# Some questions of Number Theory

I wish to know how to solve these problems. Any help will be appreciated!

## Question 1

Find the total number of natural numbers $$n$$ for which $$111$$ divides $$16^n-1$$ where $$n$$ is less than $$1000$$.

Answer: $111$

## Question 2

Find the remainder when $(1!)^2+(2!)^2+(3!)^2 \dots (100!)^2$ is divided by $1152$.

Answer: $41$

## Question 3

Find the remainder when $3^{21}+9^{21}+27^{21}+81^{21}$ is divided by $3^{20}+1$.

Answer: $60$

I have also provided the answers, but I wish to know the proper method to solve this, and if there is some trick to solve these questions in general, any opinions are welcome!

Note by Vinayak Srivastava
6 months, 3 weeks ago

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You might need to learn Modular arithmetic which is the basic requirement for such questions. Then explore number theory practice on brilliant.

• Modular Arithmetic
• Parity of Integers
• Number Theory
• Chinese Remainder Theorem
• Linear Diophantine Equations
• Fermat's Little Theorem
• Lucas' Theorem
• Sum of Squares Theorems
• Euler's Totient Function
• Euler's Theorem
• Finding the Last Digit of a Power
• Order of an Element
• Primitive Roots
• Lifting The Exponent

- 6 months, 3 weeks ago

The answer to these specific questions can be found on the internet as well

- 6 months, 3 weeks ago

Can you tell the solution of second one also? I think I understand the solution of the third problem given on Quora. Thanks for these two anyway!

- 6 months, 3 weeks ago

Umm.. I wasn't able to find for the second one. I've seen a similar question where the expression is to be divided by 100. And that is easy because after 10!, all the numbers are divisible by 100. And you can do it that way. Do tell if you find an answer to the second question

- 6 months, 3 weeks ago

OK, I'll wait for others to see this. Thanks for these two solutions!

- 6 months, 3 weeks ago

I found it! I just observed that $(6!)^2$ and so on is divisible by $1152$. So I just squared the first five terms on calculator, and got the correct answer! Your comment was a good hint on what to do! Thanks a lot!

- 6 months, 3 weeks ago

Yeah Márton has done that. Great!

- 6 months, 3 weeks ago

I did it first :( But good thinking by you both!

- 6 months, 3 weeks ago

I solved the problems as @Mahdi Raza linked them. The solution for the second question: $1152=2^7\cdot 3^2$. So from $(6!)^2$ each square is divisible by 1152. Therefore the solution is $(1+2^2+6^2+24^2+120^2)\;mod\;1152=41$.

- 6 months, 3 weeks ago

Nice!

- 6 months, 3 weeks ago