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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 number is

A
452
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B
852
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C
1252
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D
1652
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Solution

The correct option is B 852Let 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

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