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AIME I Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Complex Numbers and prime | AIME I, 2012 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Complex Numbers and prime.

Complex Numbers and primes – AIME 2012


The complex numbers z and w satisfy \(z^{13} = w\) \(w^{11} = z\) and the imaginary part of z is \(\sin{\frac{m\pi}{n}}\), for relatively prime positive integers m and n with m<n. Find n.

  • is 107
  • is 71
  • is 840
  • cannot be determined from the given information

Key Concepts


Complex Numbers

Algebra

Number Theory

Check the Answer


Answer: is 71.

AIME I, 2012, Question 6

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


Taking both given equations \((z^{13})^{11} = z\) gives \(z^{143} = z\) Then \(z^{142} = 1\)

Then by De Moivre’s theorem, imaginary part of z will be of the form \(\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}}\) where \(k \in {1, 2, upto 70}\)

71 is prime and n = 71.

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Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

Exponents and Equations | AIME I, 2010 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Exponents and Equations.

Exponents and Equations – AIME 2010


Suppose that y=\(\frac{3x}{4}\) and \(x^{y}=y^{x}\). The quantity x+y can be expressed as a rational number \(\frac{r}{s}\) , where r and s are relatively prime positive integers. Find r+s.

.

  • is 107
  • is 529
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Number Theory

Check the Answer


Answer: is 529.

AIME, 2010, Question 3.

Elementary Number Theory by Sierpinsky

Try with Hints


y=\(\frac{3x}{4}\) into  \(x^{y}=y^{x}\)  and \(x^{\frac{3x}{4}}\)=\((\frac{3x}{4})^{x}\) implies \(x^{\frac{3x}{4}}\)=\((\frac{3}{4})^{x}x^{x}\) implies \(x^{-x}{4}\)=\((\frac{3}{4})^{x}\) implies \(x^{\frac{-1}{4}}=\frac{3}{4}\) implies \(x=\frac{256}{81}\).

y=\(\frac{3x}{4}=\frac{192}{81}\).

x+y=\(\frac{448}{81}\) then 448+81=529.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and Integers | AIME I, 2007 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

Sequence and Integers – AIME I, 2007


A sequence is defined over non negetive integral indexes in the following way \(a_0=a_1=3\), \( a_{n+1}a_{n-1}=a_n^{2}+2007\), find the greatest integer that does not exceed \(\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}\)

  • is 107
  • is 224
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Inequalities

Integers

Check the Answer


Answer: is 224.

AIME I, 2007, Question 14

Elementary Number Theory by David Burton

Try with Hints


\(a_{n+1}a_{n-1}\)=\(a_{n}^{2}+2007\) then \(a_{n-1}^{2} +2007 =a_{n}a_{n-2}\) adding these \(\frac{a_{n-1}+a_{n+1}}{a_{n}}\)=\(\frac{a_{n}+a_{n-2}}{a_{n-1}}\), let \(b_{j}\)=\(\frac{a_{j}}{a_{j-1}}\) then \(b_{n+1} + \frac{1}{b_{n}}\)=\(b_{n}+\frac{1}{b_{n-1}}\) then \(b_{2007} + \frac{1}{b_{2006}}\)=\(b_{3}+\frac{1}{b_{2}}\)=225

here \(\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}\)=\(\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}\) then \(b_{2007}\)=\(\frac{a_{2007}}{a_{2006}}\)=\(\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}\)\( \gt \)\(\frac{a_{2006}}{a_{2005}}\)=\(b_{2006}\)

then \(b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}\)=225 which is small less such that all \(b_{j}\) s are greater than 1 then \(\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}\)=\(b_{2007}+\frac{1}{b_{2007}}\)=224.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Patterns and Integers | AIME I, 2001 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2001 based on Patterns and integers.

Patterns and integers – AIME I, 2001


A mail carrier delivers mail to the nineteen houses on the east side of Elm street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day, find the number of different patterns of mail delivery are possible.

  • is 107
  • is 351
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Patterns

Integers

Check the Answer


Answer: is 351.

AIME I, 2001, Question 14

Elementary Number Theory by David Burton

Try with Hints


0=house receive no mail 1=house receives mail and last two digit is not 11 then 00, 01 and 10 let \(a_{n}\)=number of n digit string ending 00,\(b_{n}\)=number of n digit string ending 01 \(c_{n}\)=number of n digit string ending 10

when nth digit ends in 00 then previous digit is 1 and the last two digits in the n-1 th substring is 10 then \(a_{n}=c_{n-1}\) when nth digit string ends in 01 then previous digit 0 or 1 and the last two digits in the n-1 digit substring is 00 or 10 then \(b_{n}=a_{n-1}+c_{n-1}\) when nth digit string ends in 10 then previous digit 0 and the last two digit of the n-1 digit substring is 01 then \(c_{n}=b_{n-1}\)

\(a_{n} b_{n} c_{n}\)

for n=2 , 1 1 1

for n=3 , 1 2 1

for n=4 , 1 2 2

for n=5 , 2 3 2

for n=6 , 2 4 3

for n=7 , 3 5 4

for n=8 , 4 7 5

for n=9 , 5 9 7

for n-10 , 7 12 9

for n=11 , 9 16 12

for n=12, 12 21 16

for n=13 , 16 28 21

for n=14 , 21 37 28

for n=15 , 28 49 37

for n=16 , 37 65 49

for n=17 , 49 86 65

for n=18 , 65 114 86

for n=19 , 86 151 114

then \(a_{19}+b_{19}+c_{19}\)=86+151+114=351.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

GCD and Sequence | AIME I, 1985 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

GCD and Sequence – AIME I, 1985


The numbers in the sequence 101, 104,109,116,…..are of the form \(a_n=100+n^{2}\) where n=1,2,3,——-, for each n, let \(d_n\) be the greatest common divisor of \(a_n\) and \(a_{n+1}\), find the maximum value of \(d_n\) as n ranges through the positive integers.

  • is 107
  • is 401
  • is 840
  • cannot be determined from the given information

Key Concepts


GCD

Sequence

Integers

Check the Answer


Answer: is 401.

AIME I, 1985, Question 13

Elementary Number Theory by David Burton

Try with Hints


\(a_n=100+n^{2}\) \(a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1\) and \(a_{n+1}-a_{n}=2n +1\)

\(d_{n}|(2n+1)\) and \(d_{n}|(100 +n^{2})\) then \(d_{n}|[(100+n^{2})-100(2n+1)]\) then \(d_{n}|(n^{2}-200n)\)

here \(n^{2} -200n=0\) then n=200 then \(d_{n}\)=2n+1=2(200)+1=401.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Planes and distance | AIME I, 2011 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and Distance.

Planes and distance- AIME I, 2011


A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled A, The three vertices adjacent to vertex A are at heights 10,11 and 12 above the plane. The distance from vertex A to the plane can be expressed as \(\frac{r-s^\frac{1}{2}}{t}\), where r and s and t are positive integers and \(r+s+t \lt 1000\), find r+s+t.

  • is 107
  • is 330
  • is 840
  • cannot be determined from the given information

Key Concepts


Plane

Distance

Algebra

Check the Answer


Answer: is 330.

AIME I, 2011, Question 13

Geometry Revisited by Coxeter

Try with Hints


Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d

\(\frac{a10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}\)=10-d and \(\frac{b10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}\)=11-d and \(\frac{c10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}\)=12-d

squaring and adding \(100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}\) then having 11-d=y, 100=3\(y^{2}\)+2then y=\(\frac{98}{3}^\frac{1}{2}\) then d=11-\(\frac{98}{3}^\frac{1}{2}\)=\(\frac{33-294^\frac{1}{2}}{3}\) then 33+294+3=330.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Logic and speed | AIME I, 2008 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Logic and Speed.

Logic and Speed – AIME I, 2008


Ed and Sue bike at equal and constant rates and they swim at equal and constant rates. The same way they jog at equal and constant rates. ed covers 74 kms after biking for 2 hrs, jogging for 3 hrs and swimming for 4 hrs while sue covers 91 kms after jogging for 2 hrs swimming for 3 hrs and biking for 4 hrs. Their biking jogging and swimming rates are whole numbers of km/hr, find the sum of the squares of Ed’s biking jogging and swimming rates.

  • is 107
  • is 314
  • is 840
  • cannot be determined from the given information

Key Concepts


Logic

Speed

Integers

Check the Answer


Answer: is 314.

AIME I, 2008, Question 3

Elementary Number Theory by David Burton

Try with Hints


Let a,b, c be biking jogging and swimming rates then 2a+3b+4c=74 first eqn and 4a+2b+3c=91 second eqn subtracting second from first eqn gives 2a-b-c=17 third eqn

third eqn multiplied by 3 + first eqn gives 8a+c=125 gives \(a \leq 15\) third eqn multiplied by 4 +first eqn gives 10a-b=142 gives \(a \gt 14\)

then a=15 and b=8, c=5 and \(a^{2} +b^{2} + c^{2}\)=225+64+25=314.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Function and symmetry | AIME I, 1984 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.

Function and Symmetry – AIME I 1984


A function f is defined for all real numbers and satisfies f(2+x)=f(2-x) and f(7+x)=f(7-x) for all x. If x=0 is root for f(x)=0, find the least number of roots f(x) =0 must have in the interval \(-1000 \leq x\leq 1000\).

  • is 107
  • is 401
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Symmetry

Number Theory

Check the Answer


Answer: is 401.

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

Try with Hints


by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots

here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)

Then least number of roots as 401.

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Equations and Integers | AIME I, 2008 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Equations and integers.

Equations and integers – AIME I, 2008


There exists unique positive integers x and y that satisfy the equation \(x^{2}+84x+2008=y^{2}\)

  • is 107
  • is 80
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Integers

Check the Answer


Answer: is 80.

AIME I, 2008, Question 4

Elementary Number Theory by David Burton

Try with Hints


\(y^{2}=x^{2}+84x+2008=(x+42)^{2}+244\) then 244=\(y^{2}-(x+42)^{2}=(y-x-42)(y+x+42)\)

here 244 is even and 244=\(2^{2}(61)\)=\( 2 \times 122\) for \(x,y \gt 0\)

(y-x-42)=2 and (y+x+42)=122 then y+x=80 and x=18 y=62.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Percentage Problem | AIME I, 2008 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Percentage.

Percentage Problem – AIME I, 2008


Of the students attending a party, 60% of the students are girls, and 40% of the students like to dance. After these students are joined by 20 more boy students, all of whom like to dance, the party is now 58% girls, find number of students now at the party like to dance.

  • is 107
  • is 252
  • is 840
  • cannot be determined from the given information

Key Concepts


Ratios

Percentage

Numbers

Check the Answer


Answer: is 252.

AIME I, 2008, Question 1

Elementary Number Theory by David Burton

Try with Hints


Let number of girls and boys be 3k and 2k, out of 3k girls, 2k likes to dance and 2k+20(boys) like to dance

here given that \(\frac{3k}{5k+20}\)=\(\frac{58}{100}\) then k=116

then 2k+20=252.

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