Preparation Marathon

[sm.joty]

আচ্ছা সবাই কম্বিনাটরিক্স দিয়ে কিভাবে ২ নং সমাধান করছো। আমাকে কেউ একটু দেখাও। আর সবচেয়ে ভালো হয় যদি কেউ অফিসিয়াল একটা solution দেয়। (একটু বিস্তারিত ব্যাখ্যা করলে ভালো হয়।)
২ নং এর সমাধান আমারটা এরকম। কোনো ভুল হলে বল

ধরি, $a=x+y$, $b=z+w$
তাহলে, $(x+y+z+w)^{2012}=(a+b)^{2012}$
এখন, $a^{2012}$ কে বিস্তৃত করলে পাওয়া যায় 2013 টা পদ,$a^{2011}$ কে বিস্তৃত করলে পাই, 2012 টা পদ।
আবার $(a^{2011}*b^{1})$ কে বিস্তৃত করলে পাই, $2012*2$ টা পদ, $(a^{2010}*b^{2})$ কে বিস্তৃত করলে পাই, $2011*3$ টা পদ।...

কাজেই এখন পদের সংখ্যা,
$S=2013+2012*2+2011*3+........+1007*1007+........2012*2+2013$
$=2013+(2013-1)*2+.....+(2013-1006)*1007+.....+(2013-2012)*2013$
$=(2013+2013*2+\cdots+2013*2013)-(1*2+2*3+\cdots+2012*2013)$
$=2013[1+2+3+.......+2013]-\frac{(2012*2013*2014)}{3}$
$=2013\times\frac{2013*2014}{2}-\frac{(2012*2013*2014)}{3}$
$=(2013*2014)\times[\frac{2013}{2}-\frac{2012}{3}]$
$=(2013*2014)\times\frac{2015}{6}$
$S=671*1007*2015$


এটাই উত্তর।

Formula : $1*2+2*3+............+n(n+1)=\frac{n(n+1)(n+2)}{3}$

[nafistiham]

my answer:

\[_{}^{2015}\textrm{C}_{3}\]


as it is like distributing $2012$ identical balls into $4$ distinct boxes.[used the theorem from COMBINATORICS by DANIEL A. MARCUS ]

আমার প্রথমটার উত্তর যেহেতু $৪$ এসেছে । সুতরাং, সম্ভবতঃ এই উত্তরটা সঠিক ।

[protik]

nice solution Joty.

[Masum]

Masum vai
page 66
103. x, y, z are positive integers such that:
$x^3 + y^3 = z^3$
then $3|xyz$

why we wont use fermat ???

No it meant that if $x^3+y^3=z^3$, then.
I think it would be better if you start a new topic about the problems with this note.

[sm.joty]

ooops,i misunderstood timing!
however,last night i solved 4.here is the solution

n is an even and has 9 divisiors.
so,$n=2^8,p^2.q^2$
here,p,q are primes.
here,$2^8$ is not acceptable for sum of its 3 divisors can't be square.
as n is even ,let p is 2
so,n will be in form of $4q^2$ ,as q is prime (not 2) then it is odd.
we can write sum of three divisors as:
$1+q+4q=5q+1$
$1+2+2q^2=2(q^2+1)+1$
$2q+2q+1=4q+1$
$2+q+2q=2(q+1)+q$
$4+q+q=2(2+q)$
$q^2+2+2=q^2+4$
$4q^2+1+1=2(2q^2+1)$
here,according to Question 1 sum should be square and even.
here,only even can be $5q+1,2(2+q),2(2q^2+1)$
now each as square,that square will be in $4x^2$ form as it is even.
let,$2(2q^2+1)=4x^2$
but,$2q^2+1$ is odd.so it can't be square.
like that,$2(2+q)$ can't be square,here (2+q) is odd
let,$5q+1=4x^2$
$5q=(2x+1)(2x-1)$,so q is serial odd of 5.so,$q=3,7$
for 3,$5q+1=16$ and for 7,$5q+1=36$
$16,36$ should be double of another sum.
let,$2a=16$,so $a=8$,then the sum is 8.but there is no sum valuing 8 for$q=3$
let,$2a=36$ so $a=18$ and for $q=7$, there is sum valuing 18=7+7+4
so,$n=4q^2=4.49=196$

too big

Nadim I can't understand the statement $n=2^8.p^2.q^2$
that implies $n$ has $(8+1)(2+1)(2+1)=162$ factors.

[nafistiham]

এইটার উত্তর পত্র কি দেওয়া হবে না

[Nadim Ul Abrar]

ooops,i misunderstood timing!
however,last night i solved 4.here is the solution

n is an even and has 9 divisiors.
so,$n=2^8,p^2.q^2$
here,p,q are primes.
here,$2^8$ is not acceptable for sum of its 3 divisors can't be square.
as n is even ,let p is 2
so,n will be in form of $4q^2$ ,as q is prime (not 2) then it is odd.
we can write sum of three divisors as:
$1+q+4q=5q+1$
$1+2+2q^2=2(q^2+1)+1$
$2q+2q+1=4q+1$
$2+q+2q=2(q+1)+q$
$4+q+q=2(2+q)$
$q^2+2+2=q^2+4$
$4q^2+1+1=2(2q^2+1)$
here,according to Question 1 sum should be square and even.
here,only even can be $5q+1,2(2+q),2(2q^2+1)$
now each as square,that square will be in $4x^2$ form as it is even.
let,$2(2q^2+1)=4x^2$
but,$2q^2+1$ is odd.so it can't be square.
like that,$2(2+q)$ can't be square,here (2+q) is odd
let,$5q+1=4x^2$
$5q=(2x+1)(2x-1)$,so q is serial odd of 5.so,$q=3,7$
for 3,$5q+1=16$ and for 7,$5q+1=36$
$16,36$ should be double of another sum.
let,$2a=16$,so $a=8$,then the sum is 8.but there is no sum valuing 8 for$q=3$
let,$2a=36$ so $a=18$ and for $q=7$, there is sum valuing 18=7+7+4
so,$n=4q^2=4.49=196$

too big

Nadim I can't understand the statement $n=2^8.p^2.q^2$
that implies $n$ has $(8+1)(2+1)(2+1)=162$ factors.

$2^8 $and $p^2.q^2$ not $2^8.p^2.q^2$

note that $9=9=(8+1)=3.3=(2+1)(2+1)$

[*Mahi*]

এইটার উত্তর পত্র কি দেওয়া হবে না

Posting a PDF official solution sheet is too much difficult. It is recommended that you discuss and find the solutions by yourselves. However, nobody seemed to solve problem 5 exam 2 right, so , the solution is $\sum^8_{k=1} k^2(8-k)$

[Nadim Ul Abrar]

এইটার উত্তর পত্র কি দেওয়া হবে না

Posting a PDF official solution sheet is too much difficult. It is recommended that you discuss and find the solutions by yourselves. However, nobody seemed to solve problem 5 exam 2 right, so , the solution is $\sum^8_{k=1} k^2(9-k)$

*Mahi* vai : $\sum^8_{k=1} k^2(9-k)$ ken hoilo ?

[nafistiham]

এইটার উত্তর পত্র কি দেওয়া হবে না

Posting a PDF official solution sheet is too much difficult. It is recommended that you discuss and find the solutions by yourselves. However, nobody seemed to solve problem 5 exam 2 right, so , the solution is $\sum^8_{k=1} k^2(9-k)$

*Mahi* vai : $\sum^8_{k=1} k^2(9-k)$ ken hoilo ?

i didnt get the $(9-k)$ part