A problem by Suhas B.R

EVERY TWO-DIGIT NUMBER CAN BE REPRESENTED IN THE FORM 10A+B. THE SUM OF DIGITS OF THIS TWO DIGIT NUMBER IS A+B. THERE THE SUM OF THE DIGITS SUBTRACTED FROM THE TWO-DIGIT NUMBER IS 10A+B-A-B=9A. THEREFORE THE SUM OF THE DIGITS OF ANY TWO-DIGIT NUMBER SUBTRACTED FROM THAT NUMBER GIVES A MULTIPLE OF 9.

$(a,b)\in\mathbb{N_0}$ , such that $(a,b)<10$ form two digit numbers $10a+b$ . So, the digit sum is $a+b$ . According to the question, we have... $10a+b-(a+b)=9a$

Hence, it's a multiple of $\fbox{9}$ .

A two digit number can be written as $10a + b$ where $a$ is the ten's place and $b$ is the one's place. The sum of digits of this number is then $a + b$ . Subtracting this from the number itself gives:

$10a + b - a - b = 9a$

So the answer is $\fbox{9}$ . (Though 3 is also a possible answer to this question).

In fact, any number at all with any amount of digits satisfies this relation. For example, for a four-digit number $1000a + 100b + 10c + d$ , and a sum of digits $a + b + c + d$ , $1000a + 100b + 10c + d - a - b - c - d = 999a + 99b + 9c$ , which is a multiple of 9 as well.