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by shovan85 » Sun Oct 10, 2010 1:14 am
Dont know man I tried all possible way but getting 2000.

Last digit can be anything except 3 so 4 ways.
10,100 and 1000th digit each can be filled in 5 ways as repitition allowed.
First didgit can not be zero so 4 ways.

Total 4*5*5*5*4 = 2000

So IMO D ;) Nearest to 2000 LOL
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by goyalsau » Sun Oct 10, 2010 2:12 am
shovan85 wrote:Dont know man I tried all possible way but getting 2000.

Last digit can be anything except 3 so 4 ways.
10,100 and 1000th digit each can be filled in 5 ways as repitition allowed.
First didgit can not be zero so 4 ways.

Total 4*5*5*5*4 = 2000

So IMO D ;) Nearest to 2000 LOL
In place of this 4 i think it should be 3 because of 5 values 3 and 4 can't be at the last place.
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by shovan85 » Sun Oct 10, 2010 4:08 am
goyalsau wrote:
In place of this 4 i think it should be 3 because of 5 values 3 and 4 can't be at the last place.
Why not 4 can be a value at last place? If it is 4 then also it would be divisible by 2. In question is either 2 or 5 or both. So 4 at the unit place satisfies as it is div by 2.
Last edited by shovan85 on Sun Oct 10, 2010 5:28 am, edited 1 time in total.
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by bitsho » Sun Oct 10, 2010 4:49 am
OA C ?
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by diebeatsthegmat » Wed Oct 13, 2010 6:20 pm
goyalsau wrote:HI! guys,
this question is complicating....
supposed the number we have to find is abcde
when abcde can be divided by 2 so a has 4 ways to choose except 0
b has 5 ways, c=5,d=5 and e=3
so total'= 4*5*5*5*4=1500
when abcde:5 so
a=4,b=5,c=5,d=5,e=2 so total =4*5*5*5*2=1000 ways
when abcde:both 2 and 5 so it must be divisible by 10
a =4,b=5,c=5,d=5,e=1 so total=4*5*5*5*1=500

because abcde cant be repeat when it divide for another number thus
i guess it must be > 3000 but dunno for sure...
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by goyalsau » Wed Oct 13, 2010 9:03 pm
diebeatsthegmat wrote:
goyalsau wrote:HI! guys,
this question is complicating....
supposed the number we have to find is abcde
when abcde can be divided by 2 so a has 4 ways to choose except 0
b has 5 ways, c=5,d=5 and e=3
so total'= 4*5*5*5*4=1500
when abcde:5 so
a=4,b=5,c=5,d=5,e=2 so total =4*5*5*5*2=1000 ways
when abcde:both 2 and 5 so it must be divisible by 10
a =4,b=5,c=5,d=5,e=1 so total=4*5*5*5*1=500

because abcde cant be repeat when it divide for another number thus
i guess it must be > 3000 but dunno for sure...
OA is
1500
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by fskilnik@GMATH » Thu Oct 14, 2010 5:18 am
Hi there, guys!

My answer: 2,000 (too)

I guess this is easy to manage if we start fixing the last digit!!

Let me try to help you on that... say the "boxes" to be filled with (single) digits are A, B, C, D , E.

FIRST scenario: ZERO is the last digit.

E -- 1 option (0)
A -- 4 options (not zero)
B -- 5 options
C -- 5 options
D -- 5 options

> So far: 1*4*5^3 = 500

Now please check that these same numbers of options apply when the last digit is FIVE, or TWO or FOUR, therefore the answer seems to me to be 4*500 = 2,000 because there are no "double-counting´s" (all scenarios are mutually exclusive).

Did I make something wrong?

Best Regards,
Fábio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by pesfunk » Mon Nov 01, 2010 5:13 pm
Even my answer came to 2000. However, i looked on internet and found this solution....hope this helps


Solution. Since it must be a 5 digit number, range is
the numbers from 20000 to 55555

we will use multiplication rule to figure this out too.
If a number is divible by two, its last(ones place) digit is even. If it
is divisible by 5 its last(ones place) digit is zero or 5.

We don't give a rat's derriere what the other digits are
(for divisibility by two or five)

The number of ways to choose the ten thousands place
digit is 4
the number of ways to choose the thousands place digit
is 5
the number of ways to choose the hundreds place digit
is 5
the number of ways to choose the tens place digit
is 5
The number of ways to choose the ones place digit is
FOUR, since it must be 0,2,4,5 and not 3

use multiplication rule and solution is
4 x 5 x 5 x 5 x 4 = 2000

So the answer MUST be 2000
fskilnik wrote:Hi there, guys!

My answer: 2,000 (too)

I guess this is easy to manage if we start fixing the last digit!!

Let me try to help you on that... say the "boxes" to be filled with (single) digits are A, B, C, D , E.

FIRST scenario: ZERO is the last digit.

E -- 1 option (0)
A -- 4 options (not zero)
B -- 5 options
C -- 5 options
D -- 5 options

> So far: 1*4*5^3 = 500

Now please check that these same numbers of options apply when the last digit is FIVE, or TWO or FOUR, therefore the answer seems to me to be 4*500 = 2,000 because there are no "double-counting´s" (all scenarios are mutually exclusive).

Did I make something wrong?

Best Regards,
Fábio.
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by diebeatsthegmat » Tue Nov 09, 2010 8:26 pm
goyalsau wrote:HI! guys,

so no explanation for this problem?
5 digit number can be abcde ( taken from 0,2,3,4,5) and these number can be repeated...
A cant be O thus A has 4 choices from above 5 numbers 0,2,3,4,5
B,c,d can be 5 combinations
e can be only 0,2,4
thus the total = 4*5^3*3=1500 ways ( which can : 2)

for the same method for numbers of combinations which can :5 is 4*5^3*2=1000 because for e, there is only 0,5 to be chosen
:10 = 4*5^3*1 =500 because only 0 = e is chosen
thus total=3000
hmhm why 1500?
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