Question

The digts of a positive no. 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 reimber.
​

Answer 1

Answer:

Original number - 852

Step-by-step explanation:

Let the digits of the number be a-d, a, a+d

The sum of the digits are - 15

a-d+a+a+d = 15

3a = 15 => a = 5

Original number - 100(a+d) + 10a + a-d

New number obtained by reversing the digits - 100(a-d)+ 10 a + a + d

100(a+d) + 10a + a-d - 594 = 100(a-d)+ 10 a + a + d

100a + 100 d + 10 a + a-d - 594 = 100 a - 100 d +10 a+ a+d

200d-2d = 594

198d = 594

d = 3

The digits are - 5-3, 5, 5+3 = 2, 5, 8

original number = 100(5+3) + 10(5) + (5-3) = 800 + 50 + 2 = 852.

New number obtained by reversing the digits - 852 - 594 = 258

Answer 2

Kyu....................