RMO-2010/3

For students of class 6-8 (age 12 to 14)
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Fatin Farhan
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RMO-2010/3

Unread post by Fatin Farhan » Sun Mar 02, 2014 12:08 pm

Find the number of $$4 \text{ digit numbers (in base 10)}$$ having non-zero digits and which are divisible by $$4$$ but not by $$8$$.
"The box said 'Requires Windows XP or better'. So I installed L$$i$$nux...:p"

Nirjhor
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Re: RMO-2010/3

Unread post by Nirjhor » Mon Mar 03, 2014 12:35 pm

Since, $[4,8]=8$, by PIE, the answer is: \[\left.\left(\left\lfloor \dfrac{9999}{4} \right\rfloor-\left\lfloor \dfrac{999}{4} \right\rfloor\right)\right/2~=~\dfrac{2250}2~ = ~\boxed{1125}\]
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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Fatin Farhan
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Re: RMO-2010/3

Unread post by Fatin Farhan » Mon Mar 03, 2014 5:16 pm

Nirjhor wrote:Since, $[4,8]=8$, by PIE, the answer is: \[\left.\left(\left\lfloor \dfrac{9999}{4} \right\rfloor-\left\lfloor \dfrac{999}{4} \right\rfloor\right)\right/2~=~\dfrac{2250}2~ = ~\boxed{1125}\]
If you kindly explain what you did. Because the answer is incorrect :?
"The box said 'Requires Windows XP or better'. So I installed L$$i$$nux...:p"

Tahmid
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Re: RMO-2010/3

Unread post by Tahmid » Mon Mar 03, 2014 10:14 pm

@fatin
i got my answer 729
is it correct??? :?:

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Thanic Nur Samin
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Re: RMO-2010/3

Unread post by Thanic Nur Samin » Thu Mar 06, 2014 10:55 am

Nirjhor wrote:Since, $[4,8]=8$, by PIE, the answer is: \[\left.\left(\left\lfloor \dfrac{9999}{4} \right\rfloor-\left\lfloor \dfrac{999}{4} \right\rfloor\right)\right/2~=~\dfrac{2250}2~ = ~\boxed{1125}\]
Nirjhor is right. Using brute force ensures us.

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Last edited by *Mahi* on Sun Mar 09, 2014 1:33 pm, edited 1 time in total.
Reason: Collapsed the list as it was making scrolling to the next post difficult.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Nirjhor
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Re: RMO-2010/3

Unread post by Nirjhor » Thu Mar 06, 2014 12:05 pm

Sorry my solution was not complete. I didn't notice the word "non-zero digits". Without that condition it's $1125$ and a python program ensures that.

Here's another one: we have $18$ multiples of $4$ in range $[1,100]$ with non-zero digits. When the last number formed by the last two digits is of the form $4(2k-1)$ (there are $9$ of them), the third digit can be any even digit (there are $4$ of them) since $100\cdot 2m+4(2k-1)=200m+8k-4\equiv 4\pmod 8$. And if the number formed by the last two digits is of the form $4\cdot 2k=8k$ (there are $9$ of them) the third digit can be any odd digit since $100(2m-1)+8k=200m+8k-100\equiv 4\pmod 8$. The fourth digit can be any non-zero one. there are $9$ of them. In total, we have: \[(9\times 4\times 9)+(9\times 5\times 9)=\boxed{729}\]
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Marzuq
Posts:9
Joined:Mon Apr 05, 2021 4:30 pm

Re: RMO-2010/3

Unread post by Marzuq » Sat Apr 17, 2021 10:37 am

My solution :
divisibility of 4 : if the last two digit is divisible by 4. Then the whole number is divisible by 4
Divisibility by 8 : if the last two digit is divisible by 4 but not by 8 and the 3rd digit is odd. Or if the last digit is divisible by 4 and 8 and the 3rd digit is even . Then the number is divisible by 8

Some non-zero 2 digits multiple of 4 but not 8 : $12,28,36,44,52,68,76,84,92$ [total 9]

Some non-zero 2 digit multiple of 4 and 8: $16,24,32,48,56,64,72,88,96$ [total 9]

Case 1: last two digit is divisible by 4 but not 8 and 3rd digit is even

Let the four digit number be $abcd$
$a$ has $9$ choice [$1,2,3,\dots,9$]
$b$ has $5$ choice [$0,2,4,6,8$]
$cd$ has $9$ choice [mentioned above]
Total = $ 9 × 5 × 9$

Case 2 : the last two digit is divisible by 4 and 8 and the 3rd digit is odd

Let the number be $abcd$
$a$ has $9$ option [$1,2,\dots,9$]
$b$ has $5$ option [$1,3,5,7,9$]
$cd$ has $9$ option [mentioned above]
Total = $9 × 5 × 9$

From the two cases , total = $9 × 5 × 9 + 9 × 5 × 9 = 9 × 10 × 9 = 81 × 10 = 810$

Am I wrong? :?

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Mehrab4226
Posts:230
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Location:Dhaka, Bangladesh

Re: RMO-2010/3

Unread post by Mehrab4226 » Sat Apr 17, 2021 10:42 am

Marzuq wrote:
Sat Apr 17, 2021 10:37 am
My solution :
divisibility of 4 : if the last two digit is divisible by 4. Then the whole number is divisible by 4
Divisibility by 8 : if the last two digit is divisible by 4 but not by 8 and the 3rd digit is odd. Or if the last digit is divisible by 4 and 8 and the 3rd digit is even . Then the number is divisible by 8

Some non-zero 2 digits multiple of 4 but not 8 : $12,28,36,44,52,68,76,84,92$ [total 9]

Some non-zero 2 digit multiple of 4 and 8: $16,24,32,48,56,64,72,88,96$ [total 9]

Case 1: last two digit is divisible by 4 but not 8 and 3rd digit is even

Let the four digit number be $abcd$
$a$ has $9$ choice [$1,2,3,\dots,9$]
$b$ has $5$ choice [$0,2,4,6,8$]
$cd$ has $9$ choice [mentioned above]
Total = $ 9 × 5 × 9$

Case 2 : the last two digit is divisible by 4 and 8 and the 3rd digit is odd

Let the number be $abcd$
$a$ has $9$ option [$1,2,\dots,9$]
$b$ has $5$ option [$1,3,5,7,9$]
$cd$ has $9$ option [mentioned above]
Total = $9 × 5 × 9$

From the two cases , total = $9 × 5 × 9 + 9 × 5 × 9 = 9 × 10 × 9 = 81 × 10 = 810$

Am I wrong? :?
The answer should be $1125$ as brute force says it.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Marzuq
Posts:9
Joined:Mon Apr 05, 2021 4:30 pm

Re: RMO-2010/3

Unread post by Marzuq » Sat Apr 17, 2021 8:13 pm

Mehrab4226 wrote:
Sat Apr 17, 2021 10:42 am
Marzuq wrote:
Sat Apr 17, 2021 10:37 am
My solution :
divisibility of 4 : if the last two digit is divisible by 4. Then the whole number is divisible by 4
Divisibility by 8 : if the last two digit is divisible by 4 but not by 8 and the 3rd digit is odd. Or if the last digit is divisible by 4 and 8 and the 3rd digit is even . Then the number is divisible by 8

Some non-zero 2 digits multiple of 4 but not 8 : $12,28,36,44,52,68,76,84,92$ [total 9]

Some non-zero 2 digit multiple of 4 and 8: $16,24,32,48,56,64,72,88,96$ [total 9]

Case 1: last two digit is divisible by 4 but not 8 and 3rd digit is even

Let the four digit number be $abcd$
$a$ has $9$ choice [$1,2,3,\dots,9$]
$b$ has $5$ choice [$0,2,4,6,8$]
$cd$ has $9$ choice [mentioned above]
Total = $ 9 × 5 × 9$

Case 2 : the last two digit is divisible by 4 and 8 and the 3rd digit is odd

Let the number be $abcd$
$a$ has $9$ option [$1,2,\dots,9$]
$b$ has $5$ option [$1,3,5,7,9$]
$cd$ has $9$ option [mentioned above]
Total = $9 × 5 × 9$

From the two cases , total = $9 × 5 × 9 + 9 × 5 × 9 = 9 × 10 × 9 = 81 × 10 = 810$

Am I wrong? :? :?
The answer should be $1125$ as brute force says it.
It says for non - zero digits. By the way, what is Brute - force

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Mehrab4226
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Re: RMO-2010/3

Unread post by Mehrab4226 » Sat Apr 17, 2021 9:06 pm

Marzuq wrote:
Sat Apr 17, 2021 8:13 pm
Mehrab4226 wrote:
Sat Apr 17, 2021 10:42 am
Marzuq wrote:
Sat Apr 17, 2021 10:37 am
My solution :
divisibility of 4 : if the last two digit is divisible by 4. Then the whole number is divisible by 4
Divisibility by 8 : if the last two digit is divisible by 4 but not by 8 and the 3rd digit is odd. Or if the last digit is divisible by 4 and 8 and the 3rd digit is even . Then the number is divisible by 8

Some non-zero 2 digits multiple of 4 but not 8 : $12,28,36,44,52,68,76,84,92$ [total 9]

Some non-zero 2 digit multiple of 4 and 8: $16,24,32,48,56,64,72,88,96$ [total 9]

Case 1: last two digit is divisible by 4 but not 8 and 3rd digit is even

Let the four digit number be $abcd$
$a$ has $9$ choice [$1,2,3,\dots,9$]
$b$ has $5$ choice [$0,2,4,6,8$]
$cd$ has $9$ choice [mentioned above]
Total = $ 9 × 5 × 9$

Case 2 : the last two digit is divisible by 4 and 8 and the 3rd digit is odd

Let the number be $abcd$
$a$ has $9$ option [$1,2,\dots,9$]
$b$ has $5$ option [$1,3,5,7,9$]
$cd$ has $9$ option [mentioned above]
Total = $9 × 5 × 9$

From the two cases , total = $9 × 5 × 9 + 9 × 5 × 9 = 9 × 10 × 9 = 81 × 10 = 810$

Am I wrong? :? :?
The answer should be $1125$ as brute force says it.
It says for non - zero digits. By the way, what is Brute - force
Oh it says non zero digits!. Sorry about that . And brute force means just finding all the such numbers and counting how many there were.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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