Problem:
The number of terms in the expression of
$latex [(a+3b)^2 (a-3b)^2]^2 $
A) 4;
B) 5;
C) 6;
D) 7;
Solution:
$latex [(a+3b)^2 (a-3b)^2]^2 $
$latex = [\{(a+3b)(a-3b)\}^2]^2 $
$latex = \{ (a^2 -9b^2)^2\}^2 = (a^2 - 9b^2)^4 $
By Binomial Theorem, the given expression contains 5 terms (since $latex (x +y)^n $ has n+1 terms in general).
Answer: B) 5;
Try this beautiful problem from TOMATO Objective no. 27 based on Closure of a set of even numbers.
Problem: Closure of a set of even numbers
S is the set whose elements are zero and all even integers, positive and negative. Consider the 5 operations- [1] addition; [2] subtraction; [3] multiplication; [4] division; and [5] finding the Arithmetic Mean. Which of these operations applied to any pair of elements of S, yield only elements of S.
Solution:
Adding, subtracting and multiplying even numbers yield even numbers.
Division may yield non-integer ($latex \frac {2}{4} = \frac{1}{2} $ ) or odd number ($latex \frac {12}{4} = 3 $) both of which are not members of S. So division is ruled out.
Finally arithmetic of two even numbers may be a odd number. For example, $latex \frac { 4 + 6}{2} = 5 $.
Thus operations [1], [2], [3] work.
Answer: (D) [1], [2], [3]
For Regional Math Olympiad, I.S.I. Entrance, Pre RMO and other Olympiads.
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url="https://cheenta.com/courses/rmo-regional-math-olympiad/" url_new_window="on" image="https://cheenta.com/wp-content/uploads/2019/03/10002.2.1.gif" _builder_version="3.27.4" link_option_url="https://cheenta.com/courses/rmo-regional-math-olympiad/" link_option_url_new_window="on" _i="0" _address="1.0.3.0"][/et_pb_blurb][et_pb_button button_url="https://cheenta.com/courses/rmo-regional-math-olympiad/" url_new_window="on" button_text="Read More" _builder_version="3.27.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#0c71c3" button_border_radius="5px" button_icon="%%3%%" button_bg_enable_color="on" _i="1" _address="1.0.3.1"][/et_pb_button][/et_pb_column][/et_pb_row][et_pb_row custom_padding_last_edited="on|phone" _builder_version="3.25" background_color="#ffffff" box_shadow_style="preset3" box_shadow_vertical="40px" box_shadow_blur="100px" box_shadow_spread="-30px" box_shadow_color="rgba(0,0,0,0.22)" max_width="1200px" custom_padding="80px|80px|27px|80px|false|true" custom_padding_phone="|30px||30px||true" use_custom_width="on" custom_width_px="1200px" _i="1" _address="1.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="1.1.0"][et_pb_text _builder_version="3.27.4" text_font="Faustina||||||||" text_font_size="24px" text_line_height="2em" link_font="|700|||||||" link_text_color="#26c699" header_font="||||||||" header_2_font="Faustina|600|||||||" header_2_text_color="#0e7162" header_2_font_size="42px" header_2_line_height="1.4em" text_orientation="center" max_width="800px" module_alignment="center" text_font_size_phone="17px" text_font_size_last_edited="on|phone" header_2_font_size_tablet="32px" header_2_font_size_phone="22px" header_2_font_size_last_edited="on|desktop" locked="off" inline_fonts="Times New Roman" _i="0" _address="1.1.0.0"]Here is the post for the Regional Mathematics Olympiad (India) RMO Number Theory Problems. These are problems from previous year papers.
(This is a work in progress. More problems will be added soon).
RMO Number Theory Problems:
Try this beautiful problem from TOMATO Objective no. 13 based on Calendar Problem. This problem is useful for BSc Maths and Stats Entrance Exams.
Problem:
June 10, 1979, was a SUNDAY. Then May 10, 1972, was a
(A) Wednesday;
(B) Friday;
(C) Sunday;
(D) Tuesday;
Solution:
In a (non-leap) year there are 365 days.
$365 \equiv 1 \mod 7 $
On a leap year, there are 366 days
$latex 366 \equiv 2 \mod 7 $
From 1972 to 1979, there are 7 years (1 of them is leap year). For each non-leap year, we have to go back 1 day and for every leap year, we have to go back 2 days. Hence in total, we have to go back 8 days for those 7 years. Also, May has 31 days. Hence we have to go back 31+8 = 39 days.
Thus $latex 39 \equiv 4 \mod 7 $ days before Sunday is a Wednesday.
Answer: (A) Wednesday.
Try this beautiful problem from TOMATO Subjective Problem no. 173 based on the Sum of Polynomials.
Problem : Sum of polynomials
Let [latex] {{P_1},{P_2},...{P_n}}[/latex] be polynomials in [latex] {x}[/latex], each having all integer coefficients, such that [latex] {{P_1}={{P_1}^{2}+{P_2}^{2}+...+{P_n}^{2}}}[/latex]. Assume that [latex] {P_1}[/latex] is not the zero polynomial. Show that [latex] {{P_1}=1}[/latex] and [latex] {{P_2}={P_3}=...={P_n}=0}[/latex]
Solution :
As [latex] {P_1},{P_2},...{P_n}[/latex] are integer coefficient polynomials so gives integer values at integer points.
Now as [latex] {P_1}[/latex] is not zero polynomial
[latex] {\displaystyle{P_1}(x)>0}[/latex] for some [latex] \displaystyle{x \in Z}[/latex]
Then [latex] {\displaystyle{P_1}(x)\ge{1}}[/latex] or [latex] P_1(x) \le -1 [/latex] as [latex] {\displaystyle{P_1}(x)}[/latex]=integer
[latex] \Rightarrow (P_1(x))^2 \ge P_1(x)[/latex] or [latex] 0 \ge P_1(x) -(P_1(x))^2 [/latex]
But it is given that [latex] {{P_1}={{P_1}^{2}+{P_2}^{2}+...+{P_n}^{2}}}[/latex]
This implies [latex] (P_2 (x))^2+...+(P_n (x) )^2 \le 0 [/latex]. This is only possible if [latex] (P_1(x))^2 = ... = (P_n(x))^2 = 0[/latex]
Hence the values of x for which [latex] P_1 (x) [/latex] is non-zero, [latex] P_2(x) , ... , P_n(x) [/latex] are all zero. The values of x for which [latex] P_1(x) = 0 [/latex], we have [latex] 0=0+(P_2 (x))^{2}+...+(P_n(x))^2[/latex] implying each is zero.
Therefore [latex] P_2(x) = ... = P_n(x) = 0 [/latex].
Finally [latex] P_1(x) = (P_1(x))^2 [/latex] implies [latex] P_1(x) = 0 \text{or} P_1(x) = 1 [/latex]. Since [latex] P_1(x) \neq 0 [/latex] hence it is 1.
(Proved)
This problem is from the Test of Mathematics, TOMATO Subjective Problem no. 172 based on the Round Robin tournament.
Problem : Suppose there are [latex] {k}[/latex] teams playing a round robin tournament; that is, each team plays against all the other teams and no game ends in a draw.Suppose the [latex] {i^{th}}[/latex] team loses [latex] {l_{i}}[/latex] games and wins [latex] {w_{i}}[/latex] games. Show that
[latex] {{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}[/latex] = [latex] {{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}[/latex]
Solution : Each team plays exactly one match against each other team.
Consider the expression [latex] \displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) } [/latex]
Since each team plays exactly k-1 matches and no match ends in a draw, hence number of wins plus numbers of loses of a particular team is k-1 (that is the number of matches it has played). In other words [latex] l_i + w_i = k-1 [/latex] for all i (from 1 to k).
Hence
[latex] \displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} } [/latex]
[latex] \displaystyle{= \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) } [/latex]
[latex] \displaystyle{= \sum_{i=1}^{k}(k-1)(l_i - w_i) } [/latex]
[latex] \displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) } [/latex]
But [latex] \displaystyle{ \sum_{i=1}^{k} l_i = \sum_{i=1}^{k} w_i } [/latex] (as total number of loses = total number of matches = total number of wins; as each match results in a win or lose of some one)
Hence [latex] \displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) = (k-1) \times 0 = 0 } [/latex]
Therefore [latex] \displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = 0 } [/latex] implying [latex] {{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}[/latex] = [latex] {{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}[/latex]
Proved.