How many four digit numbers contain the digit pattern 85 at least once and only once?

Answer 1

298 197 Assuming numbers like 0XYZ are three digit numbers, the answer is 278. =SeldomSeeN= Let's work it out.

0085, 0185... 9985 is 100 numbers, 0850, 0851... 9859 is another 100 numbers, 8500, 8501... 8599 is another 100 numbers. Now two of these sets of numbers has another 85 in it 8585 so you have to take those 2 out. So I make it 298, too.

But... if you don't allow leading zero's the lowest number in the first set is 1085 so we lose 10 numbers plus the repeated 85 making 89 in that set. The lowest number in the second set is 1850 so we only lose on number in that set, and there are no repeated 85s so we have 99 numbers there and in the third set we also only lose the repeated 85 so there are 99 in that set, making 89+99+99 = 287 Answer:

No leading zeros for four-digit numbers 85xx = 10 x 10 - 1 = 99

x85x = 9 x 10 = 90

xx85 = 9 x 10 - 1 = 89

Total = 278