Question

ABC is a three-digit number that leaves a remainder 4 when divided by 20. If ABC is divisible by 9, then find the least possible value of A.

• A. 3
• B. 4
• C. 2
• D. 1

Answer 1

The correct option is C 2

ABC is a three-digit number that leaves a remainder 4 when divided by 20.
$⟹(ABC+4)\text{is a multiple of 20.}$

$\text{Let (ABC + 4) be k.}$

$\text{For, k to be a multiple of 20, the following}\text{conditions have to be met:}$

$\text{(1) The digit in the ones place of k is 0.}$
$\text{(2) The digit in the tens place of k is even.}$

$\text{On calculating (ABC + 4) and meeting the given}\text{conditions, we get:}$


For condition 1:
$C+4=10$
$⟹C=6$

For condition 2:
$(B+1)isanevennumber.$
$⟹Bisanoddnumber.⟹A+B+C=9\left(forthelowervalueofA\right)$

$\text{Now, ABC is divisible by 9.}$
$⟹A+B+Cisamultipleof9.$
$⟹A+B+C=9\left(forthelowestA\right)$
$⟹A+B+6=9$
$⟹A+B=3$

$\text{So, A = 1; B = 2}or\text{A = 2; B = 1}$

$\text{However, B is an odd number.}$

$\text{Hence, A = 2}$