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Q: I am a three digit number. The sum of my digits is 18. When I reverse my digits and subtract the new number from the original number you get 99. What number I am?

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38

the answer is 25 and then if you reverse that it is 52

Multiply the last digit by 7. Subtract that number from the remaining digits. If that number is divisible by 23, then the original number is divisible by 23.

Reverse the digits then check of the new number is the same as the original number.

Assuming you want to reverse the digits of the number: numToReverse <- the number we want to reverse revNum <- new number as numToReverse is reversed while numToReverse is not 0 // shift digits left revNum = revNum * 10 // tack on rightmost digit of numToReverse revNum = revNum + (numToReverse modulus 10) // shift digits right numToReverse = numToReverse / 10 // numToReverse is now 0 // revNum is now the reverse of (the original value of) numToReverse

Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.Any four-digit number will do. Add it to the number 33333, to get the original number you have to subtract from.

2178 x 4 = 8712

To total 17 the two digits must be 8 and 9! The original number was 98.

83 A+

19

45

The original number is 58.

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