Dhaka Higher Secondary 2010/12

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2010/12

Unread post by BdMO » Tue Jan 18, 2011 2:07 pm

If $ \dfrac{x+2}{8}$ is an integer greater than $2$, find the remainder when $x$ is divided by $8$.

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leonardo shawon
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Re: Dhaka Higher Secondary 2010/12

Unread post by leonardo shawon » Tue Jan 18, 2011 4:20 pm

here x > 14
and the remainder is 1.
(vagsesh) is 2,3,4,5. . . . . . . .
Ibtehaz Shawon
BRAC University.

long way to go .....

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leonardo shawon
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Re: Dhaka Higher Secondary 2010/12

Unread post by leonardo shawon » Tue Jan 18, 2011 5:04 pm

{x+2}/8 > 8 so x > 14
x is divided by 8.
That means 16,24,32 . . . . . .
For x=16 the answer is 2 and remaminder 1
for x=24 answer 3 remainder 1
for x=32 answer 4 remainder 1
...... So, the remainder is 1
Ibtehaz Shawon
BRAC University.

long way to go .....

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leonardo shawon
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Re: Dhaka Higher Secondary 2010/12

Unread post by leonardo shawon » Tue Jan 18, 2011 5:08 pm

i dont think its quite a good proof... Im working on it. :-)
Ibtehaz Shawon
BRAC University.

long way to go .....

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Moon
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Re: Dhaka Higher Secondary 2010/12

Unread post by Moon » Tue Jan 18, 2011 7:09 pm

Please don't post more than once at the same time. You can always edit your post within 1 day.
BTW How did you find that the remainder is $1$ :?:. I think we have $x=8k-2$. So the remainder is?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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leonardo shawon
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Re: Dhaka Higher Secondary 2010/12

Unread post by leonardo shawon » Tue Jan 18, 2011 8:21 pm

okay i wont. I didnt find the edit option first.
Anyway... As BdMo said
x+2/8 > 2 from their i found that x > 14. And x will always be divided by 8. So, i checked for the multiple of 8 greater then 14. BTW, delete my unusal post if u r so kind.
Is there any problem bhaia?
Ibtehaz Shawon
BRAC University.

long way to go .....

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Zzzz
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Re: Dhaka Higher Secondary 2010/12

Unread post by Zzzz » Thu Jan 27, 2011 11:07 am

leonardo shawon wrote: And x will always be divided by 8.
I think, you meant 'x will be always divisible by 8' :?

But, why :?:
Every logical solution to a problem has its own beauty.
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Tahmid Hasan
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Re: Dhaka Higher Secondary 2010/12

Unread post by Tahmid Hasan » Thu Jan 27, 2011 8:43 pm

$$x+2 \equiv 0(mod 8)$$
or,$$x \equiv -2(mod8)$$
so,$$x \equiv 6(mod8)$$
so why make all that confusion?
actually the greater than 2 thingy was to draw everyone's attention into some inequality.
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