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Bunuel
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Took a similar approach as the solution during the test. Except I took out the possibilities of 0 at the start at the beginning itself.

Summation of all the digits = 17.
Each digit at the end appears:
3*3*2 = 18 times.
Thus 18*(17) = Last digit = 6.

Answer: Option D
Bunuel
What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288
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