Find the number of natural numbers which lie between 10^7 and 10^8

Question

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

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kb26 · Answer

10^7 = 10000000

10^8 = 100000000

so, in an 8 digit number product of 6 can achieved only if number contains only 1,2,3 or 1,6

case 1,6: 61111111 is a possible number and 6 can be arranged in 8 possible ways. so, 8 cases.

case 1,2,3: 23111111 is a possible number and 2,3 can be arranged as 8×7 = 56 ways.

so, total = 56+8 = 64 cases

MakB · Answer

Firstly, we need to find digits whose product equals to 6: 
 1,6 or 2,3.
 
Now, since 10^8 is highest number, so the number of digits in the required numbers will be equal to number of digits in 10^7, i.e, 8.

Possible outcomes include:
 23111111  (product of digits=6)

We have to check permutations of 2&3 in this outcome:
 8x7 = 56

Other possible outcome:
 61111111

Permutations of this outcome:
 8

Total permutations:
 8+56=64

Hence, C

Kevi · Answer

Find the number of natural numbers which lie between 10^7 and 10^8 and which have products of their digits as 6?

We can get 6 by following products: 
1*1*6
1*2*3

Number of digits in which lie between 10^7 and 10^8 are all 8 digits numbers only. 
Possible cases:
- 61111111 
In this, 6 can be arranged in any possible 8 position. Total possible ways= 8 
- 23111111 
In this, 2 can be arranged in 8 positions and 3 in remaining 7 positions. Total possible ways= 8*7=56

We have a total of (56+8= 64) possible cases where the numbers lie between 10^7 and 10^8 and have products of their digits as 6.

C

Find the number of natural numbers which lie between 10^7 and 10^8

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Find the number of natural numbers which lie between 10^7 and 10^8