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how many 4 digit number.....

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by sudhir3127 » Mon Jul 21, 2008 2:07 am
how many 4 digit number that are divisble by 4 can be formed using digits 0 to 7 if no digit is to occur more than once in each number.


a. 570
b. 370
c. 345
d. 440

OA is B
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Re: how many 4 digit number.....

by parallel_chase » Mon Jul 21, 2008 3:31 am
sudhir3127 wrote:how many 4 digit number that are divisble by 4 can be formed using digits 0 to 7 if no digit is to occur more than once in each number.


a. 570
b. 370
c. 345
d. 440

OA is B
Firstly I must say its a really good question.

The property of any number to be divisible by 4 is its last 2 digits should be divisible by 4.

There are 3 such numbers, if 0 is placed in units place the number is divisible by 4
20
40
60

Therefore, 6*5*3 = 90 (6*5*3 because we are left with only 6 digits since 0 and 2 or 4 or 6 are already taken)

There is 1 such number where 0 can be at the tens place and the number is divisible by 4
04

Therefore, 6*5*1 = 30

Now the largest number that we can have from 0-7 is 77, therefore, from 0-77 there are 19 numbers which are divisible by 4

We have to subtract the above numbers because we have already taken them into account 19-4= 15

we also have to subtract 44 since digits cannot be repeated
we also have o subtract 08, 28, 48, 68 since we dont want any digit which is beyond 7

15-5 = 10

therefore, 5*5*10 = 250 (5*5*10 because we dont want 0 at the thousands place)

250+90+30 = 370


I think there is a better way of doing this question.
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by sudhir3127 » Mon Jul 21, 2008 4:27 am
Thanks a lot..but seems to be a very lengthy method of solving the problem ..

please let me know any shortcut method of solving this...
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by reachac » Mon Jul 21, 2008 7:21 am
Not a very short method but...nehw

The property of any number to be divisible by 4 is its last 2 digits should be divisible by 4. Also a primary condition the requires numbers can end only in an even units digit. In addition 0 cant be at the 10k place.

1) Numbers ending in 0
last 2 digit combinations could be 20,40,60.
So we have 6*5*3 such nos

2) Numbers ending in 2
last 2 digit combinations could be 12,32,52,72.
So we have 5*5*4 such nos

3) Numbers ending in 6
last 2 digit combinations could be 16,36,56,76.
So we have 5*5*4 such nos

2) Numbers ending in 4
last 2 digit combinations could be 04,24,64.
Now this case is lil special wen we have 0 in the tens place, we don have to wory bout 0 occuring in the 10k place
So Nos ending in 04 we have 6*5
Nos ending 24 or 64 we have 5*5*2

Adding all we get total nos = 6*5*3 + 2*5*5*4 + 6*5 + 5*5*2 = 370
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by sudhir3127 » Mon Jul 21, 2008 7:47 am
can u please explain how did u get 6*5*4 , 5*5*4 and all such combinations...
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by sudhir3127 » Mon Jul 21, 2008 10:12 pm
thanks a lot reachac and parallel chase...

parallel chase thanks a lot for ur post on permutation and combination... it was useful..

now in this question i have a query.

"Numbers ending in 0
last 2 digit combinations could be 20,40,60.
So we have 6*5*3 such nos "

i am getting its as 6*5*4 how did u get 6*5*3.... can u please xplain

similarly with Numbers ending in 2
last 2 digit combinations could be 12,32,52,72.
So we have 5*5*4 such nos


please reply soon.. thanks for all ur efforts..
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by pepeprepa » Tue Jul 22, 2008 12:24 am
"Numbers ending in 0
last 2 digit combinations could be 20,40,60.
So we have 6*5*3 such nos "

i am getting its as 6*5*4 how did u get 6*5*3.... can u please xplain

---------------------------------------

I think there are 6 possibilities for the first digit and 5 for the second digit.
And there are 3 "last 2 digit combinations" namely 20,40,60

You think 4 because you think there is another digit to place but there are only 4 digits in the number, the 2 last one we already have and the other two with number of possibilities (6*5)

Parallel chase correct me if i'm wrong[/u]
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by parallel_chase » Tue Jul 22, 2008 1:14 am
Absolutely "pepeprepa", I have taken last 2 digits as 1 and on the basis of that combination other digits have been shown.

Sudhir I am glad you found my post helpful. If you still have any doubts let me know.
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by sudhir3127 » Tue Jul 22, 2008 1:17 am
hey thanks for that .. but then how is

with Numbers ending in 2
last 2 digit combinations could be 12,32,52,72.
So we have 5*5*4 such nos

i am getting 6*5*4 ( placing the first digit shud again be 6?)
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by parallel_chase » Tue Jul 22, 2008 1:25 am
sudhir3127 wrote:hey thanks for that .. but then how is

with Numbers ending in 2
last 2 digit combinations could be 12,32,52,72.
So we have 5*5*4 such nos

i am getting 6*5*4 ( placing the first digit shud again be 6?)
we have letters from 0,1,2,3,4,5,6,7 total 8 letters.

Numbers eneding with 2 have 2 letters each, we have to subtract 2 letters from the entire list because the numbers cannot be repeated.

In the first place we are left with 6 options (8-2) but we cannot use 0 because otherwise it wont be a 4 digit number. Therefore for first place we only have 5 options, 2nd place we also have 5 because now we can take 0. Therefore,

5*5*4

Hope this clears everything.
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by pepeprepa » Tue Jul 22, 2008 1:27 am
with Numbers ending in 2
last 2 digit combinations could be 12,32,52,72.
So we have 5*5*4 such nos

-------------------------

In that case you need a 4 digit number so you must not have 0 on the first digit or it is 0342 for example, it is not a 4 digit number.

For example _ _ 1 2
1st digit: you choose among 3 4 5 6 7 = 5 choices
2nd digit you choose among 0 3 4 5 6 7 - the digit you chose for the 1st digit= 5 choices

And there are 4 combinations of last 2 digits (12,32,52,72)
So 5*5*4
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by sudhir3127 » Tue Jul 22, 2008 1:51 am
2nd digit you choose among 0 3 4 5 6 7 - the digit you chose for the 1st digit= 5 choices

i thought it shud be 0 3 4 6 7 ( assuming the number 5 has been chosen in as the first digit ) because no digit can occur more than once.. correct if i am wrong..

Thanks a lot all for clearing my doubts ,,, i know i am pushing it a bit too long.... but want to make sure my basics are right.
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by pepeprepa » Tue Jul 22, 2008 1:58 am
When I say "2nd digit you choose among 0 3 4 5 6 7 - the digit you chose for the 1st digit= 5 choices"
I say there are 5 choices which are possible, not I choose 5.
I think it's why you ask.
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by sudhir3127 » Tue Jul 22, 2008 2:02 am
thanks a lot for all ur help guys...

Finally i can say i know how to solve the question... thanks a lot...
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by sudhir3127 » Tue Jul 22, 2008 2:48 am
hey just for my understanding i am posting another question.. please let me know if its right...

how many 4 digit number that are divisble by 2 can be formed using digits (0 1 2 3 5 6 ) if no digit is to occur more than once in each number

for a number to divisble by 2 . it should end with 0, or any multiple of 2.( even number)


ending with 0 would be

10 20 30 50 60

hence its wud be 4*3*5= 60


ending with 2

02 12 32 52 62

it would be again 4*3*5 = 60


ending with 6

06 16 26 36 56 again 60

hence it shud be 180 .. but the answer is 156..i guess he is taking the ending with 2 and ending with 6 numbers as 4*3*4= 48

hence 60+48+48=156..

please let me know where i am going wrong on this one...
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