Try this beautiful problem from the PRMO II, 2019 based on Missing Integers.
Missing Integers - PRMO II 2019
Consider the sequence of numbers [n+\(\sqrt{2n}+\frac{1}{2}\)] for \(n \geq 1\), where [x] denotes the greatest integer not exceeding x. If the missing integers in the sequence are \(n_1<n_2<n_3<...\) then find \(n_{12}\).
Real Numbers and Integers | PRMO 2017 | Question 2
Try this beautiful problem from the PRMO, 2017 based on Real Numbers and Integers.
Real Numbers and Integers - PRMO 2017
PRMO, 2017, Question 2
Suppose a, b are positive real numbers such that \(a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=183\), \(a(b)^\frac{1}{2}+b(a)^\frac{1}{2}\)=182, find \(\frac{9(a+b)}{5}\).
is 107
is 73
is 840
cannot be determined from the given information
Key Concepts
Real Numbers
Algebra
Integers
Check the Answer
Answer: 73
Book
Elementary Algebra by Hall and Knight
Try with Hints
here \(a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=183\) and \(a(b)^\frac{1}{2}+b(a)^\frac{1}{2}\)=182
This equation gives integer solutions then a and b must be squares.
Let a=\(P^{2}\)and b=\(Q^{2}\)
\(\Rightarrow a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=P^{3}+Q^{3}=183\) is first equation
and \(a(b)^\frac{1}{2}+b(a)^\frac{1}{2}\)=\(P^{2}Q+Q^{2}P\)
=\(PQ(P+Q)\)=182 is second equation
now first equation +3 second equation gives \(P^{3}+Q^{3}+3PQ(P+Q)\)=183+\(3 \times 182\)=729 \(\Rightarrow (P+Q)^{3}=9^{3}\) \(\Rightarrow(P+Q)=9\) second equation gives PQ(P+Q)=182
\(\Rightarrow PQ=\frac{182}{9}\) then \(a+b=P^{2}+Q^{2}=(P+Q)^{2}-2PQ\)=\(\frac{365}{9}\) \(\frac{9(a+b)}{5}\)=\(\frac{9}{5} \times \frac{365}{9}\)=73.
Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function.
Periodic Function (B.Stat Objective Question )
If f(x) = \(a_0+a_1cosx+a_2cos2x+....+a_ncosnx\) where \(a_0,a_1,....,a_n\) are non zero real numbers and \(a_n > |a_0|+|a_1|+....+|a_{n-1}|\), then number of roots of f(x)=0 in \( 0 \leq x \leq 2\pi\), is
0
at least 2n
53361
5082
Key Concepts
Periodic
Real Numbers
Inequality
Check the Answer
Answer:at least 2n
B.Stat Objective Problem 710
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
f is periodic with period \(2\pi\)
here \(0 <|\displaystyle\sum_{k=0}^{n-1}a_kcos(kx)| \leq \displaystyle\sum_{k=0}^{n-1}|a_k|<a_n, x\in\)set of reals
for points \(x_k=\frac{k\pi}{n}\) \(1 \leq k \leq 2n\)
\(f(x_k)=a_n[cos(k\pi)+\theta], |\theta|<1 for 1 \leq k \leq 2n\)
[ since cos \(\theta\) is periodic and f(x) is expressed for every point x=\(x_k\)]
f has at least 2n points in such a period interval where f has alternating sign.
