Question

Ayesha has a theory about two digit numbers that have a larger tens digit than ones digit. she says that reversing the digits and subtracting the reversed number from the original number will always generate a number that is a multiple of 9. Is Ayesha right ? If so why do you think this works ? If not ,why not ?
would the same patterns work for three -digit numbers in which the hundreds digit is larger than the tens digit and the tens digit is larger than the ones digit.

Answer 1

Answer:

Ayesha says that reversing the digits and subtracting the reversed number from the original number will always generate a number that is a multiple of 9.

Lets take a number 63

Now reverse of this is 36. Subtracting $63-36$ we get 27, which is a multiple of 9.

Lets take 76. The reverse is 67 and the difference is $76-67=9$, which is a multiple of 9.

So, yes Ayesha's theory is correct.

This theory works because-

Lets say the original number is 10x+y and the reverse is 10y+x

So, the difference will be

$10x+y-(10y+x)$

=$9x-9y=9(x-y)$

This shows why the resulting number is a multiple of 9.

Yes, this works for three digits as well.

Example:

$762-267=495$ (multiple of 9)

$321-123=198$ (multiple of 9)

Answer 2

Ayesha's right.  There's a good trick for knowing if a number is a multiple of nine called "casting out nines."  We just add up the digits, then add up the digits of the sum, and so on.  If the result is nine the original number is a multiple of nine.  We can stop early if we recognize if a number along the way is or isn't a multiple of nine.  The same trick works with multiples of three; we have one if we end with 3, 6 or 9.

So $1284673$ has a sum of digits 31 whose sum of digits is 4, so this isn't a multiple of nine.  It will give a remainder of 4 when divided by 9; let's check.

$1284673 = 142741 \times 9 + 4 \quad\checkmark$

Let's focus on remainders when we divide by nine. The digit summing works because 1 and 10 have the same remainder when divided by nine, namely 1.  So we see multiplying by 10 doesn't change the remainder.  So $100 \times a + 10 \times b + c$ has the same remainder as $a+b+c$.

When Ayesha reverses the digits she doesn't change the sum of the digits, so she doesn't change the remainder.  Since the two numbers have the same remainder, when we subtract them we'll get a number whose remainder is the difference, namely zero. That's why her method works.

It doesn't matter if the digits are larger or smaller or how many there are. We might want the first number bigger than the second so we get a positive difference, but even that doesn't matter; a negative difference will still be a multiple of nine. Let's pick a random number, reverse its digits, subtract, and check it's a multiple of nine:

$813219543092195 - 591290345912318 = 221929197179877 = 9 \times 24658799686653 \quad \checkmark$