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CAT 2005 | Question: 13

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The digits of a three digit number $\text{A}$ are written in the reverse order to form another three digit number $\text{B}.$ If $\text{B}$ is greater than $\text{A}$ and $\text{B-A}$ is perfectly divisible by $7,$ then which of the following is necessarily true?

  1. $100 <\text{A}< 299$
  2. $106 <\text{A}< 305$
  3. $112 <\text{A}< 311$
  4. $118 <\text{A}< 317$
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Let A = 100x + 10y + z

B = 100z + 10y + x

B – A = 99(z – x) As (B – A) is divisible by 7 and 99 is not, (z – x) is divisible by 7

z and x can have values (8, 1) or (9, 2) y can have any value from 0 to 9.

A = 1y8 or 2y9

Lowest possible value of A is 108 and the highest possible value of A is 299.

Hence, option (2)106 < A < 305 is the Answer.

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