কয়েকটা অঙ্ক পেলাম যেগুলো খুব মজার মনে হলো, তাই এখানে লিখে দিবো।

Q. If $1<a$, then show that $1<a<a^2$

Solution:

$1<a$…….[i]

$\implies 1<a^2$…….[ii]

We know for $a>1$

$(a-1)^2> 0$

$\implies a^2–2a+1> 0$

$\implies a^2-a-a+1> 0$

Since $a>1, -a+1<0$

So it must be true that

$a^2-a>0$, using dominating term technique

$\implies a^2>a$

$\implies a<a^2$……[iii]

Combining the relationships [i], [ii] and [iii]

$1<a<a^2$

Q.E.D

Q. What is the remainder when $34^{31^{301}}$ is divided by $9$?

Solution:

Euler’s Totient Theorem:

Given that $a>0, m>0, a,m\in\mathbb{Z}$ and $gcd(a,m)=1$, then $a^{\phi(m)}\equiv 1 \mod m$

$\phi(9)=\phi(3^2)=3^2–3^{2–1}=9–3=6$

$34\equiv 7\mod 9$

$\implies 34^{31}\equiv 7^{31}\mod 9$

$\implies 34^{31}\equiv 7^{6\times 5+1}\mod 9$

$\implies 34^{31}\equiv 7\mod 9$

$\implies 34^{31^{301}}\equiv 7^{301}\mod 9$

$\implies 34^{31^{301}}\equiv 7^{6\times50+1}\mod 9$

$\implies 34^{31^{301}}\equiv \boxed{7}\mod 9$

Q. What is the remainder when $11^{10}-1$ is divided by $100$?

Solution:

Euler’s Totient theorem won’t work here. I remember a rule in number theory. If I don’t know how to solve a problem, I’ll have to look for a pattern, so I’ll try something else….

$11 \equiv 11\mod 100$

$11^2 \equiv 21 \mod 100$

$11^3 \equiv 31 \mod 100$

$…….$

$…….$

$11^9 \equiv 91 \mod 100$

$11^{10} \equiv \boxed{1}01 \mod 100$

Notice that the result of $11^{10} \mod 100$ is $1$ and that is all I wanted to clarify, although the last three digits must be $101$, according to the pattern we found above.

So

$(11^{10}-1)\mod 100=11^{10}\mod 100–1\mod 100=(1-1)\mod 100$

Hence $11^{10}-1\equiv 0 \mod 100$

Q. How many $5$ digit numbers (non repeating) can be created such that they are divisible by $3,4$ and $11$?

Solution:

There are $5\times 4\times 3\times 2\times 1 = 120$ such $5$ digit numbers that can be made up of $4,5,6,7,8$ without repeating.

Now we just need to find how many among these $120$ numbers, are divisible by $lcm(3,4,11)$ or $132$

Notice that $5$ digit numbers consisting of $4,5,6,7,8$ are always divisible by $3$, since $4+5+6+7+8=30$ which is divisible by $3$.

A number is divisible by $4$, if the last two digits are divisible by $4$. So we can have

_ _ _ 4 8 $= 3! = 6$

_ _ _ 5 6 $= 3! = 6$

_ _ _ 6 4 $= 3! = 6$

_ _ _ 7 6 $= 3! = 6$

_ _ _ 8 4 $= 3! = 6$

_________________
Total numbers $= 30$

So we know that $30$ numbers out of those $120$ are divisible by 3 and 4.

Now comes the hardest part, and I see no other way out of it. Someone remember the divisibility rule by $11$? It’s this:

Take the absolute value of the difference between the sum of the digits in odd position and the sum of the digits in the even position, if it results in $0$ or $11$, then that number is a multiple of $11$.

But some of our work is done in the second step when we looked for divisibility by $4$. Still too much work, I don’t want to check $30$ numbers but I don’t think I can code up nicely that will even bring me up until here.

Let us start at random, I am checking with calculator now

_ _ _ 4 8, none of the numbers are divisible by 11.

_ _ _ 5 6, none of the numbers are divisible by 11.

_ _ _ 6 4, no numbers are divisible by 11

_ _ _ 7 6, we have 48576, 58476 that matches the criteria

_ _ _ 84, we have 57684, 67584 matching the criteria

So, there are only $4$ such numbers matching our criteria.

আরও অনেক অঙ্ক পাই সারাদিন ধরে, সব লিখলে, পড়তে পড়তে বিরক্তি ধরে যেতে পারে, এগুলো নিয়ে আবার পরে লিখব। সমস্যার খোঁজে থাকি, সমস্যার তো অভাব নেই।

ভালো থাকবেন সবাই।

Author: Awnon Bhowmik

I know very little to be proud about it. Mathematics enthusiast, possess a lust for mathematical/computational knowledge