BdMO 2017 junior/4
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
BdMO 2017 junior/4
(a) Can two consecutive numbers $n$ and $n1$ both be divisible by $3$?
(b) Determine the smallest integer $n > 1$ such that $n^2(n1)$ is divisible by $1971$. Note:
$1971 = 3^3 \times 73$.
(b) Determine the smallest integer $n > 1$ such that $n^2(n1)$ is divisible by $1971$. Note:
$1971 = 3^3 \times 73$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: BdMO 2017 junior/4
And I don't know how i missed the part (b) in the actual exam. And i not posting this solution. Left for the others.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
Re: BdMO 2017 junior/4
I also messed up part(b) during actual exam. However, I was able to find an answer using brute force then.
I got $n = 730$.
Am I correct?..
I got $n = 730$.
Am I correct?..
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
 Charles Caleb Colton
 Charles Caleb Colton
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: BdMO 2017 junior/4
By judging the 6,7,8 ..... This one is kinda hard.... Rather than 8,9.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
Re: BdMO 2017 junior/4
what is crt?ahmedittihad wrote:Apply crt to get $n=585$.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: BdMO 2017 junior/4
Chinese remainder theoremjayon_2 wrote:what is crt?ahmedittihad wrote:Apply crt to get $n=585$.
http://s3.amazonaws.com/aopscdn.artofp ... theory.pdf
Take a look at Page no.40 for further information.
Last edited by Thamim Zahin on Wed Apr 12, 2017 9:11 pm, edited 2 times in total.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: BdMO 2017 junior/4
So basically you do different cases on the primes dividing $n^2$ or $n1$. And these cases yield different divisibility conditions. And CRT is the method to find the smallest number that suppression every congruence in a system. Look it up. The notation might be scary but it's quite natural and intuitive.
Frankly, my dear, I don't give a damn.

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 Location: Sylhet
Re: BdMO 2017 junior/4
ahmedittihad can you explain how it is 585?

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 Joined: Thu Feb 01, 2018 11:56 am
Re: BdMO 2017 junior/4
n^3n=n^2(n1)
Here we have 2 cases.
Case 1:n^2 is divisible by 73 and n1 is divisible by 27.
Now n is the form of 73k.
We get 73k1 is congruent to 0(mod27)
Or,(8k+1) is congruent to 0(mod27)
Or, 8k is congruent to 26 (mod27)
Or, 4k is congruent to 13 (mod27)
Or, 4k is congruent to 40 (mod27)
Or, k is congruent to 10(mod27)
So, k=27m+10 (where m=0, 1, 2, 3, ………).Taking the smallest value of m=0 gives k=10 and n=73*10=730.
Case2: n^2 is divisible by 27 and n1 is divisible by 73.
So, k is the form of 73K+1.
We get (73K+1)^2 is congruent to 0(mod27)
Or 5329K^2+146K+1 is congruent to 0(mod27)
Or, 10 K^2+11K+1 is congruent to 0(mod27)
Middle term of it gives,(K+1)(10K+1) is congruent to 0(mod27)
If we take K+1 is congruent to 0(mod27)
We get the smallest value of K is 26.
Now
10K+1 is congruent to 0(mod27)
Or, 10K is congruent to 26(mod27)
Or, 5K is congruent to 13(mod27)
Or, 5K is congruent to 40(mod27)
Or, K is congruent to 8(mod27)
So, K=27M+8(where M=0,1,2,3,………).Taking the smallest value of M=0 gives k=8and n=73*8+1=585
Here we have 2 cases.
Case 1:n^2 is divisible by 73 and n1 is divisible by 27.
Now n is the form of 73k.
We get 73k1 is congruent to 0(mod27)
Or,(8k+1) is congruent to 0(mod27)
Or, 8k is congruent to 26 (mod27)
Or, 4k is congruent to 13 (mod27)
Or, 4k is congruent to 40 (mod27)
Or, k is congruent to 10(mod27)
So, k=27m+10 (where m=0, 1, 2, 3, ………).Taking the smallest value of m=0 gives k=10 and n=73*10=730.
Case2: n^2 is divisible by 27 and n1 is divisible by 73.
So, k is the form of 73K+1.
We get (73K+1)^2 is congruent to 0(mod27)
Or 5329K^2+146K+1 is congruent to 0(mod27)
Or, 10 K^2+11K+1 is congruent to 0(mod27)
Middle term of it gives,(K+1)(10K+1) is congruent to 0(mod27)
If we take K+1 is congruent to 0(mod27)
We get the smallest value of K is 26.
Now
10K+1 is congruent to 0(mod27)
Or, 10K is congruent to 26(mod27)
Or, 5K is congruent to 13(mod27)
Or, 5K is congruent to 40(mod27)
Or, K is congruent to 8(mod27)
So, K=27M+8(where M=0,1,2,3,………).Taking the smallest value of M=0 gives k=8and n=73*8+1=585