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  • The digit of a positive of three digits are in AP and their sum is 15  The number obtained by reversing the digits is 594 less than the original number. Find the number.  

The digit of a positive of three digits are in AP and their sum is 15 

The number obtained by reversing the digits is 594 less than the original number. Find the number.

 

 

 

 
 
 
 
 

Answers (1)

Let the no. in abc (a= hundredth place , b = tenth place , c = unit place )

GIven a,b,c are in AP 

b = a+d \\\\ c = a +2d \\\\

the sum of digits = a+ a+d +a +2d = 15 \\\\ 3a+ 3d = 15 \\\\ a +d = 5 ---- ( 1) \\\\ original \: \: no. (abc) = a \times 100 + ( a+d) \times 10 + ( a + 2d ) \\\\ = 100a +10a + 10d + a + 2d \\\\ = 111a + 12 d ---(2)

NUmber reversed (cba ) 

= ( a +2 d )\times 100 + ( a+d) \times 10 + a \\\\ = 100a + 200d + 10 aa + 10 d + a \\\\ = 111 aa + 210 d ---( 3)

According to question 

abc - cba = 594 \\\\ from \: \: ( 2) \: \:and \: \: ( 3) \\\\ ( 111a + 12d ) - ( 111 a + 210 d ) = 594 \\\\ 111a + 12 d - 111a -210 d = 594\\\\ -198 d = 594 \\\\ d = \frac{-594 }{198} \\\\d = -3

putting the value of d in eq (1) 

a+d = 5 \\\\ a-3 = 5 \\\\ a = 8 \\\\ original \: \: no. = 852

 

Posted by

Ravindra Pindel

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