Question

A $3-digit$ number has, from left to right, the digits $a,b$ and $c$ with $a>c$. When the number with the digits reversed is subtracted from the original number, the unit's digits in the difference is $4$. The next two digits, from right to left, are

• A. $5$ and $9$
• B. $9$ and $5$
• C. $5$ and $4$
• D. $4$ and $5$

Answer 1

Answer: (A)

$5$ and $9$

We know that when the reverse of a three digit number is subtracted from the same number, the difference is divisible by 9 and 11. ie. abc= 100a 10b c and cba= 100c 10b a are a reverse of eah other. Then abc- cba= 100a 10b c-100c-10b-a= 99(a-c).in this question it is mentioned that the resultant of abc-cba= any number whose unit digit is 4. Therfore, 99(a-c)= xy4. as the resultant is divisible by 9 and 11. so xy4 is such a number which is divisible by 9 and 11. now plug in the options and you will find the value of x and y as 5 and 9 that satisfy the condition of divisibility by both 9 and 11.