Question

the middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13. if the digits are reversed, the number so formed exceeds the original number by 495. find the number. ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Let x be the digit in units' place and y be the digit in hundreds' place.

Then, since the digit in tens' place is 0, the number will be represented by 100y + x.

If the digits are reversed, the number so formed will be represented by 100x +y. Therefore,

100x+y−(100y+x)=495

or 100x+y−100y−x=495

or 99x−99y=495

or x−y=5 ............. (1)

Again, since the sum of the other digits is 11, and the middle one is 0, we have

x+y=11 ......... (2)

From (1) and (2), we get x=8,y=3

Hence, the number is 308.