2006-09-26 11:35:01 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
There are 18 numbers containing one five each 5 from 1 to 100:
5, 15, 25, . . . 95 and 50, 51, 52 . . .59 (excluding 55). But 55 contains two 5's. So there are 20 5 digits.
There are 20 more from 101 to 200.
There are 6 numbers containing one 5 each from 201 to 250.
So in total we have: 20 + 20 + 6 = 46 5's
2006-09-26 11:45:56 · answer #1 · answered by Dimos F 4 · 0⤊ 0⤋
In printing the numbers 0 through 9 the number 5 occurs once. Therefore it occurs ten times in the ones place when going from zero to 100.
It occurs once in the tens place, but repeats for ten digits.
So far the total is ten plus ten, or twenty for the numbers 0 through 100.
It occurs with the same frequency for 101 through 100. Now the total is 40.
It occurs only 5 times in the ones place between 200 and 250. It occurs only once in the tens place in that region.
0 to 100 = 20 times
101 to 200 = 20 times
200 to 150 = 6 times
sum = 20 + 20 + 6 = 46 times.
2006-09-26 18:42:40 · answer #2 · answered by Curly 6 · 2⤊ 0⤋
5,15,25,35,45,55,65,75,85,95,105,115,125,135,145,150,155,165,175,185,195,205,215,225,235,245,250. ((you wanted up to 250,right?))
That's 27 numbers.
2006-09-26 18:58:23 · answer #3 · answered by Anonymous · 0⤊ 0⤋
do your own homework!
2006-09-26 18:42:41 · answer #4 · answered by Larry T 5 · 0⤊ 0⤋