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Five digit combination problem

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by Naruto » Sun Jun 14, 2009 4:34 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

In how many ways can five digit numbers be formed using digits 0, 2, 3, 4, 5 , when repetition is allowed such that the number formed is divisible by 2 or 5 or both.

A 100
B 150
C 3125
D 1500
E 125'

well I tried in all possible ways and my answer is 2000. Please explain in detail.
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by tohellandback » Sun Jun 14, 2009 5:17 am
same here i am also getting 2000
The powers of two are bloody impolite!!
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by Naruto » Sun Jun 14, 2009 5:30 am
well OA given is 1500

got this question from one of the online practise test problems
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by PAB2706 » Sun Jun 14, 2009 8:43 pm
pls explain how you got 2000
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by tohellandback » Sun Jun 14, 2009 9:01 pm
PAB2706 wrote:pls explain how you got 2000
This is how I got 2000
numbers to be divisible by 2 or 5 or both must end with 0,2,4 or 5
ending with 0
4*5*5*5=500
same for 2,4,5
total=500+500+500+500=2000
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Re: Five digit combination problem

by Vemuri » Sun Jun 14, 2009 9:11 pm
To get a 5 digit number, the first number (left most) should not be 0.

For the resulting number to be divisible by 2, the number should end with either 0,2 or 4 i.e. 4*5*5*5*3

For the resulting number to be divisible by 5, the number should end with either 0, or 5 i.e. 4*5*5*5*2

For the resulting number to be divisible by 2 & 5, the number should end with 0 i.e. 4*5*5*5*1

Since the question is saying that number formed is divisible by 2 or 5 or both, we need to add them all up.

==> 4*5*5*5*3 + 4*5*5*5*2 + 4*5*5*5*1
==> 4*5*5*5(3+2+1)
==> 500*6
==> 3000 :roll:
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by kanha81 » Sun Jun 14, 2009 9:33 pm
tohellandback wrote:same here i am also getting 2000
This is how did. Please explain how are you getting 2000 as the answer.

(1)
divisible by 2:
- - - - -
4 5 5 5 3 = 2^2 * 5^3 * 3

ie the first position from left (2,3,4, or 5)
the second position from left (0,2,3,4,5)
the third position from left (0,2,3,4,5)
the fourth position from left (0,2,3,4,5)
the fifth position from left (0,2,4)

(2)
divisible by 5:
- - - - -
4 5 5 5 2 = 2^3 * 5^3

(3)
divisible by both:
- - - - -
4 5 5 5 1 = 2^2 * 5^3

add (1) + (2) + (3)-

5^3 * 2^2 * [3 + 2 + 1] = 3000

Am I counting more than once? If so, kindly please help me rectify the mistake!

Thanks.
Last edited by kanha81 on Sun Jun 14, 2009 10:11 pm, edited 1 time in total.
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by tohellandback » Sun Jun 14, 2009 9:40 pm
kanha81 wrote:
tohellandback wrote:same here i am also getting 2000
This is how did. Please explain how are you getting 2000 as the answer.

(1)
divisible by 2:
- - - - -
4 5 5 5 3 = 2^2 * 5^3 * 3

ie the first position from left (2,3,4, or 5)
the second position from left (0,2,3,4,5)
the third position from left (0,2,3,4,5)
the fourth position from left (0,2,3,4,5)
the fifth position from left (0,2,4)

(2)
divisible by 5:
- - - - -
4 5 5 5 2 = 2^3 * 5^3

(3)
divisible by both:
- - - - -
4 5 5 5 1 = 2^2 * 5^3

add (1) + (2) + (3)-

5^3 * 2^2 * [3 + 2 + 1] = 3000

Am I counting more than once? If so, kindly please help me rectify the mistake!

Thanks.
Kanha,
of course you are counting more than once. numbers ending with 0 and 5 are being counted twice.

but i am not saying that you are wrong. may be, we do need to count them twice. I am waiting for the OA. and then i would need someone to explain the question to me.
The powers of two are bloody impolite!!
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Re: Five digit combination problem

by Vemuri » Sun Jun 14, 2009 9:58 pm
Experts....your help is required here please !!
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by PAB2706 » Sun Jun 14, 2009 10:10 pm
I think we are counting more than once...even i had got 3000 as my answer..

this is y....the question uses word "or"

so we have addition...

1st case : now when we have to divide by 2 the units digit shud be 0,2,4

2nd case :- when by 5 units digit shud be 0,5

but observe that the number of possibilities of 0 in the units has already been calculated in 1st case

thus for 5 we will have to calculate only for 5 at units place.

for division by both 2 and 5 units digit caanot be anything else but 0 ( already calculated in 1st case)

thus in my opinion the answer shud be

4x5x5x5x3 + 4x5x5x5x1 = 2000
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by tohellandback » Sun Jun 14, 2009 10:15 pm
Naruto has already posted the OA. It is 1500. Experts..is the answer wrong or are we missing something here??
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by Naruto » Mon Jun 15, 2009 12:57 am
tohellandback wrote:Naruto has already posted the OA. It is 1500. Experts..is the answer wrong or are we missing something here??
I second that, i tried its 2000.
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Re: Five digit combination problem

by Stuart@KaplanGMAT » Mon Jun 15, 2009 11:44 am
Naruto wrote:In how many ways can five digit numbers be formed using digits 0, 2, 3, 4, 5 , when repetition is allowed such that the number formed is divisible by 2 or 5 or both.

A 100
B 150
C 3125
D 1500
E 125'

well I tried in all possible ways and my answer is 2000. Please explain in detail.
The number has 5 digits, so the first digit cannot be 0.
The number must be divisible by either 2 or 5, so the last digit must be 0, 2, 4 or 5.
The 3 digits in the middle can be anything.

Therefore, the final answer is:

4*5*5*5*4 = 20*25*4 = 2000

Are you sure that the question is posted correctly? If our digits had been 0, 1, 2, 3 and 4 then the answer would be 1500 (since the last digit would have to be 0, 2 or 4 and our product would be 4*5*5*5*3 = 1500); then again, if those were our digits the "divisible by 5" portion would be irrelevant to the question.

Here's a great tip for this forum: ALWAYS post the source of your question, so we know if the question is reliable.
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by Naruto » Mon Jun 15, 2009 9:55 pm
well this was my source, i guess not a reliable one...

https://www.testpreppractice.net/GMAT/Fr ... Tests.aspx
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by Jhuff07 » Tue Oct 24, 2017 2:13 pm
So is there anyone running this site that help us with this problem?
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