$21982145917308330487013369$ is the 13^{th} power of which positive integer ?
Na..Na. Don't use calculator.
13th Power
- Fahim Shahriar
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Name: Fahim Shahriar Shakkhor
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Re: 13th Power
Let $a^{13}=21982145917308330487013369$.Fahim Shahriar wrote:$21982145917308330487013369$ is the 13^{th} power of which positive integer ?
Na..Na. Don't use calculator.
$a^{13}\equiv a\pmod {13}\Longleftrightarrow a\equiv 11\pmod{13}$
So $a=11,24,\cdots\; 76, 89$
Now prove $76^{13}<10^{25}<a^{13}$
So $a=89$.
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- Fahim Shahriar
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Re: 13th Power
How did you find $a \equiv 11 (mod 13)$ ?
Did you use divisibility rule ? or what?
Did you use divisibility rule ? or what?
Name: Fahim Shahriar Shakkhor
Notre Dame College
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Re: 13th Power
হুম, রুলই বলা যায়। তের দিয়ে ভাগের নিয়ম তৈরি করে নেন না। (সাহায্য: $1000\equiv -1\; (\bmod \;13)$)Fahim Shahriar wrote:How did you find $a \equiv 11 (mod 13)$ ?
Did you use divisibility rule ? or what?
এরপর ঐ বিশাল সংখ্যাটারে বাঁশ দেন।
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- Fahim Shahriar
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Re: 13th Power
$10^{13} < a^{13} < 100^{13}$. So $a$ is a $2$ digit number. Taking (mod 10) we can observe that $a$'s last digit is $9$. We can write $a=10k-1$ ;where $k$ is an integer & $2 \leq k \leq 10$.
$(10k-1)^{13} = (10k)^{13} -.....+ 13*10k*1-1$
We will now work on last $2$ digits.
$130k-1 = \_\_ 69$
$k=9$ satisfies it. So $a=89$
$(10k-1)^{13} = (10k)^{13} -.....+ 13*10k*1-1$
We will now work on last $2$ digits.
$130k-1 = \_\_ 69$
$k=9$ satisfies it. So $a=89$
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College