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5 digit no. divisible by 4

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5 digit no. divisible by 4

by pareekbharat86 » Sat Nov 23, 2013 11:35 pm
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

OA is E

I could conclude the the last 2 digits of the 5-digit nos. should be 04 or 12 or 20 or 24 or 32 or 40 or 52 in order for them to be divisible by 4.

Now suppose the last 2 digits are 04,20 or 40

_ _ _ _ _--> the last 2 positions can be filled in 3 ways. The first 3 however can be filled in 4C3 ways giving us 4C3*3= 12 nos.

For the remaining combinations i.e. having last 2 digits 12,24,32 or 52:

_ _ _ _ _ --> the last 2 positions can be filled in 4 ways. The first position cannot be 0. Therefore we can fill the first position in 3 ways. The remaining 2 (digit 2nd &3rd) can be filled in 3C2 ways giving us 3*3C2*4= 36 ways.

Total 36+12= 48 nos. Obviously this is wrong. Please guide.
Thanks,
Bharat.
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by GMATGuruNY » Sun Nov 24, 2013 4:26 am
pareekbharat86 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

OA is E
For the 5-digit integer to be a multiple of 4, the last two digits must form a multiple of 4.
Options:
04, 12, 20, 24, 32, 40, 52.

Case 1: Last 2 digits include 0
Number options for the last 2 digits = 3. (04, 20 or 40.)
Number of options for the ten-thousands place = 4. (Any of the 4 remaining digits.)
Number of options for the thousands place = 3. (Any of the 3 remaining digits.)
Number of options for the hundreds place = 2. (Either of the 2 remaining digits.)
To combine these options, we multiply:
3*4*3*2 = 72.

Case 2: Last 2 digits do NOT include 0
Number options for the last 2 digits = 4. (12, 24, 32, or 52.)
Number of options for the ten-thousands place = 3. (Any of the remaining 4 digits but 0.)
Number of options for the thousands place = 3. (Any of the 3 remaining digits.)
Number of options for the hundreds place = 2. (Either of the 2 remaining digits.)
To combine these options, we multiply:
4*3*3*2 = 72.

Total options = 72+72 = 144.

The correct answer is E.
Last edited by GMATGuruNY on Tue Nov 26, 2013 3:37 am, edited 1 time in total.
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by pareekbharat86 » Tue Nov 26, 2013 1:45 am
GMATGuruNY wrote:
pareekbharat86 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

OA is E
For the 5-digit integer to be a multiple of 4, the last two digits must form a multiple of 4.
Options:
04, 12, 20, 24, 32, 40, 52.

Case 1: Last 2 digits include 0
Number options for the last 2 digits = 3. (04, 20 or 40.)
Number of ways to arrange the remaining 4 digits = 4! = 24.
To combine these options, we multiply:
3*24 = 72.

Case 2: Last 2 digits do NOT include 0
Number options for the last 2 digits = 4. (12, 24, 32, or 52.)
Number of options for the ten-thousands place = 3. (Any of the remaining 4 digits but 0.)
Number of ways to arrange the remaining 3 digits = 3! = 6.
To combine these options, we multiply:
4*3*6 = 72.

Total options = 72+72 = 144.

The correct answer is E.
Hi,

We are talking of a 5-digit number. If the last 2 digits are the ones we listed, aren't we just left with 3 more blanks to fill. Then why are we considering 4! in first case (for 3 blanks) and 3! in the second case for the 2nd & 3rd blank?

Should it not be 4C3 and 3C2 in the respective cases?

Please guide.
Thanks,
Bharat.
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by theCodeToGMAT » Tue Nov 26, 2013 2:10 am
For number to be divisible by "4" we must make sure that last two digits must be divisible

So possible cases: 04, 20, 40, 12, 24, 32, 52

04, 20, 40 => (4*3*2)each = 3*(24) = 72

12, 24, 32, 52 ==> (3*3*2)*4 each = 72

Total 72+72 = 144

[spoiler]{E}[/spoiler]
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by GMATGuruNY » Tue Nov 26, 2013 3:40 am
pareekbharat86 wrote:
We are talking of a 5-digit number. If the last 2 digits are the ones we listed, aren't we just left with 3 more blanks to fill. Then why are we considering 4! in first case (for 3 blanks) and 3! in the second case for the 2nd & 3rd blank?

Should it not be 4C3 and 3C2 in the respective cases?

Please guide.
Please revisit my post above, in which I've clarified the reasoning.
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by Mathsbuddy » Tue Nov 26, 2013 4:12 am
L = 7 = the number of last digit pairs for multiples of 4: 20, 40, 12, 32, 04, 52, 24.

Each of which leaves 3 digits remaining with 4 digits to choose from.

If you include 0 as a 1st digit:
4*3*2 = 24 combinations
24 * L = 24 * 7 = 168 combinations altogether


If zero is not permitted as a 1st digit then:
there are 3 * 2 * 7 = 42 combinations to ignore


168 - 42 = 144

Answer = 144 (E)
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