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P&C - Count of numbers divisible by 3

 
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Saurabh123



Joined: 28 Jun 2006
Posts: 4
Location: Fbd

PostPosted: Fri Jul 14, 2006 10:16 am    Post subject: P&C - Count of numbers divisible by 3 Reply with quote

Hi,

how to approach for solving such problems:
- How many 4 digit numbers are there which r divisible by 3

when we know that no. divisible by 3 has sum of all its digits divisible by3?

thnks
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apgo2000



Joined: 13 Jul 2006
Posts: 15

PostPosted: Fri Jul 14, 2006 7:58 pm    Post subject: Re: P&C - Count of numbers divisible by 3 Reply with quote

[quote="Saurabh123"]Hi,

how to approach for solving such problems:
- How many 4 digit numbers are there which r divisible by 3

when we know that no. divisible by 3 has sum of all its digits divisible by3?

thnks[/quote]


The way i do such questions is that the first four digit number divisible by 3 is 1002 (999 + 3; we all know 999 is div by 3, or else u cud divide 1000/3 and multiply 3 by the next highest integer than the divisor. for eg, u get 333.something so multiply 334 with 3 and u get 1002).
the largest 4 digit number is 9999 which is very apparent.

so u have an a.p. where 1st term =a= 1002
last term=l =9999
common difference = d= 3

therefore l=a+(n-1)d
9999=1002+(n-1)*3
u get 3000

this is a very concise process requiring u to write only the equation. rest is mental cal and wont take much time.

do post a better method if anyone can.

regards,
anshu
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CZ@@@H@@@R
Serious about CAT


Joined: 14 Jan 2006
Posts: 64

PostPosted: Sun Jul 16, 2006 6:51 pm    Post subject: P&C - Count of numbers divisible by 3 Reply with quote

Quote:
Quote:
Hi,

how to approach for solving such problems:
- How many 4 digit numbers are there which r divisible by 3

when we know that no. divisible by 3 has sum of all its digits divisible by3?

thnks


The way i do such questions is that the first four digit number divisible by 3 is 1002 (999 + 3; we all know 999 is div by 3, or else u cud divide 1000/3 and multiply 3 by the next highest integer than the divisor. for eg, u get 333.something so multiply 334 with 3 and u get 1002).
the largest 4 digit number is 9999 which is very apparent.

so u have an a.p. where 1st term =a= 1002
last term=l =9999
common difference = d= 3

therefore l=a+(n-1)d
9999=1002+(n-1)*3
u get 3000

this is a very concise process requiring u to write only the equation. rest is mental cal and wont take much time.

do post a better method if anyone can.

regards,
anshu


Required Number is (9999 - 999)/3 = 3000
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