The digits of a three-digit number form an A. | vedclass

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Question

(A) Let the digits at the hundreds place,tens place,and units place of the original number be $a-d$,$a$,and $a+d$ respectively.$	herefore$ The origin

Vedclass · Accepted Answer

$258$

Vedclass · Answer

(A) Let the digits at the hundreds place,tens place,and units place of the original number be $a-d$,$a$,and $a+d$ respectively.$	herefore$ The original number $= 100(a-d) + 10(a) + (a+d) = 111a - 99d$.Given that the sum of the digits is $15$:$(a-d) + a + (a+d) = 15$$3a = 15$$a = 5$.When the digits are reversed,the new number is:$100(a+d) + 10(a) + (a-d) = 111a + 99d$.Given that the new number exceeds the original number by $594$:$(111a + 99d) - (111a - 99d) = 594$$198d = 594$$d = 3$.Therefore,the digits are:Hundreds place: $a-d = 5-3 = 2$Tens place: $a = 5$Units place: $a+d = 5+3 = 8$The original number is $258$.

The digits of a three-digit number form an $A.P.$ and their sum is $15$. The number obtained by reversing the order of the digits exceeds the original number by $594$. Find the original number.

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