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Saurabh123
Joined: 28 Jun 2006 Posts: 4 Location: Fbd
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Posted: Fri Jul 14, 2006 10:16 am Post subject: P&C - Count of numbers divisible by 3 |
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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
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Posted: Fri Jul 14, 2006 7:58 pm Post subject: Re: P&C - Count of numbers divisible by 3 |
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[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
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Posted: Sun Jul 16, 2006 6:51 pm Post subject: P&C - Count of numbers divisible by 3 |
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| 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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