Answer:
(c) 33
Step-by-step explanation:
hope it helps you can do it thanks
mark me as brainlist
The technique of polynomial factorization is the very idea we should try here. Here, I will explain it.
[tex]\;[/tex]
Let's see an example.
[tex]\text{$\rm {a}^{n}+{b}^{n}$}[/tex]
What can we substitute for it?
Let's think of an easy solution.
[tex]\text{$\rm {b}^{n}=-{a}^{n}$}[/tex]
n is natural, and the signs of powers are different.
[tex]\boxed{\Longrightarrow \rm b=-a$ where $\rm n$ is odd.$}[/tex]
It suggests that the factor is [tex]\text{$\rm (a+b)$}[/tex].
Now through synthetic division, we know that the following identity proves true.
[tex]\boxed{\text{$\rm \Longrightarrow {a}^{n}+{b}^{n}=(a+b)({a}^{n-1}-{a}^{n-2}b+\cdots-a{b}^{n-2}+{b}^{n-1})$}}[/tex]
This makes the problem solvable because the exponents may become odd.
Given,
[tex]\text{$\rm 3^{50}+5^{50}$}[/tex]
[tex]\text{$\rm ={({3}^{2})}^{25}+{({5}^{2})}^{25}$}[/tex]
[tex]\text{$\rm ={9}^{25}+{25}^{25}$}[/tex]
Consider,
[tex]\text{$\rm {9}^{25}+{25}^{25}$}[/tex]
[tex]\text{$\rm =(9+25)\cdot({9}^{24}-{9}^{23}\cdot {25}+\dots-9\cdot {25}^{23}+{25}^{24})$}[/tex]
[tex]\text{$\rm =34\cdot({9}^{24}-{9}^{23}\cdot {25}+\dots-9\cdot {25}^{23}+{25}^{24})$}[/tex]
So, the number is divisible by 34.
Have a good day, and keep learning!
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
(c) 33
Step-by-step explanation:
hope it helps you can do it thanks
mark me as brainlist
Verified answer
The technique of polynomial factorization is the very idea we should try here. Here, I will explain it.
[tex]\;[/tex]
Let's see an example.
[tex]\text{$\rm {a}^{n}+{b}^{n}$}[/tex]
What can we substitute for it?
[tex]\;[/tex]
Let's think of an easy solution.
[tex]\text{$\rm {b}^{n}=-{a}^{n}$}[/tex]
n is natural, and the signs of powers are different.
[tex]\boxed{\Longrightarrow \rm b=-a$ where $\rm n$ is odd.$}[/tex]
[tex]\;[/tex]
It suggests that the factor is [tex]\text{$\rm (a+b)$}[/tex].
Now through synthetic division, we know that the following identity proves true.
[tex]\boxed{\text{$\rm \Longrightarrow {a}^{n}+{b}^{n}=(a+b)({a}^{n-1}-{a}^{n-2}b+\cdots-a{b}^{n-2}+{b}^{n-1})$}}[/tex]
[tex]\;[/tex]
This makes the problem solvable because the exponents may become odd.
[tex]\;[/tex]
Given,
[tex]\text{$\rm 3^{50}+5^{50}$}[/tex]
[tex]\text{$\rm ={({3}^{2})}^{25}+{({5}^{2})}^{25}$}[/tex]
[tex]\text{$\rm ={9}^{25}+{25}^{25}$}[/tex]
[tex]\;[/tex]
Consider,
[tex]\boxed{\text{$\rm \Longrightarrow {a}^{n}+{b}^{n}=(a+b)({a}^{n-1}-{a}^{n-2}b+\cdots-a{b}^{n-2}+{b}^{n-1})$}}[/tex]
[tex]\;[/tex]
Given,
[tex]\text{$\rm {9}^{25}+{25}^{25}$}[/tex]
[tex]\text{$\rm =(9+25)\cdot({9}^{24}-{9}^{23}\cdot {25}+\dots-9\cdot {25}^{23}+{25}^{24})$}[/tex]
[tex]\text{$\rm =34\cdot({9}^{24}-{9}^{23}\cdot {25}+\dots-9\cdot {25}^{23}+{25}^{24})$}[/tex]
[tex]\;[/tex]
So, the number is divisible by 34.
[tex]\;[/tex]
Have a good day, and keep learning!