the digit of a positive number of three digit are in an ap and there is 15 the number obtained by reversing the digit is 594 less than the original number find the number
Answers
In your question statement is missing that Sum of Digits is 15, considering that, solution is given below
(Let three digits be a - d, a and a + d)
Let the 100 place digit be ( a - d),
10 place digit = a and
unit place digit be ( a + d)
Sum of digits is given 15
∴ a - d + a + a + d = 15
3a = 15
a = 5
We know that three digit number is of the form
100 ( 100 place digit) + 10 ( Ten place Digit) + unit place digit
∴ Original number we get by putting values of digits
100 ( a - d ) + 10a + a + d
= 100a - 100d + 10a + a + d
= 111 - 99d ------------ ( i ) ( Original Number)
Also, New Number obtained by reversing the digits ( 100 place digit at unit place and unit place digit at 100 place), we get
100 ( a + d) + 10a + a - d
= 100a + 100d + 10a + a - d
= 111a + 99d ----------- ( ii ) (New Number)
Given that
New Number is 594 less than original number
∴ 111a + 99d = 111a - 99d - 594
198d = -594
d = -3
∴ a = 5 and d = -3, putting values in original number ( i )
111a - 99d
= 111(5) - 99 (-3)
= 555 + 297
= 852
∴ Original number is 852