30) the digits of a positive integer,having three digits are in A P and their sum is 15 the - Maths - Arithmetic Progressions

Question

30) the digits of a positive integer,having three digits are in
A P and their sum is 15 the number obtained by reversing the digits
is 594 less than the original number find the number

Aakash Sharma · Accepted Answer

Let digits of the number be (a â€“ d), a and (a + d) respectively âˆ´ The required number is 100 (a â€“ d) + 10a + (a + d) Given : The sum of the digits = 15 $â‡’ (a âˆ’ d) + a + (a + d) = 15 â‡’ 3 a â¢ = â€‰ 15 â‡’ a = \frac{15}{3} = 5$ Now, the number on reversing the digits is 100 (a + d) + 10a + (a â€“ d) According to the question 100(a â€“ d) + 10a + a + d = 100 (a + d) + 10a + (a â€“ d) + 594 $â‡’ 100 a âˆ’ 100 d + 10 a + a + d = 100 a + 100 d + 10 a + a âˆ’ d + 594 â‡’ 111 a âˆ’ 99 d = 111 a + 99 d + 594 â‡’ 99 d + 99 d = âˆ’ 594 â‡’ 198 d = âˆ’ 594 â‡’ d = âˆ’ \frac{594}{198} = âˆ’ 3$ âˆ´ The digits of the number are (5 â€“ (â€“3)), 5, (5 + (â€“3) = 8, 5, 2 and the required number is 8 Ã— 100 + 5 Ã— 10 + 2 = 852

30) the digits of a positive integer,having three digits are in A.P. and their sum is 15.the number obtained by reversing the digits is 594 less than the original number.find the number.