## Some Last year Divisional Problems

For students of class 11-12 (age 16+)
sakib.creza
Posts: 26
Joined: Sat Nov 03, 2012 6:36 am

### Some Last year Divisional Problems

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)

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

### Re: Some Last year Divisional Problems

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.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

skb
Posts: 7
Joined: Fri Dec 21, 2012 8:24 pm

### Re: Some Last year Divisional Problems

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
Posts: 26
Joined: Sat Nov 03, 2012 6:36 am

### Re: Some Last year Divisional Problems

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
Posts: 7
Joined: Fri Dec 21, 2012 8:24 pm

### Re: Some Last year Divisional Problems

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
Posts: 7
Joined: Fri Dec 21, 2012 8:24 pm

### Re: Some Last year Divisional Problems

sakib.creza vai, amar kase english english hoileo english mathematical term gula amar kase hebrew, apnake personal msg e bojhanor ekta try nite partam......
Dream shall never die.

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### Re: Some Last year Divisional Problems

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
Posts: 26
Joined: Sat Nov 03, 2012 6:36 am

### Re: Some Last year Divisional Problems

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

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Some Last year Divisional Problems

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
Posts: 7
Joined: Fri Dec 21, 2012 8:24 pm

### Re: Some Last year Divisional Problems

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)
Dream shall never die.