Preparation Marathon

[Shihab]

আমি বুঝলাম না ৯ নম্বর সমস্যা কেন বাদ দেওয়া হল। এর উত্তর তো বের করা সম্ভব।
Round 3 এর সমস্যা কখন পোস্ট করা হবে?

[photon]

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

[protik]

I Solved 10 in this way

2 , 3, 6, 14, 34
1, 3, 8, 20
2, 5, 12,
3, 7,


in the first line differences between the first two is 1.
in the second line differences between the first two is 2
in the third line differences between the first two is 3
in fourth it should be 4.

so the answer should be 34..
Am I wrong??

[Nadim Ul Abrar]

@ protik
It will be enough to define $n^{th}$ term ....

what is $n^{th}$ term if the ans be 34 ?

[Nadim Ul Abrar]

vai ... bothut rokomer nth term bahir kora jaye ... tai maramari kira love nai ... Dekhi na masum vai ki koy ..

[*Mahi*]

No 9 :

$\binom{4012}{2011} \equiv 2001!\equiv a mod2011$

2011 is a prime so that via wilson $2010! \equiv -9!a \equiv -1 mod 2011$

after having a jump from $-9!a \equiv -1 mod 2011$

$5a \equiv 18^2 mod 2011$

So $5a=2011n+18^2$

now $2011n+18^2 \equiv n+4 mod 5$

$n=1$ satisfy that .

then $5a=2011+18^2$

or $a=467$

There is a mistake here. The solution is $1$.
Remember $\binom{4022}{2011} \equiv 2 (\text{mod }2011)$
Now express $\binom{4022}{2011}$ as $\frac ab \binom{4012}{2011}$
I hope you can manage the rest.

[nafistiham]

মাসুম ভাই বা মাহি কেউ অফিশিয়াল সমাধান পোস্ট করেন ।

[bristy1588]

Next Problem set Kokhon pabo?? PR 3 Kokhon dewa hobe??

[protik]

Can't wait any longer for preparation marathon-3. Masum bhai, where are you????????????????

[Masum]

Can you find a general pattern using difference table? I think here it does not work. You need to have a recursion, like $a(n)=2^n+n!,n\ge0$