An easy problem !!
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\(N\) একটি জোড় সংখ্যা ।
প্রমান করো যে , \(48 | N(N^2 + 20)\)
প্রমান করো যে , \(48 | N(N^2 + 20)\)
Last edited by Kiriti on Tue Feb 18, 2014 5:15 pm, edited 1 time in total.
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- asif e elahi
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Re: An easy problem !!
If $N=2,N(N^{2}+20)=48$ which is not divisible by $68$.
- Fatin Farhan
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Re: An easy problem !!
I think the question is wrong. It would be find all $$N$$ such that: $$N$$ একটি জোড় সংখ্যা ,$$68 | N(N^2 + 20)$$.
Now,
$$N=2m$$.
$$N(N^2+20)=4m(2m^2+10)$$.
So,$$ 17|m$$.
$$m=17k$$
$$N=34k$$
Now,
$$N=2m$$.
$$N(N^2+20)=4m(2m^2+10)$$.
So,$$ 17|m$$.
$$m=17k$$
$$N=34k$$
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Re: An easy problem !!
oh! It will be \(48 | N(N^2 +20)\)
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- Fatin Farhan
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Re: An easy problem !!
$$N=2m$$.
$$N(N^2+20)=8m(m^2+5)$$.
(1.1)If $$m \equiv 0 \pmod2$$ and $$m \equiv 0 \pmod3 $$ then $$8m(m^2+5)$$ is divisible by 48.
(1.2)If $$m \equiv 0 \pmod2$$ and $$m \equiv 1,2 \pmod3 $$ then $$m^2+5 \equiv 0 \pmod 3$$. So, $$8m(m^2+5)$$ is divisible by 48.
(2.1) If $$m \equiv1 \pmod2$$ and $$m \equiv 0 \pmod3$$ then $$8m(m^2+5)$$ is divisible by 48.
(2.2) If $$m \equiv 1 \pmod 2$$ and $$m \equiv 1,2 \pmod 3$$ then $$m^2+5 \equiv 0 \pmod 3$$, so 8m(m^2+5) is divisible by 48.
$$\therefore
N(N^2+20)$$ is divisible by 48.
$$N(N^2+20)=8m(m^2+5)$$.
(1.1)If $$m \equiv 0 \pmod2$$ and $$m \equiv 0 \pmod3 $$ then $$8m(m^2+5)$$ is divisible by 48.
(1.2)If $$m \equiv 0 \pmod2$$ and $$m \equiv 1,2 \pmod3 $$ then $$m^2+5 \equiv 0 \pmod 3$$. So, $$8m(m^2+5)$$ is divisible by 48.
(2.1) If $$m \equiv1 \pmod2$$ and $$m \equiv 0 \pmod3$$ then $$8m(m^2+5)$$ is divisible by 48.
(2.2) If $$m \equiv 1 \pmod 2$$ and $$m \equiv 1,2 \pmod 3$$ then $$m^2+5 \equiv 0 \pmod 3$$, so 8m(m^2+5) is divisible by 48.
$$\therefore
N(N^2+20)$$ is divisible by 48.
Last edited by Fatin Farhan on Tue Feb 18, 2014 5:52 pm, edited 3 times in total.
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- asif e elahi
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Re: An easy problem !!
If $m \equiv0\pmod2$,then $m^2+5 \equiv 0\pmod2$,not $3$Fatin Farhan wrote: (1)If $$m \equiv 0 \pmod2$$ then $$m^2+5 \equiv 0\pmod3$$ So, $$8m(m^2+5)$$ is divisible by 48.
.
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Re: An easy problem !!
Edited
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Re: An easy problem !!
\(N = 2m\)
Now, \(N(N^2 +20) = N^3 + 20N = 8m^3 + 40m = 8(m^3 +5m)\)
Here If \( 6 | (m^3 +5m)\) then \(8(m^3 +5m) \equiv 0 \pmod {48}\)
So, we get 5 cases ,
(1)If \(m \equiv 0\pmod 6\) then \(0^3 + 5 \times 0 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(2)If \(m \equiv 1\pmod 6\) then \(1^3 + 5 \times 1 \equiv 6 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(3)If \(m \equiv 2\pmod 6\) then \(2^3 + 5 \times 2 \equiv 18 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(4)If \(m \equiv -2\pmod 6\) then \(-2^3 + 5 \times -2 \equiv -18 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(5)If \(m \equiv -1\pmod 6\) then \(-1^3 + 5 \times -1 \equiv 6 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
So, ultimately \(6|(m^3 +5m)\) that means \(48 | N(N^2 + 20)\)
Now, \(N(N^2 +20) = N^3 + 20N = 8m^3 + 40m = 8(m^3 +5m)\)
Here If \( 6 | (m^3 +5m)\) then \(8(m^3 +5m) \equiv 0 \pmod {48}\)
So, we get 5 cases ,
(1)If \(m \equiv 0\pmod 6\) then \(0^3 + 5 \times 0 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(2)If \(m \equiv 1\pmod 6\) then \(1^3 + 5 \times 1 \equiv 6 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(3)If \(m \equiv 2\pmod 6\) then \(2^3 + 5 \times 2 \equiv 18 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(4)If \(m \equiv -2\pmod 6\) then \(-2^3 + 5 \times -2 \equiv -18 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(5)If \(m \equiv -1\pmod 6\) then \(-1^3 + 5 \times -1 \equiv 6 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
So, ultimately \(6|(m^3 +5m)\) that means \(48 | N(N^2 + 20)\)
"Education is the most powerful weapon which you can use to change the world"- Nelson Mandela
Re: An easy problem !!
$N=2n$
$N(N^{2}+20)=2n(4n^{2}+20)=8n(n^{2}+5)$
now $48\mid N(N^{2}+20)=48\mid 8n(n^{2}+5)=6\mid n(n^{2}+5)$
so,we need to prove that $6$ divides $n(n^{2}+5)$
among two integers $n$ and $n^{2}+5$ one must be even . that means $2$ divides $n(n^{2}+5)$
$n(n^{2}+5)=n^{3}+5n$
now,$n\equiv 0/1/2 (mod3)$
$n^{3}\equiv 0/1/8\equiv 0/1/2 (mod3)$..........(1)
$5n\equiv 0/5/10\equiv 0/2/1 (mod3)$..........(2)
compare (1) and (2);
$n^{3}+5n\equiv 0+0/1+2/2+1\equiv 0/0/0 (mod3)$ . that means $3$ divides $n(n^{2}+5)$
finally;
$6$ divides $n(n^{2}+5)$ $\Leftrightarrow $ $48$ divides $8n(n^{2}+5)$ $\Leftrightarrow $ $48$ divides $N(N^{2}+20)$ ...
$N(N^{2}+20)=2n(4n^{2}+20)=8n(n^{2}+5)$
now $48\mid N(N^{2}+20)=48\mid 8n(n^{2}+5)=6\mid n(n^{2}+5)$
so,we need to prove that $6$ divides $n(n^{2}+5)$
among two integers $n$ and $n^{2}+5$ one must be even . that means $2$ divides $n(n^{2}+5)$
$n(n^{2}+5)=n^{3}+5n$
now,$n\equiv 0/1/2 (mod3)$
$n^{3}\equiv 0/1/8\equiv 0/1/2 (mod3)$..........(1)
$5n\equiv 0/5/10\equiv 0/2/1 (mod3)$..........(2)
compare (1) and (2);
$n^{3}+5n\equiv 0+0/1+2/2+1\equiv 0/0/0 (mod3)$ . that means $3$ divides $n(n^{2}+5)$
finally;
$6$ divides $n(n^{2}+5)$ $\Leftrightarrow $ $48$ divides $8n(n^{2}+5)$ $\Leftrightarrow $ $48$ divides $N(N^{2}+20)$ ...
Re: An easy problem !!
A little less casework bacchasKiriti wrote: So, we get 5 cases ,
(1)If \(m \equiv 0\pmod 6\) then \(0^3 + 5 \times 0 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(2)If \(m \equiv 1\pmod 6\) then \(1^3 + 5 \times 1 \equiv 6 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(3)If \(m \equiv 2\pmod 6\) then \(2^3 + 5 \times 2 \equiv 18 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(4)If \(m \equiv -2\pmod 6\) then \(-2^3 + 5 \times -2 \equiv -18 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
(5)If \(m \equiv -1\pmod 6\) then \(-1^3 + 5 \times -1 \equiv 6 \equiv 0 \pmod 6\) . Then \(48 |N(N^2 + 20)\)
So, ultimately \(6|(m^3 +5m)\) that means \(48 | N(N^2 + 20)\)
\[m(m^2+5) \equiv m(m^2-1) \equiv m(m-1)(m+1) \equiv 0 \pmod 6\]
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