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Question

The digits of a three-digit positive integer are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Then the unit place of the number is

A
4
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B
2
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C
7
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D
5
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Solution

The correct option is B 2Let the digits at ones, tens and hundreds place be (a−d),a and (a+d) respectively. Then the number =(a+d)×100+(a×10)+(a−d)×1=111a+99d The number obtained by reversing the digits =(a−d)×100+(a×10)+((a+d)×1)=111a−99d It is given that (a−d)+a+(a+d)=15 and 111a−99d=111a+99d−594 ⇒3a=15⇒a=5 and 198d=594⇒d=3 Hence, the required number is 111×5+99×3=852, so the unit place number is 2.

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