Number Games?

Question

There are a couple of number tricks that have me intrigued.
Game 1:
Take any two digit number
Add the digits
Subtract the sum of those from the number itself
Answer is ALWAYS divisible by 9.

undertsand how that one works - essentially you're removing the UNITS - then subtracting 10% leaving 90% (a multiple of 9) of the original number

the one that still has me looking for an explanation is

GAME 2:
Take any 3 digit number
Reverse it
Subtract the smaller from the larger
Result is ALWAYS divisible by 9 (unless the rverse is the same as the original - then it's zero.)

Why is it so?

Answer 1

any 3 digit number is 100x+10y+z where x,y &z are any digit 0-9
now the revers is 100z+10y+x
subtract the second from the first one & you get
99x-99z which is always divisible by 9 for any amount of x,y &z.

Answer 2

1

Answer 3

GAME 1:
a = tens digit (1-9)
b = units digit (0-9)

Your two digit number is 10a +b
From this you are subtracting the sum of the digits:
= 10a + b - (a+b)

Simplifying:
= 10a + b - a - b
= 10a - a
= 9a

So in Game 1, you can see that you always have a power of 9 (specifically 9, ... , 72, 81).

GAME 2:
a = hundreds digit (1-9)
b = tens digit (0-9)
c = units digit (0-9)

Your 3 digit number is:
100a + 10b + c

The reversed number is:
100c + 10b + a

If a is bigger than c:
= 99a - 99c
= 99 (a-c)
= 9 * 11 (a-c)
If c is bigger than a:
= 9 * 11 (c-a)

Since a and c are integers, their difference will be an integer. So again you will have a power of 9 (actually a power of 99!). BTW, technically 0 is a power of 9 too, but it only happens if a and c are the same meaning the reversed number is the same.

So it isn't so mysterious, is it!?!

Answer 4

Both games will work for any natural number.

Let n be a natural number and b(n) be the sum of the digits of n.

Theorem: n = b(n) (mod 9). I.e. n and b(n) leave the same remainder when divided by 9.

Hence n-b(n) = 0 (mod 9) which implies that n-b(n) is divisible by 9.

In your second game, let n* be n with reversed digits. Then b(n*) = b(n), and since n* = b(n*) (mod 9), it follows that n* = b(n*) = b(n) = n (mod 9), and hence n*-n = 0 (mod 9), and similarly n-n* = 0 (mod 9).

Answer 5

game 1
97
9+7=16
97-16=81(it is divisible by 9)

game 2
099
990
9-9=0(it is divisible by 9)