Flip The Digits

Let $\overline{ABC} = 100A + 10B + C$

$\overline{ABC} - \overline{CBA} = (100A + 10B + C) - (100C + 10B + A)$

The above equation simplifies down to:

$99A - 99C = 3(33A - 33C)$

Since a $3$ can be extracted from the equation and both $A$ and $C$ are integers, the number produced is a multiple of $3$ .

Note: Since $0 = 0 \cdot 3$ , $0$ is a multiple of $3$ . In fact, $0$ is a multiple of any integer.

$\overline{ ABC } \equiv \overline{ CBA } \pmod{ 3 }$

$\therefore \quad \overline{ ABC } - \overline{ CBA } \equiv \boxed{ 0 } \pmod{ 3 } ~~ _\square$

Edit: I am adding the proof that $\overline{ ABC } \equiv \overline{ CBA } \pmod{ 3 }$

Let $dr ( n )$ denote the digital root of $n$ . Since $dr(\overline{ ABC } ) = dr( \overline{ CBA } )$ ,

$\overline{ ABC } \equiv \overline{ CBA } \pmod{ 9 }.$

Since $3 \mid 9$ , the following congruence is also true.

$\overline{ ABC } \equiv \overline{ CBA } \pmod{ 3 } ~~~~ _\square$

True. 100a + 10b + c - 100c - 10b - a = 100a - a - 100c + c 99a - 99c = 3(33) (a-c). There divisible by 3.

521-125 =396; 814 -418 = 396; abc -cba = 100a +10b + c - 100c + 10b + a = 99a = 99c = 99(a - c) is a multiple of 3

Any number subtracted fom it's reverse is a multiple of 3

Let a be the three digit number. Then a can be expressed as: 100x+10y+z

It's reverse is 100z+10y+x.

Getting the difference:

100x-100z+10y-10y + (z-x)

= 99x-99z = 99(x-z) =3(33(x-z))

If you sum any 3 numbers and the result is multiple of 3，the numbers were also multiple of 3.

Very easy kkkkk

from above question we can take any 3 digit no like 567 nd proceed.we can take other number too. reverse of 567 is 765 nd we have to subtract 567 from it so..... 765

567

198 we have 198 which is multyple of 3.

<XYZ>=100X+10Y+Z <ZYX>=100Z+10Y+X <XYZ>-<ZYX>=99X-99Z=3(33X-33Z)

Find the digital sum of ABC - CBA. Divide the digital sum by 3. If you get an integer result, it is true.

There is a problem in number theory that was introduced to me that was about bank tellers flipping and two numbers in an n-length value and the difference with be divisible by 9.

Let abc be the theree digit number cba is its revers (abc-cba)=(100a+10b+c-100c-10b-a)=99a-99c=3(33a-33c)

Subtracting any whole number from its reverse gives a multiple of 3. If the number has an odd number of digits the resulting number is divisible by 99 and if it has an even number of digits it is divisible by 9.

xyz -zyx = 99x -0y -99z = 3(33x -0y -33z).

I just went to my calculator and started punching in numbers and dividing by 3. Even works if it comes out to a negative integer.