots of 'a' take away one lot of 'a' is 9 lots (9a).

This can be written as: b x a = 9 x a.
Therefore, when we divide both sides by 'a' we get:
b = 9

This means the second digit (b) of the special number must equal 9.

Now we can check if this works.
Let's substitute 'a' for 1 and 'b' for 9:
1 x 9 + 1 + 9 = 10 x 1 + 9
19 = 19
It works.

Now substitute 'a' for 2:
2 x 9 + 2 + 9 = 10 x 2 + 9
29 = 29

Does 3 work?
3 x 9 + 3 + 9 = 10 x 3 + 9
39 = 39

And so on.
4, 5, 6, 7, 8 and 9 work too.

Now I will provide a reason for why the special numbers go up by 10:

If 'a' is 1, and 'b' is 9, then the number is 19.
The substitutions above show that this works.

If 'a' is 2, then the number is 29 (10 more than 19).
The difference between 1 x 9 and 2 x 9 is +9,
and the difference between 1 + 9 and 2 + 9 is +1.
Add the differences to get a total difference of +10.

Repeat this paragraph with the successive two numbers in the pattern and you will find that it works.


Ben Page from Hethersett High School also used some algebra to explain his results:
 
1x9 + 1 + 9 = 19
2x9 + 2 + 9 = 29
3x9 + 3 + 9 = 39
4x9 + 4 + 9 = 49
5x9 + 5 + 9 = 59
6x9 + 6 + 9 = 69
7x9 + 7 + 9 = 79
8x9 + 8 + 9 = 89
9x9 + 9 + 9 = 99

ab + a + b = 10a + b
ab + a = 10a
ab = 9a
b= 9
So all the two digit "special numbers" end in 9.  
 
Jamie from Mold Alun School explained why the special numbers increased by 10:

Any 2-digit number ending in 9 is one of the special numbers.
Starting with 19 you have
1 + 9 = 10 and 1x9 = 9
10 + 9 = 19

If you increase 19 by 10, the sum increases by 1 and the product increases by 9, giving you the extra 10 you need to make up the next number, 29.

Every time you increase the previous special number by 10, the sum increases by 1 and the product increases by 9, giving you the extra 10 you need to make up the next number.

e.g. for 29:
2 + 9 = 11, 1 more than the sum of the digits of 19,
and 2 x 9 = 18, 9 more than the product of the digits in 19.

 
Lena from SHHS found two sets of special numbers:
 
The expression for a two digit number is 10a+b.
To solve the first special number you create an algebraic equation.
You need to add the sum of the two digits to the product of the two digits to get the original number.

The equation looks like this:
10a + b = a + b + ab
9a + b = b + ab
9a = ab
b = 9

So now you know the units part of the two digit number will always be 9.
 
A different special number could be to add twice the tens digit to the units digit, then add this to the product of the digits and this gives back to the original number.

The equation would look like this:
10a + b = 2a + b + ab
8a + b = b + ab
8a = ab
b = 8

So you know the the second digit in the two digit number will always be 8.
For example 28 is a special number because
2x2 + 8 + 2x8 = 12 + 16 = 28

Joseph, from Wilson's Grammar School, chose ab to represent a two digit number and showed how to find special numbers that satisfied the different criteria: 
 
a + b + ab = 10a + b
a + b - b + ab = 10a
a + ab = 10a
ab = 10a - a
ab = 9a
b = 9
So the special numbers are 19, 29, 39, 49, 59, 69, 79, 89, 99

2a + b + ab =10a + b
2a + ab = 10a
ab = 10a - 2a
ab = 8a
b = 8
So in this case the special numbers are 18, 28, 38, 48, 58, 68, 78, 88, 98

3a + b + ab = 10a + b
3a + ab = 10a
ab = 10a - 3a
ab = 7a
b = 7
So in this case the special numbers are 17, 27, 37, 47, 57, 67, 77, 87, 97

4a + b + ab = 10a + b
4a + ab = 10a
ab = 10a - 4a
ab = 6a
b = 6
So in this case the special numbers are 16, 26, 36, 46, 56, 66, 76, 86, 96

5a + b + ab = 10a + b
5a + ab = 10a
ab = 10a - 5a
ab = 5a
b = 5
So in this case the special numbers are 15, 25, 35, 45, 55, 65, 75, 85, 95


Rajeev from Haberdashers' Aske's Boys' School found a general rule:

When the tens digit is multiplied n times, the units digit of the special numbers will be 10-n.

Max from Twyford C of E High School came to the same conclusion:
 
When it's 2 times the tens digit you take 2 away from 10 and that gives you the value of the units digit of the special numbers.
When its 3 times, or 4 times, etc, you just subtract it from 10 to give you the value of the units digit of the special numbers.  
 
Well done to you all.