$42a+b=204$

$423a+b=2109$

Solving the simultaneous equations, we get $a=5, b=-6$.

Therefore, if we give $2017$, we will get $10094$.

## Search found 62 matches

- Thu Oct 31, 2019 8:57 pm
- Forum: Divisional Math Olympiad
- Topic: Feni Higher Secondary 2017 P9
- Replies:
**2** - Views:
**49608**

- Thu Oct 31, 2019 8:46 pm
- Forum: Divisional Math Olympiad
- Topic: Narayanganj Higher Secondary 2014 P5
- Replies:
**1** - Views:
**35114**

### Re: Narayanganj Higher Secondary 2014 P5

$1+...+10=55$ $11+...+20=100+55$ $21+...+30=200+55$ $31+...+40=300+55$ $41+...+50=400+55$ $51+...+60=500+55$ $61+...+70=600+55$ Adding=> $100(1+...+6)+55\times7=2485$ Now, $2610-2485=125$, which is not possible. Adding $71$ to $2485$, the summation of all the coupon numbers become $2556$. Therefore,...

- Thu Oct 31, 2019 8:30 pm
- Forum: Divisional Math Olympiad
- Topic: Narayanganj Higher Secondary 2014 P9
- Replies:
**2** - Views:
**36003**

### Re: Narayanganj Higher Secondary 2014 P9

There is a much more formal solution to this problem using Modular Arithmetic.

But I'm too lazy to type that long a solution!

But I'm too lazy to type that long a solution!

- Thu Oct 31, 2019 8:29 pm
- Forum: Divisional Math Olympiad
- Topic: Narayanganj Higher Secondary 2014 P9
- Replies:
**2** - Views:
**36003**

### Re: Narayanganj Higher Secondary 2014 P9

At least $8$ numbers are needed to be taken to ensure that the difference between at least two numbers is divisible by $7$. Now, if we take one more number, the difference between this and one of the "original" $8$ numbers is divisible by $7$. Going on, we can see that at least $17$ numbers are requ...

- Thu Oct 31, 2019 7:56 pm
- Forum: Higher Secondary Level
- Topic: Chittagong-Higher secondary 2016
- Replies:
**3** - Views:
**51846**

### Re: Chittagong-Higher secondary 2016

The solution is as such=> Let $ABCD$ be a rectangle where $AB$ is $10$ and $BC$ is $11$ Take the base of an isosceles triangle to be $BC$. Let the longer sides be $x$. So, $x^2=(5\frac{1}{2})^2 + 100$ $x^2= 130\frac {1}{4}$ Therefore, $x > 11$ Now, it is confirmed that $BC$ can be used as the base o...

- Thu Oct 31, 2019 7:26 pm
- Forum: Higher Secondary Level
- Topic: Chittagong-Higher secondary 2016
- Replies:
**3** - Views:
**51846**

### Re: Chittagong-Higher secondary 2016

$(221\sqrt 3 - 400)$ equals to a negative value. So your solution is wrong.

Go back to the drawing boards!

Go back to the drawing boards!

- Thu Oct 31, 2019 7:16 pm
- Forum: Higher Secondary Level
- Topic: Probablity of coin flip
- Replies:
**2** - Views:
**38328**

### Re: Probablity of coin flip

But then again, I'm not entirely sure about the conditions of the problem.

Maybe it's different from the way I have understood and structured the solution.

More details (if any) will be helpful.

Maybe it's different from the way I have understood and structured the solution.

More details (if any) will be helpful.

- Thu Oct 31, 2019 7:11 pm
- Forum: Higher Secondary Level
- Topic: Probablity of coin flip
- Replies:
**2** - Views:
**38328**

### Re: Probablity of coin flip

Probability of the Head appearing thrice consecutively is $\frac{1}{8}$.

So, if the coin is tossed $24 (3 \times 8)$ times, once the Head will appear thrice consecutively.

So, if the coin is tossed $24 (3 \times 8)$ times, once the Head will appear thrice consecutively.

- Thu Oct 17, 2019 3:16 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2019/P3
- Replies:
**5** - Views:
**58076**

### Re: IMO 2019/P3

DAMN!!!!!!! This forum has got so dead... It needs a serious waking call... Maybe we should consider calling Thanos... to invade this forum. :twisted: Maybe then the "Avengers" (admins, moderators and former-active members) might turn their focus to this forum, again. :lol: (P.S: I don't know whethe...

- Thu Oct 17, 2019 3:08 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2019/P3
- Replies:
**5** - Views:
**58076**

### Re: IMO 2019/P3

BTW, why haven't anyone responded with their solutions?