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Genet multiplied a 3-digit number by 1002 and got AB007C, where A, B, and C stand for digits. What was Genet's original 3-digit number?

Respuesta :

sqdancefan

Answer:

  539

Step-by-step explanation:

The 007C requires the least significant two digits be in the range 35-39. In order for the 10-thousands digit to be zero, the sum of 1000 times the least digit of Genet's number and 2 times the hundreds digit must result in a sum with no thousands. About the only way to do that is to make the least digit 9 and the hundreds digit 5.

Then you have ...

  539 × 1002 = 540078 . . . . . ABC =548

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