Little tricky one.. Here is my approach..
Easy part --> 700 to 899. # of integers containing 7 or 8 = 899-700+1=200
Now # of integers from 600 to 699.
Let integers be XYZ format. X always has to be 6. No. of ways X can be filled =>1
Number of integers where 7 or 8 is in unit's digit:
Z can be any of the two digits 7 or 8. So Z can be filled in 2 ways =>2
Y can vary from 0 to 9( since Z contains only 7 or 8) =>10
number of ways(where 7 or 8 is in units digit) = 1*10*2=20.
Now determine number of integers where 7 or 8 is in ten's digit.
X can be filled in 1 way
Y can be filled in 2 ways
Z can be filled in 0 to 9 ways. But we have already accounted for 7 and 8 in the above calculation. So we need to vary Y from 0-6 and 9 only. Y can be filled in 8 ways=>8
number of ways ( where or 8 is in ten's digit) = 1*2*8=16
Adding combinations = 200+20+16 = 236.
The tricky part was to determine that there are only 16 ways to fill ten's digit with 7 or 8 after counting for unit's digit. I was easily tricked into 240 because of double counting.
May be there is even an easier way to solve this
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