Question

The digit of a three digit positive number are in AP and the sum of the digit is 15 on subtracting 594 from number the digits are reversed find number

Answer 1

Hi there!
Here's the answer:

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Let No.s be (a-d) , a, (a+d)
°•° Given that the three digits are in A.P (have common difference 'd' b/w them)


3-digit No. = (100×face value of hundredths place) + (10 × face value of tens place) + (1 × face value of units place)


•°• 3- digit No. = [100×(a+d)] + [10×a] + [1×(a-d)]
= 100a + 100d + 10a + a - d
= 111a + 99d


If the Digits are reversed, New No. is Obtained

New No. = [100×(a-d)] + [10×a] + [1×(a+d)]
= 100a - 100d + 10a + a + d
= 111a - 99d

Given,
Sum of the digits = 15
a + d + a + a - d = 15
3a = 15
=> a = 5
& d = 3


Hence the No. is 111(5) + 99(3) = 852


•°• Required No. is 852


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