## 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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Joined: Tue Jan 18, 2011 1:31 pm

### Dhaka Higher Secondary 2010/12

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

leonardo shawon
Posts: 169
Joined: Sat Jan 01, 2011 4:59 pm
Location: Dhaka

### Re: Dhaka Higher Secondary 2010/12

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

long way to go .....

leonardo shawon
Posts: 169
Joined: Sat Jan 01, 2011 4:59 pm
Location: Dhaka

### Re: Dhaka Higher Secondary 2010/12

{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 .....

leonardo shawon
Posts: 169
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Location: Dhaka

### Re: Dhaka Higher Secondary 2010/12

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

long way to go .....

Moon
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### Re: Dhaka Higher Secondary 2010/12

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?
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leonardo shawon
Posts: 169
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Location: Dhaka

### Re: Dhaka Higher Secondary 2010/12

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 .....

Zzzz
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### Re: Dhaka Higher Secondary 2010/12

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
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm
$$x+2 \equiv 0(mod 8)$$
or,$$x \equiv -2(mod8)$$
so,$$x \equiv 6(mod8)$$