How Many 4 digits numbers contain the number pattern 75 at least once?

No repeats..4 digit numbers only and the pattern has to contain the number 75 and nothing elseHow Many 4 digits numbers contain the number pattern 75 at least once?
Let's see...





There are 9000 four-digit numbers.





Each hundred, there is one number where the last two digits will be 75. There are 90 groups of 100.





Each thousand, there is a group of numbers where the middle two digits will be 75 ten times in a row. There are 9 groups of 1000.





Finally, there are 100 numbers between 7500 and 7599 which all start with 75.





We have 90 + 90 + 100. But we must eliminate duplicates. I think 7575 is the only number that falls into two categories.





The total is therefore 90 + 90 + 100 - 1 = 279.How Many 4 digits numbers contain the number pattern 75 at least once?
7500, 7501, 7502...


1750, 1751, 1752...


1075, 1175, 1275...





100 + 90 + 90 = 280?


is that right?


no wait.. minus the repeats...


1775, 7575


so 278?
Hmm... everybody is taking a different meaning from your ';no repeats'; condition. Here's mine, if the other 2 digits must not be 7 or 5, nor the same as each other.





75xy - 8 digits for x, then 7 digits for y, total 56.


x75y - 7 digits for x (not 0), then 7 digits for y, total 49.


xy75 - same again, another 49.





Total 56 + 49 + 49 = 154.





But ';75 at least once'; should mean that it can be there twice - which can only be in 7575 - but ';no repeats'; means it can't, so your conditions contradict each other, and there will be different answers depending on how people resolve the contradiction.
There are three possibilities:





75_ _


_75_


_ _ 75





In each case, any blank can have 8 possible numbers and the second blank seven possible numbers (since you said no repeats).





Thus, the total is 8*7 + 8*7 + 8*7 = 168 numbers
299.


75xx = 100 of them


x75x = another hundred


xx75 = another hundred


subtract 1 because 7575 got counted twice.