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November 27, 2014

November 27, 2014

Posted by **alex** on Saturday, May 28, 2011 at 1:38pm.

a-b+c

(I came up with 252 integers so when I did the computation it didn't make sense)

- math -
**MathMate**, Saturday, May 28, 2011 at 11:00pmHere we have 10 digits, of which the first one cannot be zero.

If the three digits are distinct, there are 9 choices for the first digit, still 9 (including zero) for the second, and 8 for the third.

So 9*9*8=648 numbers.

There are therefore 900-648=252 numbers which have at least one repeated digit.

So a=2, b=5, c=2, and a-b+c=2-5+2=-1

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