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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


First hint

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 \(n \lt 1000\)

Second Hint

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

\(16 \leq n \lt 32\)

\(64 \leq n \lt 128\)

\(256 \leq n \lt 512\)

Final Step

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

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

Integer – AIME I, 1993


Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

  • is 107
  • is 870
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Algebra

Check the Answer


Answer: is 870.

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

Second Hint

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

Final Step

taking first two solutions a<b<c<d<500

or,\(1 \leq c-93, c+1 \leq 499\)

or, \(94 \leq c \leq 498 \) gives 405 solutions

and \(1 \leq c-31, c+3 \leq 499\)

or, \(32 \leq c \leq 496\) gives 465 solutions

or, 405+465=870 solutions.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Trapezoid Problem | AIME I, 1992 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on trapezoid.

Trapezoid – AIME I, 1992


Trapezoid ABCD has sides AB=92, BC=50,CD=19,AD=70 with AB parallel to CD. A circle with centre P on AB is drawn tangent to BC and AD. Given that AP=\(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 164
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Trapezoid

Angle Bisectors

Check the Answer


Answer: is 164.

AIME I, 1992, Question 9

Coordinate Geometry by Loney

Try with Hints


First hint

Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Trapezoid Problem

Second Hint

Let F be point on CD where it meets

Taking angle bisector theorem,

let YB=z(92-y), YA=zy for some z

YD=zy-70, YC=z(92-y)-50

\(\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}\)

Final Step

solving we get 120y=(70)(92)

or, AP=y=\(\frac{161}{3}\)

or, 161+3=164.

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Function Problem | AIME I, 1988 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on function.

Function Problem – AIME I, 1988


For any positive integer k, let \(f_1(k)\) denote the square of the sum of the digits of k. For \(n \geq 2\), let \(f_n(k)=f_1(f_{n-1}(k))\), find \(f_{1988}(11)\).

  • is 107
  • is 169
  • is 634
  • cannot be determined from the given information

Key Concepts


Functions

Equations

Algebra

Check the Answer


Answer: is 169.

AIME I, 1988, Question 2

Functional Equation by Venkatchala

Try with Hints


First hint

\(f_1(11)=4\)

or, \(f_2(11)=f_1(4)=16\)

or, \(f_3(11)=f_1(16)=49\)

Second Hint

or, \(f_4(11)=f_1(49)=169\)

or, \(f_5(11)=f_1(169)=256\)

or, \(f_6(11)=f_1(256)=169\)

or, \(f_7(11)=f_1(169)=256\)

Final Step

This goes on between two numbers with this pattern, here 1988 is even,

or, \(f_1988(11)=f_4(11)=169\).

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Problem on Fibonacci sequence | AIME I, 1988 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.

Fibonacci sequence Problem – AIME I, 1988


Find a if a and b are integers such that \(x^{2}-x-1\) is a factor of \(ax^{17}+bx^{16}+1\).

  • is 107
  • is 987
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Sets

Check the Answer


Answer: is 987.

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

Try with Hints


First hint

Let F(x)=\(ax^{17}+bx^{16}+1\)

Let P(x) be polynomial such that

\(P(x)(x^{2}-x-1)=F(x)\)

constant term of P(x) =(-1)

now \((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\) where \(c_{i}\)=coefficient

Second Hint

comparing the coefficients of x we get the terms

since F(x) has no x term, then \(c_{15}\)=1

getting \(c_{14}\)

\((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\)

=terms +\(0x^{2}\) +terms

or, \(c_{14}=-2\)

proceeding in the same way \(c_{13}=3\), \(c_{12}=-5\), \(c_{11}=8\) gives a pattern of Fibonacci sequence

Final Step

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=\(c_1=F_{16}\) where \(F_{16}\) is 16th Fibonacci number

or, a=987.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Reflection Problem | AIME I, 1988 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988, Question 14, based on Reflection.

Reflection Problem – AIME I, 1988


Let C be the graph of xy=1 and denote by C’ the reflection of C in the line y=2x. let the equation of C’ be written in the form \(12x^{2}+bxy +cy^{2}+d=0\), find the product bc.

  • is 107
  • is 84
  • is 840
  • cannot be determined from the given information

Key Concepts


Geometry

Equation

Algebra

Check the Answer


Answer: is 84.

AIME I, 1988, Question 14

Coordinate Geometry by Loney

Try with Hints


First hint

Let P(x,y) on C such that P'(x’,y’) on C’ where both points lie on the line perpendicular to y=2x

slope of PP’=\(\frac{-1}{2}\), then \(\frac{y’-y}{x’-x}\)=\(\frac{-1}{2}\)

or, x’+2y’=x+2y

also midpoint of PP’, \((\frac{x+x’}{2},\frac{y+y’}{2})\) lies on y=2x

Second Hint

or, \(\frac{y+y’}{2}=x+x’\)

or, 2x’-y’=y-2x

solving these two equations, x=\(\frac{-3x’+4y’}{5}\) and \(y=\frac{4x’+3y’}{5}\)

putting these points into the equation C \(\frac{(-3x’+4y’)(4x’+3y’)}{25}\)=1

Final Step

which when expanded becomes

\(12x’^{2}-7x’y’-12y’^{2}+25=0\)

or, bc=(-7)(-12)=84.

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AIME I Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Problem on Complex plane | AIME I, 1988| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Complex Plane.

Problem on Complex Plane – AIME I, 1988


Let w_1,w_2,….,w_n be complex numbers. A line L in the complex plane is called a mean line for the points w_1,w_2,….w_n if L contains points (complex numbers) z_1,z_2, …..z_n such that \(\sum_{k=1}^{n}(z_{k}-w_{k})=0\) for the numbers \(w_1=32+170i, w_2=-7+64i, w_3=-9+200i, w_4=1+27i\) and \(w_5=-14+43i\), there is a unique mean line with y-intercept 3. Find the slope of this mean line.

  • is 107
  • is 163
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 163.

AIME I, 1988, Question 11

Elementary Algebra by Hall and Knight

Try with Hints


First hint

\(\sum_{k=1}^{5}w_k=3+504i\)

and \(\sum_{k-1}^{5}z_k=3+504i\)

Second Hint

taking the numbers in the form a+bi

\(\sum_{k=1}^{5}a_k=3\) and \(\sum_{k=1}^{5}b_k=504\)

Final Step

or, y=mx+3 where \(b_k=ma_k+3\) adding all 5 equations given for each k

or, 504=3m+15

or, m=163.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Problem on Real Numbers | AIME I, 1990| Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Real Numbers.

Problem on Real Numbers – AIME I, 1990


Find \(ax^{5}+by^{5}\) if real numbers a,b,x,y satisfy the equations

ax+by=3

\(ax^{2}+by^{2}=7\)

\(ax^{3}+by^{3}=16\)

\(ax^{4}+by^{4}=42\)

  • is 107
  • is 20
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 20.

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let S=x+y, P=xy

\((ax^{n}+by^{n})(x+y)\)

\(=(ax^{n+1}+by^{n+1})+(xy)(ax^{n-1}+by^{n-1})\)

Second Hint

or,\( (ax^{2}+by^{2})(x+y)=(ax^{3}+by^{3})+(xy)(ax+by)\) which is first equation

or,\( (ax^{3}+by^{3})(x+y)=(ax^{4}+by^{4})+(xy)(ax^{2}+by^{2})\) which is second equation

or, 7S=16+3P

16S=42+7P

or, S=-14, P=-38

Final Step

or, \((ax^{4}+by^{4})(x+y)=(ax^{5}+by^{5})+(xy)(ax^{3}+by^{3})\)

or, \(42S=(ax^{5}+by^{5})+P(16)\)

or, \(42(-14)=(ax^{5}+by^{5})+(-38)(16)\)

or, \(ax^{5}+by^{5}=20\).

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Digits and Integers | AIME I, 1990 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.

Digits and Integers – AIME I, 1990


Let T={\(9^{k}\): k is an integer, \(0 \leq k \leq 4000\)} given that \(9^{4000}\) has 3817 digits and that its first (leftmost) digit is 9, how many elements of T have 9 as their leftmost digit?

  • is 107
  • is 184
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Sets

Check the Answer


Answer: is 184.

AIME I, 1990, Question 13

Elementary Number Theory by David Burton

Try with Hints


First hint

here \(9^{4000}\) has 3816 digits more than 9,

Second Hint

or, 4000-3816=184

Final Step

or, 184 numbers have 9 as their leftmost digits.

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AIME I Complex Numbers Math Olympiad USA Math Olympiad

Complex numbers and Sets | AIME I, 1990 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Complex Numbers and Sets.

Complex Numbers and Sets – AIME I, 1990


The sets A={z:\(z^{18}=1\)} and B={w:\(w^{48}=1\)} are both sets of complex roots with unity, the set C={zw: \(z \in A and w \in B\)} is also a set of complex roots of unity. How many distinct elements are in C?.

  • is 107
  • is 144
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Complex Numbers

Sets

Check the Answer


Answer: is 144.

AIME I, 1990, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

18th and 48th roots of 1 found by de Moivre’s Theorem

=\(cis(\frac{2k_1\pi}{18})\) and \(cis(\frac{2k_2\pi}{48})\)

Second Hint

where \(k_1\), \(K_2\) are integers from 0 to 17 and 0 to 47 and \(cis \theta = cos \theta +i sin \theta\)

zw= \(cis(\frac{k_1\pi}{9}+\frac{k_2\pi}{24})=cis(\frac{8k_1\pi+3k_2\pi}{72})\)

Final Step

and since the trigonometric functions are periodic every period \({2\pi}\)

or, at (72)(2)=144 distinct elements in C.

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