The ranges 700-799, 800-899, 900-999 all have the same number of integers that match this condition. However, the question asks for integers greater than 700. Since 700 itself is not included in the range, the number of digits will be a multiple of 3 minus 1. 80 is the only answer that satisfies this.
Assuming the above was unknown, I'll write out the results for the range 700-799, multiple the number by 3, then subtract by 1 (700 cannot be included).
1st and 3rd digits
707
717
727
737
747
757
767
787
797
1st and 2nd digits
770
771
772
773
774
775
776
778
779
2nd and 3rd digits
700
711
722
733
744
755
766
788
799
Total: 27
27*3-1 = 80
Another way to do this is a less tedious method using combinatorics to examine the range 700-799.
The notation I am using is C(n,r) = n C r = n! / (r!*(n-r)!)
The 1st digit must be 7.
If the 2nd digit is 7, the 3rd digit must not be 7: C(1,1)*C(9,1) = 9, where C(1,1) is "the 2nd digit has one option to choose from (7)", and C(9,1) are the "third digit can choose from 9 digits (any but 7)".
If the 2nd digit is not 7, the 3rd digit must be 7 or whatever the 2nd digit is: C(9,1)*C(2,1) = 18, where C(2,1) is for the "3rd digit can choose from 2 digits: 7, or the 2nd digit value".
9+18 = 27 integers for this range, multiple by 3 for 700-999, subtract by 1 for 701-999. 80 again is the answer.
