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# total number of 4 digits

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total number of 4 digits [#permalink]

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20 Jun 2011, 07:43
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what is the total 4 digit numbers that can be formed using the digits 0 to 5 without repetition , such that number is divisible by 9?
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Rahul

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Re: total number of 4 digits [#permalink]

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20 Jun 2011, 07:56
ITMRAHUL wrote:
what is the total 4 digit numbers that can be formed using the digits 0 to 5 without repetition , such that number is divisible by 9?

I think its 36.
for divisibility by 9 .. sum of all the digits must be divisible by 9.
first set of numbers can 0,2,3,4
3*3*2*1 = 18 ways

second set of numbers
0,1,3,5
3*3*2*1= 18 ways

total 36.
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Re: total number of 4 digits [#permalink]

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20 Jun 2011, 08:16
sudhir18n wrote:
ITMRAHUL wrote:
what is the total 4 digit numbers that can be formed using the digits 0 to 5 without repetition , such that number is divisible by 9?

I think its 36.
for divisibility by 9 .. sum of all the digits must be divisible by 9.
first set of numbers can 0,2,3,4
3*3*2*1 = 18 ways

second set of numbers
0,1,3,5
3*3*2*1= 18 ways

total 36.

I am getting 124. Not too sure though.
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Re: total number of 4 digits [#permalink]

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20 Jun 2011, 08:24
Second thought- 100. What's the OA, ITMRAHUL.
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Re: total number of 4 digits [#permalink]

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20 Jun 2011, 09:55
we have two patterns:
$$3, 4, 2, 0$$
$$5, 3, 1, 0$$
$$0$$ can't be on the first place.

For the first place we choose among 3 digits, for the second among three digits (now we can use zero) as well, for the third between 2 digits, for the fourth - 1.
$$3*3*2*1=18 18*2=36$$
Re: total number of 4 digits   [#permalink] 20 Jun 2011, 09:55
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# total number of 4 digits

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