Problem Teachers' Notes Hint Solution

"all of the numbers can be divided by 11."

1) Let's take a number 'abcd'

Value of this number in decimal notation = a*1000 + b*100 + c*10 + d

2) After moving the first digit to the last position, the number becomes 'bcda'.

Value of this number in decimal notation = b*1000 + c*100 + d*10 + a

Adding up the values of two numbers in decimal notation

(a*1000 + b*100 + c*10 + d) + (b*1000 + c*100 + d*10 + a)

= a(1000+1) + b(100+1000) + c(10+100) + d(1+10)

= 1001a + 1100b + 110c + 11d

= 11(91a + 100b + 10c + d)

As can be seen, the total has a factor of 11 and is divisible by 11.

Mikey from Tadcaster Grammar School offered a very similar explanation, added that this only works for numbers with an even number of digits and gave a few examples of numbers with an odd number of digits where it does not work. Click here to see his solution.

They are all multiples of eleven. It works for all numbers with even digits because it will be adding up multiples of 11 together.

Take 8937 for example:

8937+9378 = 8008+9900+330+77 = 18315

and 8008, 9900, 330 and 77 are all multiples of eleven.

It won't work for odd digits because numbers that have odd numbers of 0 in the middle would not be a multiple of 11:

Take 135 for example:

135+351 = 101+330+55 = 486

which is not a multiple of 11 because 101 is not a multiple of 11 (and the others are).

First digit = a

Second digit = b

If a + b = c

a b + b a = c c

Any four-digit number,can be rewritten as 1000A+100B+10C+D.

Moving the A to the back gives 1000B+100C+10D+A, and the sum of the two is 1001A+1100B+110C+11D.

It became evident that the numerical coefficient for every term had a common factor of 11 (1001=11*7*13).

Therefore, every such four digit number ABCD added to BCDA will be divisible by eleven (as suggested by the title).

It works because as you move a digit to the back, all the other digits (except for the first) will be added to its tenfold (since they are moved one place to the left) thus giving 11, 110, 1100...

The first digit will increase by its unit-place value as in (3000+3) for 3456, and generally (K000+K for KBCD).

Further, with any two digit number

(10A+B+10B+A) = 11A+11B is also divisible by eleven,

while any three digit number

(100A+10B+C+100B+10C+A) =101A+110B+11C isn't divisible by eleven as 11 isn't a factor of 101.

A pattern begins to emerge: all values other than the first: B, C, D... will always become multiples of eleven as they are added to their tenfold:

1=> 11 20=> 220, 500=> 5500.

Therefore only the value of the first digit in the sum will determine the divisibility of 11.

Two-digit: (10+1)A=11A

Three-digit: (100+1)=101A

Four digit: (1000+1)A=1001A

It becomes clear that any number with an EVEN number of digits added to its counterpart will be divisible by 11 while any number with an ODD number of digits will not.

So it will work for a 38 digit number as well as a 100...(Billion) digit number!