## Dhaka Higher Secondary 2010/12

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**don't post problems (by starting a topic)**in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.

### 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. . . . . . . .

and the remainder is 1.

(vagsesh) is 2,3,4,5. . . . . . . .

Ibtehaz Shawon

BRAC University.

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

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.

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

i dont think its quite a good proof... Im working on it.

Ibtehaz Shawon

BRAC University.

BRAC University.

*long way to go .....*### 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?

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

Please

Please

**install LaTeX fonts**in your PC for better looking equations,**learn****how to write equations**, and**don't forget**to read Forum Guide and Rules.- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**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?

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.

BRAC University.

*long way to go .....*### Re: Dhaka Higher Secondary 2010/12

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

But, why

Every logical solution to a problem has its own beauty.

(

(

**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: Dhaka Higher Secondary 2010/12

$$x+2 \equiv 0(mod 8)$$

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.

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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