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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


\(f_1(11)=4\)

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

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

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

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 Functions Math Olympiad USA Math Olympiad

Arranging in column | AIME I, 1990 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Arranging in column.

Arranging in column – AIME I, 1990


In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marks man is to break all the targets according to the following rules

1 ) The marksman first chooses a column from which a target is to be broken,

2 ) the marksman must then break the lowest remaining target in the chosen column. If the rules are followed, in how many different orders can the eight targets be broken?

  • is 107
  • is 560
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Arrangement

Algebra

Check the Answer


Answer: is 560.

AIME I, 1990, Question 8

Combinatorics by Brualdi

Try with Hints


Let the columns be labelled A,B and C such that first three targets are A, A and A the next three being B, B and B and the next being C and C in which we consider the string AAABBBCC.

Since the arrangement of the strings is one-one correspondence and onto to the order of shooting for example first A is shot first, second A is shot second, third A is shot third, first B is shot fourth, second B is shot fifth, third B is shot sixth, first C is shot seventh, second C is shot eighth,

or, here arrangement of the strings is bijective to the order of the shots taken

the required answer is the number of ways to arrange the letters which is \(\frac{8!}{3!3!2!}\)=560.

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Digits and Order | AIME I, 1992 | Question 2

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

Digits and order – AIME I, 1992


A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. Find number of ascending positive integers are there.

  • is 107
  • is 502
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Order

Check the Answer


Answer: is 502.

AIME I, 1992, Question 2

Elementary Number Theory by David Burton

Try with Hints


There are nine digits that we use 1,2,3,4,5,6,7,8,9.

Here each digit may or may not be present.

\(\Rightarrow 2^{9}\)=512 potential ascending numbers, one for subset of {1,2,3,4,5,6,7,8,9}

Subtracting empty set and single digit set

=512-10

=502.

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Remainders and Functions | AIME I, 1994 | Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.

Remainders and Functions – AIME I, 1994


The function f has the property that, for each real number x, \(f(x)+f(x-1)=x^{2}\) if f(19)=94, find the remainder when f(94) is divided by 1000.

  • is 107
  • is 561
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Remainder

Functions

Check the Answer


Answer: is 561.

AIME I, 1994, Question 7

Elementary Number Theory by David Burton

Try with Hints


f(94)=\(94^{2}-f(93)=94^{2}-93^{2}+f(92)\)

=\(94^{2}-93^{2}+92^{2}-f(91)\)

=\((94^{2}-93^{2})+(92^{2}-91^{2})\)

\(+….+(22^{2}-21^{2})+20^{2}-f(19)\)

=94+93+…..+21+400-94

=4561

\(\Rightarrow\) remainder =561.

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AIME I Algebra Complex Numbers Functions Math Olympiad USA Math Olympiad

Function of Complex numbers | AIME I, 1999 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex Numbers and Integers.

Function of Complex Numbers – AIME I, 1999


Let f(z) =(a+bi)z where a,b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin given that |a+bi|=8 and that \(b^{2}\)=\(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 259
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Integers

Complex Numbers

Check the Answer


Answer: is 259.

AIME I, 1999, Question 9

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


Let z=1+i f(1+i)=(a+bi)(1+i)=(a-b)+(a+b)i The image point must be equidistant from (1,1) and(0,0) then the image point lie on the line with slope -1 and which passes through \((\frac{1}{2},\frac{1}{2})\) that is x+y=1

putting x=(a-b) and y=(a+b) gives 2a=1 and \(a=\frac{1}{2}\)

and \((\frac{1}{2})^{2} +b^{2}=8^{2}\) then \(b^{2}=\frac{255}{4}\) then 255+4=259.

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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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