[Guest]

A contributor to one of K. Sengupta's recent puzzles recently stated a result, and assumed that any intelligent observer would have known it as a matter of course.

To the unintelligencia like myself, it is not so obvious, even though I think it to be undoubtably true. Effectively the statement was that, if one picked any integer in the decimal notation, then randomized the integers to form a separate number, and further subtracted the smaller number from the other, then the result would be divisible by 9.

Would the contributor be agreeable to furnish a simple proof of the statement?

[Guest]

Let N1 = a₀ + 10.a₁ + 10¹.a₂ + .. + 10^x.a_x

where a_i is an integer from 0 to 9.

A permutation of the digits of N1 is an other such expression, where the exponents of 10 are permutated.

N2 = 10^b1.a₀ + 10^b2.a₁ + 10^b3.a₂ + .. + 10^bx.a_x

Subtracting one from the other,

N2 - N1 = (10^something - 10^{somethingelse}).a₀ + (10^something - 10^{somethingelse}).a₁ + .. + (10^something - 10^{somethingelse}).a_x

Each of these terms have a factor that is a difference between two powers of 10. Difference between two powers of 10 is always a multiple of 9. Hence the whole expression is a multiple of 9.

[Guest]

Bleeding bloody obvious is it not? Many thanks.

[Guest]

A simpler way to think of it

The first number has a digital root of x . The second because it contains the same digits has the same digital root.

When the two numbers are subtracted the result must therefore have a digital root of 0. Any number with a digital root of 0 or 9 is divisible by 9.

[Guest]

Another method

I had done this sort of problem before- ended up using an excel spreadsheet and found patterns, etc.

Now I think the easiest way to explain it to a resilient brain like mine is with modular math.

Take original number, then find out what it is in mod base 9

Rearrange the digit and then take mod base 9 of that new number- it's the same as the mod 9 of the first number

Or, you could say, the remainders when divided by 9 are equal. Another way of writing each number would be

(number of times 9 goes in evenly) + remainder

_number above is a multiple of 9_

So if you subtract [(number of times 9 goes in) + R] - [(number of times 9 goes in second number) + R]

You get the difference of two multiples of 9, which is definitely divisible by 9 itself. Or, if you stay in mod 9 for both numbers and take the difference, you get 0. 0 in mod 9 means the number is divisible by 9