Last visit was: 25 Apr 2026, 16:19 It is currently 25 Apr 2026, 16:19
Home
Messages
Tests
Schools
Events
Close
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
Your Progress

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 25 Apr 2026
Posts: 109,830
Own Kudos:
811,280
 [6]
Given Kudos: 105,886
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,830
Kudos: 811,280
 [6]
Find the number of natural numbers which lie between 10^7 and 10^8 and 28 Feb 2024, 02:30
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

53% (02:32) correct 47% (02:46) wrong based on 87 sessions

History

Date
Time
Result
Not Attempted Yet
­­Find the number of natural numbers which lie between 10^7 and 10^8 and which have products of their digits as 6?

A. 36
B. 55
C. 64
D. 81
E. 105


ShowHide Answer
Official Answer
C
User avatar
Shubhradeep
User avatar
Joined: 11 Jan 2024
Last visit: 07 Feb 2026
Posts: 120
Own Kudos:
247
 [2]
Given Kudos: 108
Location: India
Posts: 120
Kudos: 247
 [2]
Find the number of natural numbers which lie between 10^7 and 10^8 and 28 Feb 2024, 05:12
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6

we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
User avatar
unraveled
User avatar
Joined: 07 Mar 2019
Last visit: 10 Apr 2025
Posts: 2,706
Own Kudos:
2,329
 [1]
Given Kudos: 763
Location: India
WE:Sales (Energy)
Posts: 2,706
Kudos: 2,329
 [1]
Find the number of natural numbers which lie between 10^7 and 10^8 and 28 Feb 2024, 09:04
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Find the number of natural numbers which lie between 10^7 and 10^8 and which have products of their digits as 6?

A. 36
B. 55
C. 64
D. 81
E. 105

10^7 means eight digits in the number.
Thus, ways in which 6 is product of the eitht digits are 
A. Only digits 1,2 and 3 are used
A1. 1 2 3 _ _ _ _ _ = 6 numbers wherein the blanks are filled with 1 and 3 can take 6 places.
A2. 1 3 2 _ _ _ _ _ = 6 numbers wherein the blanks are filled with 1 and 2 can take 6 places.
A3. 2 1 3 _ _ _ _ _ = 6 numbers wherein the blanks are filled with 1 and 3 can take 6 places.
A4. 3 1 2 _ _ _ _ _ = 6 numbers wherein the blanks are filled with 1 and 2 can take 6 places.
A5. 1 1 2 3 _ _ _ _ = 5 numbers wherein the blanks are filled with 1 and 3 can take 5 places.
A6. 1 1 3 2 _ _ _ _ = 5 numbers wherein the blanks are filled with 1 and 2 can take 5 places.
A7. 1 1 1 2 3 _ _ _ = 4 numbers wherein the blanks are filled with 1 and 3 can take 4 places.
A8. 1 1 1 3 2 _ _ _ = 4 numbers wherein the blanks are filled with 1 and 2 can take 4 places.
A9. 1 1 1 1 2 3 _ _ = 3 numbers wherein the blanks are filled with 1 and 3 can take 3 places.
A10. 1 1 1 1 3 2 _ _ = 3 numbers wherein the blanks are filled with 1 and 2 can take 3 places. 
A11. 1 1 1 1 1 2 3 _ = 2 numbers wherein the blanks are filled with 1 and 3 can take 2 places.
A12. 1 1 1 1 1 3 2 _ = 2 numbers wherein the blanks are filled with 1 and 2 can take 2 places
A13. 1 1 1 1 1 1 2 3 = 2 numbers wherein the 2 and 3 between each other i.e. 2 places.
A14. 2 3 1 _ _ _ _ _ = 1 number wherein the blanks are filled with 1 and 1 can take 6 places.
A15. 3 2 1 _ _ _ _ _ = 1 numbers wherein the blanks are filled with 1 and 1 can take 6 places.

Total = 56 numbers

B. Only digits 1 and 6 are used
6 _ _ _ _ _ _ _ = 8 numbers wherein 1 takes the rest of the blank places and 6 moves in those 8 places.

Total = 56 + 8 = 64 numbers

IMO Answer C.
Gmatguy007
Joined: 31 Dec 1969
Last visit: -
Kudos: 0
Find the number of natural numbers which lie between 10^7 and 10^8 and 28 Feb 2024, 10:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shubhradeep
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6

we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
Show more
­Hey!
Very helpful your answer but I need your help to clarify something because I'm confused. I also think it's C, but regarding the fractions, in the first one the denominator is 6! because in the first set only 2 out of the 8 numbers (2 and 3) have product equal to 6?

If it is so, then why isn't it also in the second denominator?­
User avatar
Shubhradeep
User avatar
Joined: 11 Jan 2024
Last visit: 07 Feb 2026
Posts: 120
Own Kudos:
247
 [2]
Given Kudos: 108
Location: India
Posts: 120
Kudos: 247
 [2]
Find the number of natural numbers which lie between 10^7 and 10^8 and Updated on: 28 Feb 2024, 13:56
Originally posted by Shubhradeep on 28 Feb 2024, 13:27.
Last edited by Shubhradeep on 28 Feb 2024, 13:56, edited 2 times in total.
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Gmatguy007

Shubhradeep
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6

we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
­Hey!
Very helpful your answer but I need your help to clarify something because I'm confused. I also think it's C, but regarding the fractions, in the first one the denominator is 6! because in the first set only 2 out of the 8 numbers (2 and 3) have product equal to 6?

If it is so, then why isn't it also in the second denominator?­
Show more
­
It's not actually that way which you are assuming. It seems You are assuming that 2x3 makes 6, so 2 digits are gone that's why we make 6! in the denominator and same should be the case for the second one.

We have to find number of natural numbers which have products of their digits as 6.
We can make it this way - number of arrangements of digits possible with the factors of 6 - which is eventually 2, 3 in first case and 1,6 in second case.

Now it boils down to this problem-
Case 1: Number of possible arrangements with digits 1,1,1,1,1,1,2,3
Case 2: Number of possible arrangements with digits 1,1,1,1,1,1,1,6

I shall give an example so that you understand
Suppose, we have few letters L,E,T,T,E,R,S
How many ways you can arrange these letters( or how many words can you make with these leters)

Frequency of each letter:
L-> 1
E-> 2
T-> 2
R-> 1
S-> 1
Total 7 letters

So we can make \(\frac{7!}{(1!*2!*2!*1!*1!)}\) = \(\frac{7!}{(2!*2!)}\) words

This 2!x2! in the denominator is for the repeating letters E and T. So If we would have written 7! possible arrangements possible that would have been wrong because  'EE' or 'TT' is 1 combination, so in each case we have to divide by 2!

Same is with the cases in the given question.
There are 6 repeating 1's in the first case so divide by 6! and in the second case 7 repeating 1's in the second case so divide by 7!
 ­­­­­
Gmatguy007
Joined: 31 Dec 1969
Last visit: -
Own Kudos:
0
 [1]
Kudos: 0
 [1]
Find the number of natural numbers which lie between 10^7 and 10^8 and 28 Feb 2024, 13:44
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shubhradeep

Gmatguy007

Shubhradeep
Option C should be correct

\(10^{7}\)­ = 10000000 , so, we need 8 digit numbers which has 6 as the product of their digits.

6 = 2x3
6 = 1x6

we can have S1 = {1,1,1,1,1,1,2,3} and S2 = {1,1,1,1,1,1,1,6}

We can arrange the dogits of S1 in \(\frac{8!}{6!}\) = 7x8 = 56
We can arrange the dogits of S2 in \(\frac{8!}{7!}\) = 8x1 = 8

So, total number of natural numbers = (56+8) = 64

Hence, option C­
­Hey!
Very helpful your answer but I need your help to clarify something because I'm confused. I also think it's C, but regarding the fractions, in the first one the denominator is 6! because in the first set only 2 out of the 8 numbers (2 and 3) have product equal to 6?

If it is so, then why isn't it also in the second denominator?­
­
It's not actually that way which you are assuming. It seems You are assuming that 2x3 makes 6, so 2 digits are gone that's why we make 2! in the denominator and same should be the case for the second one.

We have to find number of natural numbers which have products of their digits as 6.
We can make it this way - number of arrangements of digits possible with the factors of 6 - which is eventually 2, 3 in first case and 1,6 in second case.

Now it boils down to this problem-
Case 1: Number of possible arrangements with digits 1,1,1,1,1,1,2,3
Case 2: Number of possible arrangements with digits 1,1,1,1,1,1,1,6

I shall give an example so that you understand
Suppose, we have few letters L,E,T,T,E,R,S
How many ways you can arrange these letters( or how many words can you make with these leters)

Frequency of each letter:
L-> 1
E-> 2
T-> 2
R-> 1
S-> 1
Total 7 letters

So we can make \(\frac{7!}{(1!*2!*2!*1!*1!)}\) = \(\frac{7!}{(2!*2!)}\) words

This 2!x2! in the denominator is for the repeating letters E and T. So If we would have written 7! possible arrangements possible that would have been wrong because  'EE' or 'TT' is 1 combination, so in each case we have to divide by 2!

Same is with the cases in the given question.
There are 6 repeating 1's in the first case so divide by 6! and in the second case 7 repeating 1's in the first case so divide by 7!
 ­
Show more
­
Yeah, I was assuming that since 2*3 = 6, so 2 digits are gone that's why we make 6! in the denominator and same should be the case for the second one. But through the example it's now crystal clear, thanks Shubhradeep
Moderators:
Bunuel
Math Expert
109830 posts
Krunaal
Tuck School Moderator
852 posts