GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 05 Dec 2019, 07:20 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # How many five-digit numbers exist such that the product of their digit

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Aug 2009
Posts: 8282
How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

3
10 00:00

Difficulty:   75% (hard)

Question Stats: 56% (02:32) correct 44% (02:39) wrong based on 136 sessions

### HideShow timer Statistics

How many five-digit numbers exist such that the product of their digits is 400?

a) 60
b) 120
c) 180
d) 30
e) 75

_________________

Originally posted by chetan2u on 12 Feb 2015, 20:27.
Last edited by chetan2u on 13 Feb 2015, 02:08, edited 2 times in total.
RENAMED THE TOPIC.
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15636
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

2
2
Hi chetan2u,

This question is a "twist" on prime factorization. Since we're limited to 5-digit numbers, we have to consider ALL of the different possible 5-digit permutations.

We're asked to name all of the 5-digit numbers whose product-of-digits is 400. We can use prime-factorization to figure out the digits, BUT this comes with a "twist".....NOT ALL of the digits have to be primes....

400 = (25)(16)

The 25 = (5)(5) and there's no way around that. Every option MUST have two 5s as two of the digits.

The 16 can be broken down in a few different ways though:

16 = (1)(2)(8)
16 = (1)(4)(4)
16 = (2)(2)(4)

Each of those options has to be accounted for.

So, we have 3 sets of 5 digits:

1) 1, 2, 8, 5, 5 --> This gives us 5!/2! arrangements = 120/2 = 60 different 5-digit numbers.

2) 1, 4, 4, 5, 5 --> This gives us 5!/(2!2!) arrangements = 120/4 = 30 different 5-digit numbers.

3) 2, 2, 4, 5, 5 --> This gives us 5!/(2!2!) arrangements = 120/4 = 30 different 5-digit numbers.

60 + 30 + 30 = 120 different 5-digit numbers

GMAT assassins aren't born, they're made,
Rich
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 8282
How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

2
1
chetan2u wrote:
How many five-digit numbers exist such that the product of their digits is 400?

a) 60
b) 120
c) 180
d) 30
e) 75

OA in a while

hi,

OE-

in questions, especially dealing with combinations and probability, it is very important to break down the question and look for clues without wasting time..
question gives us following info..
1) it consists of five digits so 0 to9..
2) since product is 400, these digits have to be factors of 400..

$$400=2^4*5^2$$..
following points emerge..
1) out of 5 digits, two have to be 5.. as 5 multiplied by other number 2 or 5 will give a two digit number..
2) so the remaining 3 digits will have product equal to 400/25=16...
3) out of the remaining 3 digits, it can be 1,2,4,8..
4) so the three digits can be fixed in following way to get a product of 16..
a)1,2,8
b)1,4,4
c)2,2,4

total ways
a)$$\frac{5!}{2!}= 60$$..
b)$$\frac{5!}{2!2!}=30$$..
c)$$\frac{5!}{2!2!}=30$$
total=120...
ans B..
_________________
Intern  Joined: 22 Aug 2014
Posts: 38
Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

chetan2u wrote:
How many five-digit numbers exist such that the product of their digits is 400?

a) 60
b) 120
c) 180
d) 30
e) 75

OA in a while

Option A

Lets assume the five digit no to be abcde

And the a*b*c*d*e = 400

Factoring 400 = 2^4*5^2

The possible sets can be

2,2,4,5,5
1,4,4,5,5
1,2,8,5,5

For each set the # of 5 digit nos come upto = 5!/3! = 5*4 = 20

The total set is 3 and therefore the # 5 digits nos whose a*b*c*d*e = 400 is 3*20 = 60

I did a mistake while clicking on the answer. Did not consider the third set and also wrote the combination formula. Math Expert V
Joined: 02 Aug 2009
Posts: 8282
Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

aj13783 wrote:
chetan2u wrote:
How many five-digit numbers exist such that the product of their digits is 400?

a) 60
b) 120
c) 180
d) 30
e) 75

OA in a while

Option A

Lets assume the five digit no to be abcde

And the a*b*c*d*e = 400

Factoring 400 = 2^4*5^2

The possible sets can be

2,2,4,5,5
1,4,4,5,5
1,2,8,5,5

For each set the # of 5 digit nos come upto = 5!/3! = 5*4 = 20

The total set is 3 and therefore the # 5 digits nos whose a*b*c*d*e = 400 is 3*20 = 60

I did a mistake while clicking on the answer. Did not consider the third set and also wrote the combination formula. hi,
you are bang on with your explanation till you have found the possible sets ..
however each possible set will have different number of ways as we have in 'ways to make different words from say 'school',which is 6!/2!..
so please go through each set again and i am sure you will come up with correct answer...
_________________
Intern  Joined: 22 Aug 2014
Posts: 38
Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

chetan2u wrote:
aj13783 wrote:
chetan2u wrote:
How many five-digit numbers exist such that the product of their digits is 400?

a) 60
b) 120
c) 180
d) 30
e) 75

OA in a while

Option A

Lets assume the five digit no to be abcde

And the a*b*c*d*e = 400

Factoring 400 = 2^4*5^2

The possible sets can be

2,2,4,5,5
1,4,4,5,5
1,2,8,5,5

For each set the # of 5 digit nos come upto = 5!/3! = 5*4 = 20

The total set is 3 and therefore the # 5 digits nos whose a*b*c*d*e = 400 is 3*20 = 60

I did a mistake while clicking on the answer. Did not consider the third set and also wrote the combination formula. hi,
you are bang on with your explanation till you have found the possible sets ..
however each possible set will have different number of ways as we have in 'ways to make different words from say 'school',which is 6!/2!..
so please go through each set again and i am sure you will come up with correct answer...

Sure will do. This is where i make mistakes. Need to improve on it by solving problems on a regular basis.
Non-Human User Joined: 09 Sep 2013
Posts: 13703
Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: How many five-digit numbers exist such that the product of their digit   [#permalink] 22 Aug 2017, 09:11
Display posts from previous: Sort by

# How many five-digit numbers exist such that the product of their digit  