A three-digit natural number $X$ is such that the middle digit is equal to the sum of the other two digits. Let $Y$ be the number obtained by reversing the digits of $X .$
Which of the following CAN NOT be the value of $X + Y ?$

968

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526

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605

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1452

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Solution

The correct option is B $526$
Let $X = 100 a_{2} + 10 (a_{2} + a_{0}) + a_{0}$

Then $Y = 100 a_{0} + 10 (a_{2} + a_{0}) + a_{2}$

$X + Y = 100 (a_{2} + a_{0}) + 20 (a_{2} + a_{0}) + (a_{2} + a_{0})$
$\Rightarrow X + Y = 121 (a_{2} + a_{0})$
i.e., $X + Y$ is a multiple of $121.$

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A three-digit natural number X is such that the middle digit is equal to the sum of the other two digits. Let Y be the number obtained by reversing the digits of X. Which of the following CAN NOT be the value of X+Y ?

The correct option is B 526 Let X=100a2+10(a2+a0)+a0 Then Y=100a0+10(a2+a0)+a2 X+Y=100(a2+a0)+20(a2+a0)+(a2+a0) ⇒X+Y=121(a2+a0) i.e., X+Y is a multiple of 121.

A three-digit natural number $X$ is such that the middle digit is equal to the sum of the other two digits. Let $Y$ be the number obtained by reversing the digits of $X .$
Which of the following CAN NOT be the value of $X + Y ?$

The correct option is B $526$
Let $X = 100 a_{2} + 10 (a_{2} + a_{0}) + a_{0}$

Then $Y = 100 a_{0} + 10 (a_{2} + a_{0}) + a_{2}$

$X + Y = 100 (a_{2} + a_{0}) + 20 (a_{2} + a_{0}) + (a_{2} + a_{0})$
$\Rightarrow X + Y = 121 (a_{2} + a_{0})$
i.e., $X + Y$ is a multiple of $121.$