Bunuel
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 90
(B) 82
(C) 80
(D) 45
(E) 36
We can solve this problem by analyzing the digits (0 to 9) that are being “doubled”.
If 0 is doubled, then the numbers can only be 800 and 900. So we have 2 numbers.
If 1 is doubled, then the numbers can only be 711, 811 and 911. So we have 3 numbers.
If 2, 3, 4, 5, or 6 is doubled, then we should have 3 numbers for each case since it’s analogous to 1 being doubled.
If 7 is doubled, then the numbers can only be 770, 771, 772, 773, 774, 775, 776, 778, 779; 707, 717, 727, 737, 747, 757, 767, 787, 797; 877, and 977. So we have a total of 20 numbers.
If 8 or 9 is doubled, then we should have 20 numbers for each case since it’s analogous to 7 being doubled.
Thus, altogether we have 2 + 6 x 3 + 3 x 20 = 2 + 18 + 60 = 80 such numbers.
Answer: C