2006-11-09 12:49:32 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
no of even numbers=2006/2=1003
no fo multiplesof 5
t1=5 and tn=2005
2005=5+(n-1)*5
2000=(n-1)*5
n-1=400
n=401
so the no of numbers divisible by 2 or 5
as kindly pointed out by math_kp,god bless him,from this i must subtract the no of 10s as 10 is divisible by both 5 and 2
no of multiples of 10
t1=10 and tn=2000
2000=10+(n-1)*10
1990=10n-10
2000=10n
n=200
so the answer is 1404-200=1204
thank you math_kp once again
2006-11-09 12:58:53 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Well, every set from 1-10 has 7 (2, 4, 6, 8, 5, 10)
There are 200 of these sets, plus the additional 2, 4, 5 and 6 in that final six.
So yeah, my best guess is 200*7 + 4 or 1404
2006-11-09 20:54:12 · answer #2 · answered by UMRmathmajor 3 · 0⤊ 0⤋
1003 divisible by 2
401 divisible by 5 of which 200 are divisible by 2, also
so 1204 are divisible by 2 or 5
2006-11-09 21:16:47 · answer #3 · answered by bob h 3 · 0⤊ 0⤋
The ones that are even...and end in 5
2006-11-09 20:57:35 · answer #4 · answered by Rob 2 · 0⤊ 0⤋
All of them, but if are talking about evenly then
1003 are divisible by2 and 401 are divisible by 5
2006-11-09 21:01:03 · answer #5 · answered by kidd 4 · 0⤊ 0⤋
the even ones and the ones that end in 5
2006-11-09 20:51:27 · answer #6 · answered by davidosterberg1 6 · 0⤊ 0⤋
Think about it, and hey, here's a concept...do the math.
2006-11-09 20:52:27 · answer #7 · answered by T Time 6 · 0⤊ 0⤋