Some Last year Divisional Problems

For students of class 11-12 (age 16+)
sakib.creza
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Some Last year Divisional Problems

Unread post by sakib.creza » Thu Jan 24, 2013 8:09 pm

Guys, I need help for these problems. Couldn't solve these. Please help me out.

1) Consider the sequence $2^0, 2^1, 2^2 … 2^k$. You have to choose some of these numbers and
their product will be the numerator of a fraction. The product of the remaining numbers
will be the denominator. You want the fraction to be equal to $1$. What is the $2012$-th value
of $k$ for which this can be done?

2) A number has $180$ factors. What is the maximum number of distinct primes that can
divide that number?

3) Senjuti drew a circle that is inscribed in two different squares. One of the squares has an
area of $32$ square units. What is the difference of the area of those two squares?

4) The number $abbcca$ is divisible by $7$. When $abc$ is divided by $7$ it yields a remainder of $4$.
What will be the remainder when $aca$ is divided by $7$?

5) If a number with $2012$ divisors is multiplied with a number with $2011$ divisors then what
is the maximal number of distinct prime divisors the product can have?

6) The number $ababab$ has $60$ divisors and the sum of the divisors is $678528$. Find
$\displaystyle \frac {b}{a}$

7) $N$ is a number of $2012$ digits. If you take any consecutive $m$ digits ($m\leq 2012$) from $N$
starting from any position in that number, there’ll be another position in $N$ so that the $m$ consecutive digits starting from that position will be in the reverse order of the former one. Total number of possible values of $N$ can be written as $a \times 10^b$ where $a$ and $b$ are
positive integers, $a$ is not divisible by $10$. What is the value of $a + b$?

8) Find the final decimal digit of $1! + 2! + 3! + ... + 10!$. (it is from a book, not last year problem. I did solve it but I found my solution very uncool)

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SANZEED
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Re: Some Last year Divisional Problems

Unread post by SANZEED » Thu Jan 24, 2013 10:58 pm

Firstly problem $6$ has a wrong statement.
For problem $2$, The diameter of the circle is the side of the squares(think why),and they have the same area.
For problem $2$, Remember that if the prime factorization of $n$ is $p_{1}^{a_{1}}\times p_{2}^{a_{2}}\times ...\times p_{k}^{a_{k}}$ then the number of divisors of $n$ is $(a_{1}+1)\times (a_{2}+1)\times ...\times (a_{k}+1)$. Since $180=2\times 2\times 3\times 3\times 5$, the number can be of the form $p_{1}\times p_{2}\times p_{3}^{2}\times p_{4}^{2}\times p_{5}^{4}$. So the number can have $5$ distinct prime factors at most.
For problem $5$,see the previous one,and remember that $2011$ is a prime. If I am not wrong then the answer here is $4$.
For problem $8$,remember that a number is divisible by $10$ if and only if both $2$ and $5$ divide it. The factorial numbers from $5!$ and above are all thus divisible by $10$. Thus if we right $S=1!+2!+...+10!$ then $S\equiv 1!+...+4!\equiv 33\equiv 3 (mod 10)$ and the answer is $3$.
For problem $1$,The first possible value of $k$ is $3$, simply take numerator $2^{0}\times 2^{3}$ and denominator $2^{1}\times 2^{2}$. Using induction it can easily be proved that possible values for $k$ appears after every $4$ integers. So the general form of possible $i$th $k$ is $4i-1$. Thus the $2012$th value is $8047$.
For problem $7$,Prove it palindromic. Then the total number of possible N's will be the number of ways to choose the first $1006$ digits,where the first digit can't be zero.

Sorry for not maintaining sequence. :oops:
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

skb
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Re: Some Last year Divisional Problems

Unread post by skb » Fri Jan 25, 2013 12:00 am

২ নাম্বারটার ক্ষেত্রে,
ধরি, সংখ্যাটার মৌলিক উৎপাদকগুলার সেট $\{P_1, P_2, P_3.......P_n\}$ তাহলে, সংখ্যাটার মোট উৎপাদক হবে সেটটার সব উপসেট অর্থাৎ পাওয়ার সেটের উপাদানের সমান ( যেমন উৎপাদকগুলা হবে $P_1 \times P_2, P_1 \times P_2 .......... P_2 \times P_3 \times P_4$.......) অর্থাৎ $2^n$
(যদি মৌলিক উৎপাদকগুলা একবার করে থাকে)............ $180$ এর নিচে নিকটবর্তী $2$ এর ঘাত $128$ যা $2^7$....... তাই সর্বাধিক মৌলিক উৎপাদক হবে $7$ টা
পাওয়ার সেটের একটা উপাদান ফাঁকা সেট, তাই $2^n-1$ নেওয়া উচিত ছিল, কিন্তু $180$ টা উৎপাদকের একটা হল $1$, যেটাকে অন্তর্ভুক্ত করার জন্য আবার $+1$ করা হয়েছে
(এই লজিকে কোন ভুল আছে????)
Dream shall never die.

sakib.creza
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Joined: Sat Nov 03, 2012 6:36 am

Re: Some Last year Divisional Problems

Unread post by sakib.creza » Fri Jan 25, 2013 12:33 am

Thnx Sanzeed Bhai for your help. But cudn't understand solution to problem 2. Ar ektu khule explain korle bhalo hoto plz.(only understood the first line). R skb bhaia, sorry but tomar solutionta amar mathar upor die gese. This is because ami english medium background theke, so the bangla mathematical terms are literally Chinese to me.

skb
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Re: Some Last year Divisional Problems

Unread post by skb » Fri Jan 25, 2013 1:05 am

Problem 4
$abbcca = 100001a + 11000b + 110c \equiv 0\pmod {7} .........(1)$
$abc = 100a + 10b + c \equiv 4\pmod{7}$
So, $110000a + 11000b + 1100c \equiv 4\pmod{7}$ , $[\times 1100] ...........(2)$
$(2) - (1)$
$9999a + 990c \equiv 4\pmod{7}$
Or, $99(101a + 10c) \equiv 4\pmod{7}$
So, $99 \times aca \equiv 4\pmod{7}$
We know that, if $a_1 \equiv b_1\pmod {n}$ and $a_2 \equiv b_2\pmod {n}$ , then $a_1a_2 \equiv b_1b_2 \pmod {n}$
Using this formula, $aca \equiv 4 \pmod{7}$
Dream shall never die.

skb
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Re: Some Last year Divisional Problems

Unread post by skb » Fri Jan 25, 2013 1:15 am

sakib.creza vai, amar kase english english hoileo english mathematical term gula amar kase hebrew, apnake personal msg e bojhanor ekta try nite partam......
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Phlembac Adib Hasan
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Re: Some Last year Divisional Problems

Unread post by Phlembac Adib Hasan » Fri Jan 25, 2013 12:09 pm

skb wrote:২ নাম্বারটার ক্ষেত্রে,
ধরি, সংখ্যাটার মৌলিক উৎপাদকগুলার সেট $\{P_1, P_2, P_3.......P_n\}$ তাহলে, সংখ্যাটার মোট উৎপাদক হবে সেটটার সব উপসেট অর্থাৎ পাওয়ার সেটের উপাদানের সমান ( যেমন উৎপাদকগুলা হবে $P_1 \times P_2, P_1 \times P_2 .......... P_2 \times P_3 \times P_4$.......) অর্থাৎ $2^n$
(যদি মৌলিক উৎপাদকগুলা একবার করে থাকে)............ $180$ এর নিচে নিকটবর্তী $2$ এর ঘাত $128$ যা $2^7$....... তাই সর্বাধিক মৌলিক উৎপাদক হবে $7$ টা
পাওয়ার সেটের একটা উপাদান ফাঁকা সেট, তাই $2^n-1$ নেওয়া উচিত ছিল, কিন্তু $180$ টা উৎপাদকের একটা হল $1$, যেটাকে অন্তর্ভুক্ত করার জন্য আবার $+1$ করা হয়েছে
(এই লজিকে কোন ভুল আছে????)
আছে। $P_1$ উৎপাদক হলে $P_1^2$-ও উৎপাদক হতে পারে। তবে সেটা নিশ্চিত করার জন্য আমাদের আগে জানতে হবে ঐ সংখ্যায় $P_1$-এর ঘাত কত ছিল। তুমি এটাকে গোনায় ধরছ না।
কোন সংখ্যার মোট উৎপাদক সংখ্যা বের করার জন্য একটা সূত্র আছে। সানজিদের পোস্ট দেখ। যদি সংখ্যাটা $P_1^{a_1}\cdot P_2^{a_2}...P_k^{a_k}$ হয় তবে তার মোট উৎপাদক আছে $(a_1+1)(a_2+1)...(a_k+1)$ টি। এটা সহজ কম্বিন্যাটরিক্স দিয়ে প্রমাণ করতে পারবে। এই জিনিসটা জানলে প্রশ্নটা আসলে দাঁড়ায় এক অপেক্ষা বড় সর্বাধিক কয়টি সংখ্যা গুণ করে $180$ পাওয়া সম্ভব? খুব সহজ প্রশ্ন যার সঠিক উত্তর হচ্ছে পাঁচ।
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sakib.creza
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Re: Some Last year Divisional Problems

Unread post by sakib.creza » Fri Jan 25, 2013 2:09 pm

Adib Bhai, thank you very much. এই প্রবলেমটা ভালোই জ্বালাচ্ছিল । এখন বুঝছি ।
Few more problems-

$1)$Find all prime numbers $p$ and integers $a$ and $b$ (not necessarily positive) such that $p^{a} + p^{b}$
is the square of a rational number.

$2)$Find the number of odd coefficients in expansion of $(x + y)^{2010}$.

$3)a_{1}, a_{2}, . . . , a_{k}, . . . , a_{n}$ is a sequence of distinct positive real numbers such that $a_{1} < a_{2} < . . . . < a_{k}$ and $a_{k} > a_{k+1} > . . . > a_{n}$. A Grasshopper is to jump along the real axis, starting at
the point $O$ and making $n$ jumps to the right of lengths $a_{1}, a_{2}, . . . , a_{n}$ respectively. Prove
that, once he reaches the rightmost point, he can come back to point O by making n jumps
to the left of lengths $a_{1}, a_{2}, . . . , a_{n}$ in some order such that he never lands on a point which
he already visited while jumping to the right. (The only exceptions are point $O$ and the
rightmost point)

P.S. - thanks for latexing, sourav da :)
Last edited by sourav das on Fri Jan 25, 2013 4:36 pm, edited 1 time in total.
Reason: the last line is also edited :D

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Tahmid Hasan
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Re: Some Last year Divisional Problems

Unread post by Tahmid Hasan » Fri Jan 25, 2013 7:23 pm

SANZEED wrote:For problem $1$,The first possible value of $k$ is $3$, simply take numerator $2^{0}\times 2^{3}$ and denominator $2^{1}\times 2^{2}$. Using induction it can easily be proved that possible values for $k$ appears after every $4$ integers. So the general form of possible $i$th $k$ is $4i-1$. Thus the $2012$th value is $8047$.
All positive integers of the form $4k$ satisfy the given requirements also ;)
বড় ভালবাসি তোমায়,মা

skb
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Re: Some Last year Divisional Problems

Unread post by skb » Fri Jan 25, 2013 8:38 pm

Problem 1 - I solved it in a lengthy (also complex) way, I guess
Suppose, $a < b$ and $b = a+m$
and $p^a + p^b = k^2$
or, $p^a + p^{a+m} = k^2$
or, $p^a (1 + p^m) = k^2$
Now, we have two subcases, either p is even or odd
Subcase 1 - When $p$ is even, $(p=2)$
$1 + 2^m$ cannot be divided by $2$ and there should be pairs of integers in the dividers of a perfect square
so, both $2^a$ and $(1 + 2^m)$ are perfect squares
let, $1 + 2^m = n^2$
so, $2^m = (n+1)(n-1)$
as, $2^m$ cannot be divided by any other prime, so $n+1$ and $n-1$ both are power of $2$
it is possible only if $n=3$
so, here, $p=2$ , $a =\; $ any even number, $b = a+3$

Subcase -2 When $p$ is odd
using the same logic,
$p^a$ and $(1 + p^m)$ both are perfect squares
$1 + p^m = f^2$
$p^m = (f+1)(f-1)$
again, $f+1$ and $f-1$ both are power of $p$
but it is not possible in any case
so here should be no answer

(I am confused whether I'm right or wrong) :!: :!:
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