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Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

Basically

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

There were a number of correct solutions to this problem which either involved trial and improvement or the solution of a quadratic equation.

Correct solutions were received from Michael Brooker, Yatir Halvi (Maccabim-Reut High School), Ross Haines, Robert Haynes and Andre Lazanu. Well done!

Ross Haines first explained how number bases work:

We use base ten, which is this:

Thousands Hundreds Tens Ones

This can also be written as

10 cubeds 10 squareds 10s 1s

The second way is the way you write bases. It is the same format for all of them. For example, base 4 would be written as:

4 cubeds 4 squareds 4s 1s

To work a number from a base to base 10 you times the number by the column it is in. eg, using the number 321 in base four

4 cubeds 4 squareds 4s 1s
3
2
1

Which gives:

1 x 1 equals 1
2 x 4 equals 8
3 x ( 4 squared) equals 48

so 321 in base 4 would be 1+8+48=57 in base 10

Here is the trial and improvement solution submitted by Michael:

I worked it out like this: I decided to begin with bases much higher than 10, as 123 is much smaller than 3723. By trial and error I discovered that the answer was somewhere between 40 and 80.

So I tried translating "123" in Base 60 into Base 10. This time I did get 3723. Using powers I worked out the value of each base-60 digit in base 10 and added the figures together:

"3", the digit on the right, is the same in decimal

"2" x 60' gives 120

"1" x 60² gives 3600

Total = 3723

Yatir and Andrei found the solution by solving the quadratic equation 3x^2+2x+3=3723.

This works because the values of the columns in base x are:

x 2 x 1
So
1 2 3 = 1x 2 + 2x + 3
= 3723