Categories
AIME II Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

Sequence and permutations – AIME II, 2015


Call a permutation \(a_1,a_2,….,a_n\) of the integers 1,2,…,n quasi increasing if \(a_k \leq a_{k+1} +2\) for each \(1 \leq k \leq n-1\), find the number of quasi increasing permutations of the integers 1,2,….,7.

  • is 107
  • is 486
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Permutations

Integers

Check the Answer


Answer: is 486.

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

Try with Hints


While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 \(\times\) number of permutations of (n-1) elements

or, number of permutations for n elements=\(3^{2}\) number of permutations of (n-2) elements

……

or, number of permutations for n elements=\(3^{n-2}\) number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 \(\times\) \(3^{n-2}\)

forming recurrence relation as the number of permutations =2 \(\times\) \(3^{n-2}\)

for n=3 all six permutations taken and go up 18, 54, 162, 486

for n=7, here \(2 \times 3^{5} =486.\)

Header text

as

Header text

sds

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Math Olympiad USA Math Olympiad

Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

Numbers of positive integers – AIME 2012


Find the number of positive integers with three not necessarily distinct digits, \(abc\), with \(a \neq 0\) and \(c \neq 0\) such that both \(abc\) and \(cba\) are multiples of \(4\).

  • is 107
  • is 40
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Number Theory

Algebra

Check the Answer


Answer: is 40.

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

Try with Hints


Here a number divisible by 4 if a units with tens place digit is divisible by 4

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

So for 10 b’s with 2 a’s and 2 c’s for every b gives \(10 \times 2 \times 2\)

Then number of ways \(10 \times 2 \times 2\) = 40 ways.

Subscribe to Cheenta at Youtube


Categories
Algebra Functional Equations Math Olympiad USA Math Olympiad

Algebraic Equation | AIME I, 2000 Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

Algebraic Equation – AIME 2000


Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, \(x+\frac{1}{z}=5\) and \(y+\frac{1}{x}=29\) then \(z+\frac{1}{y}\)=\(\frac{m}{n}\) where m and n are relatively prime, find m+n

  • is 107
  • is 5
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Integers

Check the Answer


Answer: is 5.

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

Try with Hints


 here \(x+\frac{1}{z}=5\) then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and \(y=(29-\frac{1}{x}\)) together gives 5-x=x\((29-\frac{1}{x}\)) then x=\(\frac{1}{5}\)

then y=29-5=24 and z=\(\frac{1}{5-x}\)=\(\frac{5}{24}\)

\(z+\frac{1}{y}\)=\(\frac{1}{4}\) then 1+4=5.

.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

Sequence and fraction – AIME I, 2000


A sequence of numbers \(x_1,x_2,….,x_{100}\) has the property that, for every integer k between 1 and 100, inclusive, the number \(x_k\) is k less than the sum of the other 99 numbers, given that \(x_{50}=\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 173
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 173.

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

Try with Hints


Let S be the sum of the sequence \(x_k\)

given that \(x_k=S-x_k-k\) for any k

taking k=1,2,….,100 and adding

\(100S-2(x_1+x_2+….+x_{100})=1+2+….+100\)

\(\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050\)

\(\Rightarrow S=\frac{2525}{49}\)

for \(k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}\)

\(\Rightarrow x_{50}=\frac{75}{98}\)

\(\Rightarrow m+n\)=75+98

=173.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Finding smallest positive Integer – AIME I, 1996


Find the smallest positive integer solution to \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).

  • is 107
  • is 159
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Trigonometry

Integers

Check the Answer


Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints


\(\frac{cos96+sin96}{cos96-sin96}\)

=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)

=\(\frac{sin186+sin96}{sin186-sin96}\)

=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)

=\(\frac{2sin141cos45}{2cos141sin45}\)

=tan141

here \(tan(180+\theta)\)=\(tan\theta\)

\(\Rightarrow 19x=141+180n\) for some integer n is first equation

multiplying equation with 19 gives

\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]

\(\Rightarrow x=159\).

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

Amplitude and Complex numbers – AIME 1996


Let P be the product of the roots of \(z^{6}+z^{4}+z^{2}+1=0\) that have a positive imaginary part and suppose that P=r(costheta+isintheta) where \(0 \lt r\) and \(0 \leq \theta \lt 360\) find \(\theta\)

  • is 107
  • is 276
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Complex Numbers

Integers

Check the Answer


Answer: is 276.

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


here\(z^{6}+z^{4}+z^{2}+1\)=\(z^{6}-z+z^{4}+z^{2}+z+1\)=\(z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}\)=\(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\) then \(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\)=0

gives \(z^{5}=1 for z\neq 1\) gives \(z=cis 72,144,216,288\) and \(z^{2}-z+1=0 for z \neq 1\) gives z=\(\frac{1+-(-3)^\frac{1}{2}}{2}\)=\(cis60,300\) where cis\(\theta\)=cos\(\theta\)+isin\(\theta\)

taking \(0 \lt theta \lt 180\) for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.

.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

Triangle and integers – AIME I, 1995


Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and \(\angle BDC\)=3\(\angle BAC\). then the perimeter of \(\Delta ABC\) may be written in the form \(a+\sqrt{b}\) where a and b are integers, find a+b.

Triangle and integers
  • is 107
  • is 616
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Triangle

Trigonometry

Check the Answer


Answer: is 616.

AIME I, 1995, Question 9

Plane Trigonometry by Loney

Try with Hints


Let x= \(\angle CAM\)

\(\Rightarrow \angle CDM =3x\)

\(\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}\)=11 [by trigonometry ratio property in right angled triangle]

\(\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx\)

solving we get, tanx=\(\frac{1}{2}\)

\(\Rightarrow CM=\frac{11}{2}\)

\(\Rightarrow 2(AC+CM)\) where \(AC=\frac{11\sqrt {5}}{2}\) by Pythagoras formula

=\(\sqrt{605}+11\) then a+b=605+11=616.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

Sequence and greatest integer – AIME I, 2000


Let S be the sum of all numbers of the form \(\frac{a}{b}\),where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed \(\frac{S}{10}\).

  • is 107
  • is 248
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 248.

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

Try with Hints


We have 1000=(2)(2)(2)(5)(5)(5) and \(\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3\)

sum of all numbers of form \(\frac{a}{b}\) such that a and b are relatively prime positive divisors of 1000

=\((2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})\)

\(\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times\) \(\frac{(5^{-3})(5^{7}-1)}{5-1}\)

=2480 + \(\frac{437}{1000}\)

\(\Rightarrow [\frac{s}{10}]\)=248.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

Lcm and Integer – AIME I, 1998


Find the number of values of k in \(12^{12}\) the lcm of the positive integers \(6^{6}\), \(8^{8}\) and k.

  • is 107
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Lcm

Algebra

Integers

Check the Answer


Answer: is 25.

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


here \(k=2^{a}3^{b}\) for integers a and b

\(6^{6}=2^{6}3^{6}\)

\(8^{8}=2^{24}\)

\(12^{12}=2^{24}3^{12}\)

lcm\((6^{6},8^{8})\)=\(2^{24}3^{6}\)

\(12^{12}=2^{24}3^{12}\)=lcm of \((6^{6},8^{6})\) and k

=\((2^{24}3^{6},2^{a}3^{b})\)

=\(2^{max(24,a)}3^{max(6,b)}\)

\(\Rightarrow b=12, 0 \leq a \leq 24\)

\(\Rightarrow\) number of values of k=25.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

Positive Integer – AIME I, 1996


For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that \(n \lt 1000\) and that \([log_{2}n]\) is a positive even integer.

  • is 107
  • is 340
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequality

Greatest integer

Integers

Check the Answer


Answer: is 340.

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

Try with Hints


here Let \([log_{2}n]\)=2k for k is an integer

\(\Rightarrow 2k \leq log_{2}n \lt 2k+1\)

\(\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}\) and \(n \lt 1000\)

\(\Rightarrow 4 \leq n \lt 8\)

\(16 \leq n \lt 32\)

\(64 \leq n \lt 128\)

\(256 \leq n \lt 512\)

\(\Rightarrow 4+16+64+256\)=340.

Subscribe to Cheenta at Youtube