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The digits of a three digit positive integer are in .A.P. and their sum is 15. The number obatined by reversing the figits is 594 less than original number. Find the number
Asked by rushabhjain.av | 20 Nov, 2018, 11:01: AM
Let unit place digit = a, tenth place digit = a+d and hundredth place digit = a+2d ;

Sum of digits = a + (a+d ) + ( a +2d ) = 3a +3d = 15 ;  hence a+d = 15  ......................(1)

Original number = (a+2d)×100 + (a+d)×10 + a = 111a +210 d ................(2)

number reversed by digits = a×100 + (a+d)×10 + a+2d  = 111a + 12d ..............(3)

it is given that original number is greater than the number formed by reversing digits by 594

hence 111a + 210d = 111a + 12d + 594
198×d = 594  or d =3
by substituting d=3 in eqn.(1), we get a = 2

hence the number is 8×100 + 5×10 + 2×1 = 852
Answered by Thiyagarajan K | 22 Nov, 2018, 02:12: AM

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