Fast way to do this?

range of number is 4-49 ! when multiplied by 3 gets a remainder of 1
tn=a+(n-1)d
49=4+(n-1)3
45/3=n-1
n=16

Vishu

I wonder how 17????

4,7,10,13, 16,19,22,25, 28,31,34,37 ,40,43,46,49 again 16 digits !!

Vishu

vishubn wrote:4,7,10,13, 16,19,22,25, 28,31,34,37 ,40,43,46,49 again 16 digits !!

Vishu

hey vishnu .. u missed 1.

50/3 = 16 , hence we have 16 integers which are divisible by 3

for each of the integer (n) divisible by 3 there would be the next integer (n+1) which gives a reminder 1.

hence we have 16 integers starting from 4 upto 49

and 1 divided by 3 gives reminder 1 (and inclusive in our range)

so the total number of digits would be 16+1 = 17.

fastest method i could come up with was this,

1,4,7,10....49

(49-1)/3 +1= 16+1 =17.since inclusive.

hey vishnu .. u missed 1.

OOPS
Thanks sudhir

Vishu

vikz_316 wrote:fastest method i could come up with was this,

1,4,7,10....49

(49-1)/3 +1= 16+1 =17.since inclusive.

Fastest method I could come up was this:
50/3 = 16 + remainder 2.
Every third number is divisible by 3. So every 4th number
will have a remainder of 1.
This division inherently assumes numbers are >= 3.
So to get the remainder for values < 3, add 1 coz there
is only 1 number < 3 which has remainder of 1.

hope this helps

16 multiples of 3 between 0 and 50. 16*3 = 48. Add 1 to each of these .. you have 16 integers that give a remainder 1.. and the digit 1 .. so 16+1 = 17.


- pradeep

Honestly, I think the best/easiest way to do this is simply to count them. It took me about 30 seconds, and I was 100% sure of my answer. The tricks and shortcuts might work fine, but counting them is foolproof. There's many problems like this, where if you simply dive in and start counting, you'll get the answer as fast as if you had tried to use a trick, and you'll be much more confident in the answer.

The only problem might be if you forget to include 1.