Numbers divisible by 3

Attempt to a solution?

 

I don't get 136. How did you come up with that answer?

 

I don't get 136 either. How did you get that?

 

There are 'roughly' 20 numbers in every 100 that are divisible by 5. There are 'roughly' 900 three digit numbers. So there are 'roughly' 180 three digit numbers divisible by 5. That's enough off from 136 to make me agree with Defennder and Tedjin. You must be wrong.

 

5, 10, 15, 20, 25, 30

Its all the numbers that end in 5 or zero, and then ignore the last one (1000) because it had the temerity to posses 4 digits.

As you start building the sequence, you see the rule easy enough.

 

a_n=nth term of an AP (Arithmetic progression)
a=1st term
d=common difference

Since the 1st 3 digit no. divisible by 5 is 100
a=100
the last 3 digit no. divisible by 5 is 995
a_n=995
common difference (d)=5
n=no. of 3 digit nos. divisible by 5

a_n=a+(n-1)d
995=100+(n-1)5
995-100=(n-1)5
895=(n-1)5
895/5=n-1
179=n-1
179+1=n
n=180

Therefore,
exactly (not roughly) 180 3 digit numbers are divisibe by 5

Absolutely correct Dick!!!

@Amith2006
How did you get 136?!

 

Aha that made me laugh, insane bump