Always a Multiple?

Michelle and Cathy, from Mount Waverley, started off by checking a few examples:

21: reverse 12
   21+12= 33
   33 = a multiple of 11.

25: reverse 52
  25+52=77
  77 = a multiple of 11.

18: reverse 81
  18+81=99
  99 = amultiple of 11.

31: reverse 13
  31+13=44
  44 = a multiple of 11.

That's a positive start - it looks like all two-digit numbers are going to behave like this! Hana, from Avenue House School, did a different kind of check:

I arranged multilink to show four tens and two units for 42, and two tens and four units for 24. I then put the four units with the four tens, and the two units with the two tens, giving six lots of eleven.

Interesting method - does it work for Michelle and Cathy's examples above too? Can we explain why this seems to be true for all two-digit numbers? Mariella, from Westwood Girls' College, wrote the following:

The number will always be a multiple of 11 because if (for example) I have the number 13, and I switch the digits, I will then have the number 31. Then I would add the numbers in the tens column, and get the number 4. Then I would add the numbers in the units column, and get the number 4 - again. The reason for this is that the two numbers in the tens and units columns are exactly the same, just swapped around. The first few numbers of the 11 times tables, have the same numbers in the tens and units column making it a member of the 11 times table.

If the two numbers add to make a number bigger than 100, they will still make a number in the 11 times tables because; if (for example) I have the number 89 when I switch the digits I will have the number 98, and when I add the numbers I get the answer 187.
187 is made up of 17 units (17) and 17 tens (170):

   17x10
+17x1
=17x11

I can see that 17 x 11 (187) will be in the 11 times tables.

Great - I'm convinced! What about the next problems?

Take any two-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?
    
Rupa, from Garden International School, solved this using algebra:

Using algebra, let a be the tens number and b be the units number.

For the number 'ab', the number is worth 10a+b.
For 'ba', the number is worth 10b+a.
So, (10a+b) - (10b+a) = 9a-9b.
Then you can factorize it to get 9(a-b).
This shows that 9 is the common factor.

For example: 54.
Reverse the number, you get 45.
And then 54-45= 9.

Another example, just to check: 76.
Reverse, 67.
Then 76-67=9.

Great - well spotted!
    
Take any two-digit number. Add its digits, and subtract your answer from your original number. What do you notice?

Monel and Rasel, from Globe Academy, gave us their thoughts on this problem:

If you have 36 (for example) and you take away the units digit first, then you will always get a multiple of 10.
If you then take away the tens digit you will get a multiple of 9 (because 3x10 - 3x1 = 3x9).

Nice! Maisy, from England, also solved this using algebra:

The number ab (tens: a, units: b) can be written as 10a + b.
The digits are a and b, so when you add them you get (a + b).
Then (10a + b) - (a + b) = 9a + 0b = 9a (always a multiple of 9).

Take any three-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?

Rupa wrote:

If a is the hundreds digit, b is the tens digit and c is the units digit, the number would look like this: 100a + 10b + c.
After reversing the digits, it would be 100c + 10b + a.
Subtracting gives: (100a + 10b + c) - (100c + 10b + a).
Simplifying gives: 99(a-c)
which means that, for every three digit number, you always get a multiple of 99.
   
Take any five-digit number. Reverse the digits, and subtract your answer from your original number. What do you notice?

Maisy wrote:

We want to do abcde - edcba:
(10,000a + 1000b + 100c + 10d + 1e) - ( 1a + 10b + 100c + 1000d + 10,000e)
= 9999a + 990b + 0c - 990d - 9999e.
Simplifying gives:
99(101a + 10b - 10d - 101e)
which is always a multiple of 99.

Rupa also had a go at a couple of the other tricks:

Special numbers
If the tens unit is replaced with a, and the units digit is replaced with b, the value of the original number is 10a+b.
The challenge is to arrive at the original number when adding (a+b) to ab.
The equation would look like this:
(a+b)+ab=10a+b
So ab=9a
So b=9

Think Of two Numbers
Think of two whole numbers under 10.

Take one of them and add 1.
Multiply by 5.
Add 1 again.

Double your answer.

Subtract 1.

Add your second number.

Add 2.

Double again.

Subtract 8.

Halve this number and tell me your answer.

If the two numbers were a and b, each step of the sequence would look like this:

a and b
a+1
5(a+1)=5a+5
5a+6
10a+12
10a+11
10a+11+b
10a+13+b
20a+26+2b
20a+18+2b
10a+9+b

When you are told the answer, subtract 9. You will be left with a two-digit number; the digits of this number will be the numbers that your friend thought of.


Fantastic. Thanks!