How many positive integers less than 10,000 are such that the product

Question

How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12 
B. 24
C. 36
D. 38
E. 50

Bunuel · Accepted Answer

OFFICIAL SOLUTION:

30 = 2*3*5 = 6*5 (only 2*3 gives single digit number 6).

So, we should count the number of positive integers less than 10,000 with the digits {2, 3, 5} and {5, 6} and any number of 1's with each set.

2-digit numbers: 
{5, 6} - the number of combinations = 2: 56 or 65.

3-digit numbers: 
{1, 5, 6} - the number of combinations = 3! = 6: 156, 165, 516, 561, 615, or 651.
{2, 3, 5} - the number of combinations = 3! = 6.

4-digit numbers: 
{1, 1, 5, 6} - the number of combinations = 4!/2! = 12.
{1, 2, 3, 5} - the number of combinations = 4! = 24.

Total = 2 + 6 + 6 + 12 + 24 = 50.

Answer: E.

askhere · Answer

30 = 1 * 30, 2 * 15, 3* 10, 5 *6

We have the option of using single digits 1,2,3,5 and 6
 No : of four digit numbers such that the product is 30 
The possible combinations are using digits (1,2,3,5) & (1,1, 5, 6) such that the product 30.
No: of different combinations possible using (1,2,3,5) = 4! = 24
No: of different combinations possible using (5,6,1,1) = 4!/2! = 12 ('1' is repeated twice)
Total  4 digit numbers  so that the product is 30 is 24+12 =  36

No: of three digit numbers such that the product is 30 
The possible combinations are using digits (2,3, 5) & (1,5,6)
No: of different combinations possible using (2,3,5) = 3! = 6
No: of different combinations possible using (1,5,6) = 3! = 6
Total  3 digit numbers  so that the product is 30 is 6+6 =  12

No: of  2 digit numbers  such that the product is 30 is  2  (56 and 65)

Total numbers under 10000 such that the product is 30 is 36+12+2 =  50

Answer :  E

Ambarish

Jackal · Answer

My attempt:

We need a number from one digit to four digits whose product is 30. None of the digits can be zero then.
The single digit factors of 30 are 1,2,3,5,6 (rest are more than 1 digit)

So we form the combinations as -

Starting with 4 digit numbers
1 _ _ _ 1st combination of 2,3,5 has 3! and 2nd combination of 5,6,1 has 3!
2 _ _ _ combination of 1,3,5 has 3!
3 _ _ _ combination of 1,2,5 has 3!
5 _ _ _ 1st combination of 1,2,3 has 3! and 2nd combination of 6,1,1 has 3!/2!
6 _ _ _ combination of 5,1,1 has 3!/2!

So total 36 ways

Now 3 digit numbers
156 combinations gives 3! ways
235 combination gives 3! ways

So total 12 ways

Now 2 digit numbers
56 combination gives us 2! ways

So total 2 ways

Grand total of ways = 36+12+2=50

I am kind of confused now as none of the options is 50

chetan2u · Answer

hi, 
you are forgetting 3 digits and 2 digits number....

chetan2u · Answer

hello,
we can have 2,3 or 4 digits number satisfying the condition..

4 digits number: it can consist of 1,2,3,5 or 1,1,6,5..
 1,2,3,5- 4! ways
 1,1,5,6- 4!/2 ways
total-36 ways

3 digits number: it can have 2,3,5 or 1,5,6..
 2,3,5 - 3! ways
 1,5,6 - 3! ways
total - 12 ways

2 digits- 5,6
 total - 2 ways

overall- 50 ways...

Bunuel · Answer

Option E should have been 50. Edited.

askhere · Answer

Thanks for pointing out. I edited my answer.

UJs · Answer

Q integers >10,000 are such that the product of their digits is 30

possible factors of 30 that can be digits (5,6,1,1) and (2,3,5,1), so two sets to choose from 
 Set 1 :  (2,3,5,1)
 Set 2 :  (5,6,1,1)

4 Digit numbers   _ _ _ _   (4 spots to be filled by 4 digits )

Set 1 : (2,3,5,1)   _ _ _ _   4.3.2.1 = 24
Set 2 : (5,6,1,1)   _ _ _ _   (4.3.2.1)/2 = 12

3 Digit numbers   _ _ _   (3 spots to be filled by 3 digits )
 
 Set 1 : (2,3,5)   _ _ _   3.2.1 = 6
Set 2 : (5,6,1)   _ _ _   3.2.1 = 6

2 Digit numbers   _ _   (2 spots to be filled by 2 digits )

Set 1 : not possible
Set 2 : (5,6) _ _ _ _ 2.1 = 2

total = 36 + 12 + 2 = 50   Ans : E

RudeyboyZ · Answer

Correct me if I'm wrong but you have chosen 1,1,5,6 in 4!/2, is it because integer 1 is being repeated twice in set (1,1,5,6)?

chetan2u · Answer

Hi  RudeyboyZ ,
you are correct .. 
say the set was 1,2,1,1, the ways would have been 4!/3!...here out of 4 positions , there will be 3! will be common .. basically the ways 1 can be placed within themselves...
if the set was 1,1,2,2, the ways would have been 4!/2!2!..

Kinshook · Answer

30 = 1*2*3*5 = 1*6*5 = 5*6

2-digit numbers consisting of 5*6 = 2
3-digit numbers consisting of 1,5,6 = 3! = 6
3-digit numbers consisting of 2,5,3 = 3! = 6
4-digit numbers consisting of 1,2,3,5 = 4! = 24
4-digit numbers consisting of 1,1,6,5 = 4!/2! = 12

Total numbers = 2 + 6 + 6 + 24 + 12 = 50

IMO E

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How many positive integers less than 10,000 are such that the product

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