How many 5 digit numbers can be created if the following : GMAT Problem Solving (PS)
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# How many 5 digit numbers can be created if the following

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How many 5 digit numbers can be created if the following [#permalink]

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13 Aug 2007, 05:10
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How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500

[Reveal] Spoiler:
In my way:C(4,1)C(5,1)C(3,1)A(7,2)+C(4,1)C(5,1)C(3,1)A(8,2)

But there is no correct answer in my way...
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13 Aug 2007, 07:26
3360?

Left most can be chosen from 2,4,6,8 ( 4 ways)
2nd one
Can be split into
a)When 9 is chosen then 1 ways is how 2nd digit is chosen and 3rd digit then can be chosen from 1,3,5,7 (4ways)

b)and when 9 is not chosen then 2nd digit can be chosen from ( 4 ways)
and 3rd in 4 ways

and the other two are random so 7.6

Answer is 4.1.4.7.6 + 4.4.4.7.6= 4.5.7.6.4

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13 Aug 2007, 09:43
zero is considered even as well, isnt it ?

For the first three terms (from the left), there are 5 ways to pick each ... the last two terms are throwing me off...
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13 Aug 2007, 09:53
hmm..i get

4.5.3.7.6??

4 for (2, 4, 6, 8)

5 (5 odd numbers)

3 (3, 5, 7)

7 (cause we have used 3 digits already)

6 ( cause we have used 4 digits already)..
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13 Aug 2007, 10:49
fresinha12 wrote:
hmm..i get

4.5.3.7.6??

4 for (2, 4, 6, 8)

5 (5 odd numbers)

3 (3, 5, 7)

7 (cause we have used 3 digits already)

6 ( cause we have used 4 digits already)..

I think it's more complex than that.

First digit: (2, 4, 6, 8)
Second : (1, 3, 5, 7, 9)
Third : (1, 3, 5, 7)

then the last two digits have to be digits that haven't be used.

BUT, we can't assume we're used three digits already. The second and third digit can be the same one. We could have it startout as 233 and that would fall within the rules.

I don't know how to come to the correct answer, but I'm fairly positive that we can't say 4*5*4*7*6
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13 Aug 2007, 11:25
If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

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13 Aug 2007, 11:31
VAGMAT wrote:
If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

I agree with the 4*5*3 (forgot about 1 not being prime) but what about after that? don't you have to take into account that under some circumstances you won't have 8 choices left? what if the first digits are 235...this leaves us with only 7 choices for the 4th digit and 5 for the 5th. don't we have to take this into account at all?
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13 Aug 2007, 12:10
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[quote="eschn3am"][quote="VAGMAT"]If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

I agree with the 4*5*3 (forgot about 1 not being prime) but what about after that? don't you have to take into account that under some circumstances you won't have 8 choices left? what if the first digits are 235...this leaves us with only 7 choices for the 4th digit and 5 for the 5th. don't we have to take this into account at all?[/quote]

The question says "How many numbers..." I understood the question to read that is the max number of possibilities that are possible. So, it's possible that to get the max number of possibilities, out of ten possibile numbers for the fourth and fifth digits, the first number and the third numbers could be the same, and the second be an odd number so that leaves 8 possibilities for the fourth number and 7 possibilities for the fifth number.
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13 Aug 2007, 12:25
can you explain how you get 8 and 7 for the last 2 digits??

In my understanding, we have already taken 3 of the 10 digits in the first three digits, correct??

VAGMAT wrote:
If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

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13 Aug 2007, 12:28
eschn3am wrote:
VAGMAT wrote:
If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

I agree with the 4*5*3 (forgot about 1 not being prime) but what about after that? don't you have to take into account that under some circumstances you won't have 8 choices left? what if the first digits are 235...this leaves us with only 7 choices for the 4th digit and 5 for the 5th. don't we have to take this into account at all?

I am getting 2688
4*3*8*7+4*12*7*6
there are 3 combinations when 2nd and 3rd digit are same and 12 combinations when they are different, accordingly available choices for 4th digit are 8 and 7 respectively, similarly for 5th digit, they are 7 and 6 respectively.
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13 Aug 2007, 17:52
I am not following..either its too late in the day or my brain is just plain ol..fried

beckee529 wrote:
racha24 wrote:
eschn3am wrote:
VAGMAT wrote:
If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

I agree with the 4*5*3 (forgot about 1 not being prime) but what about after that? don't you have to take into account that under some circumstances you won't have 8 choices left? what if the first digits are 235...this leaves us with only 7 choices for the 4th digit and 5 for the 5th. don't we have to take this into account at all?

I am getting 2688
4*3*8*7+4*12*7*6
there are 3 combinations when 2nd and 3rd digit are same and 12 combinations when they are different, accordingly available choices for 4th digit are 8 and 7 respectively, similarly for 5th digit, they are 7 and 6 respectively.

Great catch! Of course this question is not that straightforward! What is the OA?
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14 Aug 2007, 07:15
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racha24 wrote:
eschn3am wrote:
VAGMAT wrote:
If we consider the five digits to be {a,b,c,d, &e}

a is even, so that leaves us with four options [2,4,6,or 8]

b is odd, so that leaves us with five options [1,3,5,7,or 9]

c is a non-even prime, so that leaves us with three options [3,5, or 7]

and d & e are random digits that are not repeated which leaves us with the possibility of eight and seven random possibilities;

with that said, if we multiply the various options together we get the total outcomes possible.

4*5*3*8*7 = 3360

I agree with the 4*5*3 (forgot about 1 not being prime) but what about after that? don't you have to take into account that under some circumstances you won't have 8 choices left? what if the first digits are 235...this leaves us with only 7 choices for the 4th digit and 5 for the 5th. don't we have to take this into account at all?

I am getting 2688
4*3*8*7+4*12*7*6
there are 3 combinations when 2nd and 3rd digit are same and 12 combinations when they are different, accordingly available choices for 4th digit are 8 and 7 respectively, similarly for 5th digit, they are 7 and 6 respectively.

Just perfect.. It took time for me to before the bulb could lighten up !

For the benefit of those who could not understand, let me try and explain:

1st digit can be chosen in 4 ways : 2,4,6, 8
2nd digit can be choosen in 5 ways : 1 3 5 7 9
3rd digit can be choosen in 3 ways 3 5 7
4th in 7
5th in 6
But as per question 4th and 5th digit should have been used at 1st, 2nd and 3rd place. But there are no such restriction for 2nd and 3rd digit.
And depending on whether 2nd and 3rd digits are same or different, it can be how many digits we have for 4th and 5th digit.
2 and 3rd digit can be as follows ;
13 15 17
33 35 37
53 55 57
73 75 77
93 95 97
So, we have 3 combinations of same number and 12 combination for different number.
When we have same numbers, number of digits available is 1 more for 4th and 5th places.

Hence we have , 4*3*8*7 + 4* 12* 7*6 = 2688.
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14 Aug 2007, 07:30
yongyang wrote:
How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520
b) 3150
c) 3360
d) 6000
e) 7500

In my way:C(4,1)C(5,1)C(3,1)A(7,2)+C(4,1)C(5,1)C(3,1)A(8,2)

But there is no correct answer in my way...

I got D, but not really sure.

Say the number is ABCDE
Possible A = 2,4,6,8 => Total = 4
Possible B = 1,3,5,7,9 => Total = 5
Possible C = 3,5,7 => Total =3
Possible D & E = Anything hasn't used before, say x and x
Together, we have total possibility of 4*5*3*x*x = 60*x^2

Since we know the answer must be divisible 60, B is out.
We also know that the answer must be divisible by x^2
So for A, 252/6 = 42
for C, 336 / 6 = 56
for D, 600/6 = 100
For E, 750/6 = 125

Looks like on D is possible answer since it is in term of 10*10 square. However, 10 is too high to use here, I think, so I am not sure.
I need to point out that it is unclear to me fron the question if we can use the same number for the last two digits.
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14 Aug 2007, 12:55
It's not x^2, because once you use a digit for the 4th space, you only have x-1 choices left for the 5th space.
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14 Aug 2007, 15:58
I am getting B.

1st digit is an even integer hence 5 choices
2nd digit is an odd integer hence 5 choices
3rd digit is non-even prime hence 3 choices
4th digit should be a random number which is not one of the other 3 hence 7 choices
5th digit should be a random number which is not one of the other 4 hence 6 choices

Therefore no. of choices =5*5*3*7*6=3150
_________________

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Subhen

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14 Aug 2007, 16:10
Hi All,

This is the first time I've ever posted on this site before so bear with me if I do something retarded. Anyway... here is how I see the problem:

There are 3 cases:

1) The 2nd and 3rd digit are equal.
2) The 2nd and 3rd digit are different and the 2nd digit is prime.
3) The 2nd and 3rd digit are different and the 2nd digit is prime.

In the first case you have 5*3*1*8*7 = 840 choices because if the 2nd and third digit are equal then the second digit must be chosen from the same set as the third and the third digit must be the same as the second.

In the second case you have 5*3*2*7*6 = 1260 choices because if the 2nd digit is prime you only can choose 3, 5, or 7 and since the third digit must be different you only get 2 choices.

In the 3rd case you have 5*2*3*7*6 = 1260 choices because if the 2nd digit is not prime you can only choose 1 or 9. But you still can choose any of the three for the third digit.

So, 840 + 1260 + 1260 = 3360, which is C.

I'm not positive about this answer but I think it is correct. Does anyone know what the official guide says? What problem number from what book is it?
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14 Aug 2007, 16:14
Hi All,

This is the first time I've ever posted on this site before so bear with me if I do something retarded. Anyway... here is how I see the problem:

There are 3 cases:

1) The 2nd and 3rd digit are equal.
2) The 2nd and 3rd digit are different and the 2nd digit is prime.
3) The 2nd and 3rd digit are different and the 2nd digit is NOT prime.

In the first case you have 5*3*1*8*7 = 840 choices because if the 2nd and third digit are equal then the second digit must be chosen from the same set as the third and the third digit must be the same as the second.

In the second case you have 5*3*2*7*6 = 1260 choices because if the 2nd digit is prime you only can choose 3, 5, or 7 and since the third digit must be different you only get 2 choices.

In the 3rd case you have 5*2*3*7*6 = 1260 choices because if the 2nd digit is not prime you can only choose 1 or 9. But you still can choose any of the three for the third digit.

So, 840 + 1260 + 1260 = 3360, which is C.

I'm not positive about this answer but I think it is correct. Does anyone know what the official guide says? What problem number from what book is it?
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14 Aug 2007, 17:07
Hi all,

I wrote a program to figure out the answer. The answer is C, 3360.

Here is my code (if you care).

#include <stdio>
#include <stdlib>

int main()
{
int count = 0;

for(int i = 0; i < 100000; i++)
{
char num[6];

sprintf(num, "%05d", i);

if((((int) num[0] - 48) % 2) != 0)
{
continue;
}
else if((((int) num[1] - 48) % 2) == 0)
{
continue;
}
else if((num[2] != '3') && (num[2] != '5') && (num[2] != '7'))
{
continue;
}
else if((num[3] == num[2]) || (num[3] == num[1]) || (num[3] == num[0]))
{
continue;
}
else if((num[4] == num[3]) || (num[4] == num[2]) ||
(num[4] == num[1]) || (num[4] == num[0]))
{
continue;
}

count += 1;
}

printf("Count: %d\n", count);

return 0;
}[/code]
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15 Aug 2007, 14:25
Well, the problem states that the first digit must be even.

According to the arithmetic review of the official guide as well as wikipedia, 0 is an even number, which makes sense based upon the definition of an even number.

So, the choices for the first number are { 0, 2, 4, 6, 8 }
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16 Aug 2007, 06:35
rrglover wrote:
Well, the problem states that the first digit must be even.

According to the arithmetic review of the official guide as well as wikipedia, 0 is an even number, which makes sense based upon the definition of an even number.

So, the choices for the first number are { 0, 2, 4, 6, 8 }

If the first digit were 0, it would not be a five-digit number
16 Aug 2007, 06:35

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