Question 254429
every 10 numbers, you have a last digit of 6.


300 - 100 = 200 numbers in between.


divide that by 10 and you get 20.


you will have 20 numbers that have a last digit of 6.


those number will be:

106
116
126
136
146
156
166
176
186
196
206
216
226
236
246
256
266
276
286
296


every 100 numbers, you will have 10 numbers that have a middle digit of 6.


those will be:


160 to 169
260 to 269


that makes a total of 20.


only half of them will be even.


those numbers will be:


160
162
164
166
168
260
262
264
266
268


this makes a total of 10 even number with a middle digit of 6.


within 2 of those numbers, however, you will have a number that has a middle digit of 6 and a last digit of 6.


those number will be:


166
266


we have to subtract these so they won't be counted twice.


the total number of numbers that have at least 1 digit of 6 would be 20 + 10 - 2 = 28


the total of even numbers between 100 and 300 = (300-100)/2 + 1 = 101


the total of even numbers between 100 and 300 that do not contain a digit of 6 either in the last digit or the middle digit would then be equal to:


101 - 28 = 73


that would be selection B.


here are the total even numbers that contain a digit of 6.


106
116
126
136
146
156
160
162
164
166
168
176
186
196
206
216
226
236
246
256
260
262
264
266
268
276
286
296


that's a total of 28


the total number of even digits is (300-100)/2 + 1 = 200/2 + 1 = 101


that 1 has to be added because the difference doesn't take it into account, and it is even.


example:


(30-20)/2 + 1 = 10/2 + 1 = 6


even numbers are:


20
22
24
26
28
30