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

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Math Expert
Joined: 02 Aug 2009
Posts: 7592
How many five-digit numbers exist such that the product of their digit  [#permalink]

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Updated on: 13 Feb 2015, 02:08
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Difficulty:

75% (hard)

Question Stats:

56% (02:32) correct 44% (02:42) wrong based on 131 sessions

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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

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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.
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Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

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12 Feb 2015, 23:30
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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

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Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

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12 Feb 2015, 23:50
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
Joined: 02 Aug 2009
Posts: 7592
Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

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13 Feb 2015, 01:00
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...
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Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

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13 Feb 2015, 01:16
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.
Math Expert
Joined: 02 Aug 2009
Posts: 7592
How many five-digit numbers exist such that the product of their digit  [#permalink]

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13 Feb 2015, 02:06
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..
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Re: How many five-digit numbers exist such that the product of their digit  [#permalink]

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22 Aug 2017, 09:11
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Re: How many five-digit numbers exist such that the product of their digit   [#permalink] 22 Aug 2017, 09:11
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