What is the Relationship Between Factorials and Unit Digits?

[Saitama]

Homework Statement

Let $100!=N\cdot 10^n$. If N is relatively prime with 10 and unit digit of N is d, then n+d is equal to
A)26
B)28
C)30
D)32

Homework Equations

The Attempt at a Solution

I don't think it would be a good idea to expand the factorial and separately write out the factors of 10. I have got no idea about how to approach this problem.

Any help is appreciated. Thanks!

 

[sankalpmittal]

Homework Statement

Let $100!=N\cdot 10^n$. If N is relatively prime with 10 and unit digit of N is d, then n+d is equal to
A)26
B)28
C)30
D)32

Homework Equations

The Attempt at a Solution

I don't think it would be a good idea to expand the factorial and separately write out the factors of 10. I have got no idea about how to approach this problem.

Any help is appreciated. Thanks!

I may not be an expert but can you find index power of 10 in 100! ? You must be knowing the method of finding that. Many good course materials like "Vidyamandir" and RSM illustrate that.

 

[Ray Vickson]

Homework Statement

Let $100!=N\cdot 10^n$. If N is relatively prime with 10 and unit digit of N is d, then n+d is equal to
A)26
B)28
C)30
D)32

Homework Equations

The Attempt at a Solution

I don't think it would be a good idea to expand the factorial and separately write out the factors of 10. I have got no idea about how to approach this problem.

Any help is appreciated. Thanks!

In 100!, the multiples of 10 are 10,20,30,...,90, 100, so figuring out what power of 10 these produce is easy. Next, look at all the remaining even factors (2,4,6,8,12,...,18,22...,98) and all the remaining multiples of 5 (5,15,25,35,...,95). Some of these combine to give you factors 2*5 = 10, etc.

 

[Saitama]

In 100!, the multiples of 10 are 10,20,30,...,90, 100, so figuring out what power of 10 these produce is easy. Next, look at all the remaining even factors (2,4,6,...,98) and all the remaining multiples of 5 (5,15,25,35,...,95). Some of these combine to give you factors 2*5 = 10, etc.

I had thought of this earlier but wouldn't that be a bit tedious? This is a question from my test paper and I suppose there is a easier way. If I go by this method and if I miss even a single factor, I will end up with a wrong answer.

 

[ehild]

The numbers from 1-9 produce one zero at the end (2*5). So do the numbers from 11 to 19 (12*15), and so on: 10 zeros.
The product of 10, 20 ... 100 produce 12 zeros. (100 and 20*50 have to extra zeros)

Can you increase the number of zeros further at the end of 100! ?

The remaining numbers are 1,3,46,7,8,9 (mod 10) What is the last number of their product?
edit: 4 has been left out.

You have 10 such groups...

ehild

 

[Saitama]

The numbers from 1-9 produce one zero at the end (2*5). So do the numbers from 11 to 19 (12*15), and so on: 10 zeros.
The product of 10, 20 ... 100 produce 12 zeros. (100 and 20*50 have to extra zeros)

Can you increase the number of zeros further at the end of 100! ?

The remaining numbers are 1,3,6,7,8,9 (mod 10) What is the last number of their product?

You have 10 such groups...

ehild

From the 10 groups we have 2 zeroes from the groups: 21-29 and 71-79. This gives us a total of 24 zeroes.

The last digit of the product of numbers is 2.
This gives n+d=26.

Thanks a lot ehild!

EDIT: Woops, the unit digit of product of those numbers is 2. There are 10 groups so the unit digit of N is the unit digit of 2^(10) i.e. 4. Hence, the answer is 28.

 

[ehild]

From the 10 groups we have 2 zeroes from the groups: 21-29 and 71-79. This gives us a total of 24 zeroes.

The last digit of the product of numbers is 2.
This gives n+d=26.

Thanks a lot ehild!

EDIT: Woops, the unit digit of product of those numbers is 2. There are 10 groups so the unit digit of N is the unit digit of 2^(10) i.e. 4. Hence, the answer is 28.

I left out 4 from the remaining numbers... So the last digit of the product 1*3*4*6*7*8*9 is 8, but it is -2 mod 10, and its 10th power is 4 mod 10, so the result is the same.

ehild

 

[Saitama]

I left out 4 from the remaining numbers... So the last digit of the product 1*3*4*6*7*8*9 is 8, but it is -2 mod 10, and its 10th power is 4 mod 10, so the result is the same.

ehild

Thanks once again ehild!


Although I don't understand that modulo arithmetic notation. :tongue2:

 

[ehild]

Although I don't understand that modulo arithmetic notation. :tongue2:

_{4+10k, k integer}