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How many five-digit numbers can be formed from the digits 0,

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How many five-digit numbers can be formed from the digits 0, [#permalink]

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How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jan 2014, 03:29, edited 1 time in total.
Edited the question and added the OA.

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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Bunuel wrote:
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.


Similar question to practice: how-many-five-digit-numbers-can-be-formed-using-digits-91597.html
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How many five-digit numbers can be formed from the digits 0, [#permalink]

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 10 May 2014, 06:32
Bunuel wrote:
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.


Dear Bunnel,

i dont understand the highlighted part... when the last 2 digits are 04, 20 or 40 then we have only 3 digits left for the 1st 3 to choose from... as 0,2 or 4 are gone???

same for the next highlighted part.

please explain.
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 10 May 2014, 06:39
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nandinigaur wrote:
Bunuel wrote:
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.


Dear Bunnel,

i dont understand the highlighted part... when the last 2 digits are 04, 20 or 40 then we have only 3 digits left for the 1st 3 to choose from... as 0,2 or 4 are gone???

same for the next highlighted part.

please explain.


Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case.

Hope it's clear.
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 10 May 2014, 11:20
Bunuel wrote:
nandinigaur wrote:
Dear Bunnel,

i dont understand the highlighted part... when the last 2 digits are 04, 20 or 40 then we have only 3 digits left for the 1st 3 to choose from... as 0,2 or 4 are gone???

same for the next highlighted part.

please explain.


Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case.

Hope it's clear.


no what i mean is that we have 0, 1, 2, 3, 4, 5 to choose from. so, if the last two digits are 04, 20, or 40... then 3 of the 6 digits to choose from are gone. we will be left with 2,3,5??? how 4?
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 11 May 2014, 04:29
nandinigaur wrote:
Bunuel wrote:
nandinigaur wrote:
Dear Bunnel,

i dont understand the highlighted part... when the last 2 digits are 04, 20 or 40 then we have only 3 digits left for the 1st 3 to choose from... as 0,2 or 4 are gone???

same for the next highlighted part.

please explain.


Yes, we are told that no digits can repeat. So, if the last two digits are 04, 20, or 40, we are left with 4 digits for the first, 3 digits for the second and 2 digits for the third one. The same applies to the second case.

Hope it's clear.


no what i mean is that we have 0, 1, 2, 3, 4, 5 to choose from. so, if the last two digits are 04, 20, or 40... then 3 of the 6 digits to choose from are gone. we will be left with 2,3,5??? how 4?


For EACH case of 04, 20, or 40 we used 2 digits and we are left with 4. Isn't it?
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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To be divisible by four the number needs to end in 04, 40, 20, 12, 32 or 52.

Now then, we have 4 numbers remaining and 3 slots. 4C3 * 3! ways to order them.
We do this for each combination so: 4C3*3!*6=24*6=144

Answer: E

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 03 Jul 2014, 11:48
Bunuel wrote:
Bunuel wrote:
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.


Similar question to practice: how-many-five-digit-numbers-can-be-formed-using-digits-91597.html

Hi Bunuel,

When I try to use the same method in the link you sent, I still don't get the wanted result.
If we choose all numbers except for "0", we have either 12, 24, 32,52 for the endings, and 3*2*1 for the first place. This gives 6*4.
If we include "0", we get, for the ending, 20,40,04. and for the first digits, 4*3*2 - > 24*2 = 48.

all in all: 48 + 24 = 72.... Still not the right answer... What is wrong with this?

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 03 Jul 2014, 12:02
ronr34 wrote:
Bunuel wrote:
Bunuel wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144

A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.


Similar question to practice: how-many-five-digit-numbers-can-be-formed-using-digits-91597.html

Hi Bunuel,

When I try to use the same method in the link you sent, I still don't get the wanted result.
If we choose all numbers except for "0", we have either 12, 24, 32,52 for the endings, and 3*2*1 for the first place. This gives 6*4.
If we include "0", we get, for the ending, 20,40,04. and for the first digits, 4*3*2 - > 24*2 = 48.

all in all: 48 + 24 = 72.... Still not the right answer... What is wrong with this?


That other question is about numbers divisible by 3, not 4. You cannot use the same approach.
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 03 Jul 2014, 12:56
Bunuel wrote:
ronr34 wrote:
Bunuel wrote:

Hi Bunuel,

When I try to use the same method in the link you sent, I still don't get the wanted result.
If we choose all numbers except for "0", we have either 12, 24, 32,52 for the endings, and 3*2*1 for the first place. This gives 6*4.
If we include "0", we get, for the ending, 20,40,04. and for the first digits, 4*3*2 - > 24*2 = 48.

all in all: 48 + 24 = 72.... Still not the right answer... What is wrong with this?


That other question is about numbers divisible by 3, not 4. You cannot use the same approach.

I realize that.
My question is why not?

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 03 Jul 2014, 13:09
ronr34 wrote:
Bunuel wrote:
ronr34 wrote:
That other question is about numbers divisible by 3, not 4. You cannot use the same approach.

I realize that.
My question is why not?


Strange question...

Because numbers divisible by 4 are those whose last two digits are divisible by 4, while numbers divisible by 3 are those whose sum of the digits is divisible by 3. Different patterns.
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Re: How many 5 digit number combos are divisible by 4? [#permalink]

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How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A) 36
B) 48
C) 72
D) 96
E) 144

Answer is E. Unsure of how to solve this??


You have 6 digits and you need a 5 digit number.

In how many ways can the number be divisible by 4?
If it ends with 04 or 12 or 20 or 24 or 32 or 40 or 52, it will be divisible by 4.

All 3 combinations that have a 0 as one of the last two digits can be formed by using basic counting principle.
4 * 3 * 2= 24. (First digit in 4 ways, second in3 ways and third in 2 ways)
Since there are 3 such combinations, you get 24*3 = 72

The other 4 combinations which do not have a 0 in the last two digits can be formed by 3 * 3 * 2 = 18
(First digit in 3 ways (no 0), second in 3 ways (leftover 2 digits and 0) and third in 2 ways.
Since there are 4 such combinations, you get 18*4 = 72

Total number of ways = 72 + 72 = 144
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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 21 Dec 2015, 23:08
Yep, got it right. to do it faster, we need to combine cases where in 0 occurs in the last 2 digits and where it doesnt occur.

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 30 Dec 2015, 03:21
Hi Karishma. I understand 3 * 3 * 2 = 18. 1st digit has 3 options, 2nd has 3 options, then 2, and 1 finally.
However, instead of starting with the 1st digit, can I think this way : the 3rd digit has 4 options, then 2nd digit has 3 options, then 1st has only 1 option? Then it is 4*3 = 12. Why thinking in these two ways should two different numbers of possible arrangement? Can you help please?

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How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 10 Nov 2016, 06:45
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


My approach:
Total of 600 possible number: 5x5x4x3x2x1 (zero can't be the 1st digit)
Of those 600 number, 300 are even, because you have 3 odd and 3 even numbers. Of those 300 even numbers, about half are multiple of 4. So answer choice E.

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 29 Nov 2017, 11:40
Bunuel wrote:
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.



Sir what i dont understand is how can the 3 numbers be arranged in 4 ways in case 2? (highlighted text)

Thanks

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Re: How many five-digit numbers can be formed from the digits 0, [#permalink]

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New post 29 Nov 2017, 20:01
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srishti201996 wrote:
Bunuel wrote:
guerrero25 wrote:
How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144


A number to be divisible by 4 its last two digit must be divisible by 4 (similarly a number to be divisible by 2 its last digit must be divisible by 2; to be divisible by 8, last three digits must be divisible by 8 and so on).

Thus the last two digit must be 04, 12, 20, 24, 32, 40, or 52.

If the last two digits are 04, 20, or 40, the first three digits can take 4*3*2= 24 values.
Total for this case: 24*3 = 72.

If the last two digits are 12, 24, 32, or 52, the first three digits can take 3*3*2= 18 values (that's because the first digit in this case cannot be 0, thus we are left only with 3 options for it not 4, as in previous case).
Total for this case: 18*4 = 72.

Grand total 72 +72 =144.

Answer: E.



Sir what i dont understand is how can the 3 numbers be arranged in 4 ways in case 2? (highlighted text)

Thanks


In that case, the last two digit can take 4 different values: 12, 24, 32, or 52. The first three digits can take 18 values. Total = 4*18.
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