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!

#NumberTheory

Easy Math Editor

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Comments

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

Mahdi Raza - 11 months, 4 weeks ago

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

Mahdi Raza - 11 months, 4 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!

Vinayak Srivastava - 11 months, 4 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

Mahdi Raza - 11 months, 4 weeks ago

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

Vinayak Srivastava - 11 months, 4 weeks ago

@Mahdi Raza – 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!

Vinayak Srivastava - 11 months, 4 weeks ago

@Vinayak Srivastava – Yeah Márton has done that. Great!

Mahdi Raza - 11 months, 3 weeks ago

@Mahdi Raza – I did it first :( But good thinking by you both!

Vinayak Srivastava - 11 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 $(1+2^2+6^2+24^2+120^2)\;mod\;1152=41$ .

A Former Brilliant Member - 11 months, 4 weeks ago

Nice!

Mahdi Raza - 11 months, 3 weeks ago

@David Vreken, @Pi Han Goh, @Finnley Paolella, @Mahdi Raza, @Páll Márton , @Zakir Husain

Vinayak Srivastava - 11 months, 4 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$