venmic wrote:Of the three-digit integers greater than 600, how many have
two digits that are equal to each other and the remaining digit
di¤erent from the other two?
(A) 120
(B) 116
(C) 108
(D) 107
(E) 72
Is there a faster way to this - algebrically instead of actually forming sequences
any suggestions
D
Good = total - bad.
Total:
Given consecutive integers, the total number = (biggest-smallest) + 1:
999-601+1 = 399.
Bad case 1: all 3 digits are different
Number of options for the hundreds digit = 4. (6,7,8,or 9)
Number of options for the tens digit = 9. (Any digit 0-9 other than the digit already used.)
Number of options for the units digit = 8. (Any digit 0-9 other than the 2 digits already used.)
To combine these options, we multiply:
4*9*8 = 288.
Bad case 2: all 3 digits are the same
Number of options = 4. (666,777,888,999).
Good integers = 399-288-4 = 107.
The correct answer is
D.
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