Number Problem 18

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Problem: a, b and c are three digits from left to right of a three digit number.
If the digits are reversed and the new number subtracted from the original one,
we get 4 in the units place. What will be other two digits from left to right
in the resulting number?

Solution: If a, b, c is a three digit number with digits a, b, c from left to right

Hence, its value = 100 X a + 10 X b + c

= 100a + 10b + c

When the digits are reversed, the value will be 100c +10b + a

The remainder got when the second number is subtracted from the first

= 99a – 99c

= 99(a - c)

∴ The remainder is a multiple of 99.

We get 4 in the units place

Hence,, the remainder is 99 X 6 = 594

∴ The required numbers are 5 and 9.

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