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Algebra Arithmetic Math Olympiad PRMO

Nearest value | PRMO 2018 | Question 14

Try this beautiful problem from the PRMO, 2018 based on Nearest value.

Nearest Value – PRMO 2018


If x=cos1cos2cos3…..cos89 and y=cos2cos6cos10….cos86, then what is the integer nearest to \(\frac{2}{7}log_2{\frac{y}{x}}\)?

  • is 107
  • is 19
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Numbers

Multiples

Check the Answer


Answer: is 19.

PRMO, 2018, Question 14

Higher Algebra by Hall and Knight

Try with Hints


First hint

\(\frac{y}{x}\)=\(\frac{cos2cos6cos10…..cos86}{cos1cos2cos3….cos89}\)

=\(2^{44}\times\sqrt{2}\frac{cos2cos6cos10…cos86}{sin2sin4…sin88}\)

[ since cos\(\theta\)=sin(90-\(\theta\)) from cos 46 upto cos 89 and 2sin\(\theta\)cos\(\theta\)=sin2\(\theta\)]

Second Hint

=\(\frac{2^{\frac{89}{2}}sin4sin8sin12…sin88}{sin2sin4sin6…sin88}\)

[ since sin\(\theta\)=cos(90-\(\theta\))]

=\(\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}\)

[ since cos\(\theta\)=sin(90-\(\theta\))]

Final Step

=\(\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}\)

[since \(cos4cos8cos12…cos88\)

\(=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60\)

\(=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)\)

\(=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)\)

\(=(1/2)^{20}(cos36cos72)\)

\(=(1/2)^{20}(cos36sin18)\)

\(=(1/2)^{22}(4sin18cos18cos36/cos18)\)

\(=(1/2)^{22}(sin72/cos18)\)

\(=(1/2)^{22}\)]

=\(2^\frac{133}{2}\)

\(\frac{2}{7}log_2{\frac{y}{x}}\)=\(\frac{2}{7} \times \frac{133}{2}\)=19.

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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


First hint

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

Second Hint

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

Final Step

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

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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.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Number of points and planes – AIME I, 1999


Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 489
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of points

Plane

Probability

Check the Answer


Answer: is 489.

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

\(10 \choose 3\) sets of 3 points which form triangles,

Second Hint

fourth distinct segment excluding 3 segments of triangles=45-3=42

Final Step

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

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Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

Arithmetic Sequence Problem – AIME 2012


The terms of an arithmetic sequence add to \(715\). The first term of the sequence is increased by \(1\), the second term is increased by \(3\), the third term is increased by \(5\), and in general, the \(k\)th term is increased by the \(k\)th odd positive integer. The terms of the new sequence add to \(836\). Find the sum of the first, last, and middle terms of the original sequence.

  • is 107
  • is 195
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Number Theory

Algebra

Check the Answer


Answer: is 195.

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

Try with Hints


First hint

After the adding of the odd numbers, the total of the sequence increases by \(836 – 715 = 121 = 11^2\).

Second Hint

Since the sum of the first \(n\) positive odd numbers is \(n^2\), there must be \(11\) terms in the sequence, so the mean of the sequence is \(\frac{715}{11} = 65\).

Final Step

Since the first, last, and middle terms are centered around the mean, then \(65 \times 3 = 195\)

Hence option B correct.

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Algebra Arithmetic Math Olympiad PRMO

Smallest value | PRMO 2018 | Question 15

Try this beautiful problem from the PRMO, 2018 based on Smallest value.

Smallest Value – PRMO 2018


Let a and b natural numbers such that 2a-b, a-2b and a+b are all distinct squares. What is the smallest possible value of b?

  • is 107
  • is 21
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Numbers

Multiples

Check the Answer


Answer: is 21.

PRMO, 2018, Question 15

Higher Algebra by Hall and Knight

Try with Hints


First hint

2a-b=\(k_1^2\) is equation 1

a-2b=\(k_2^2\) is equation 2

a+b=\(k_3^2\) is equation 3

Second Hint

adding 2 and 3 we get

2a-b=\(k_2^2+k_3^2\)

or, \(k_2^2+k_3^2\)=\(k_1^2\) \((k_2<k_3)\)

Final Step

For least ‘b’ difference of \(k_3^2\) and \(k_2^2\) is also least and must be multiple of 3

or, \(k_2^2\)=a-2b=\(9^2\) and \(k_3^2\)=a+b=\(12^2\)

or, \(k_3^2-k_2^2\)=3b=144-81=63

or, b=21

or, least b is 21.

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Algebra Arithmetic Math Olympiad PRMO

Digits of number | PRMO 2018 | Question 3

Try this beautiful problem from the PRMO, 2018 based on Digits of number.

Digits of number – PRMO 2018


Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.

  • is 107
  • is 70
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Numbers

Multiples

Check the Answer


Answer: is 70.

PRMO, 2018, Question 3

Higher Algebra by Hall and Knight

Try with Hints


First hint

abccba (b is odd)

=a(\(10^5\)+1)+b(\(10^4\)+10)+c(\(10^3\)+\(10^2\))

=a(1001-1)100+a+10b(1001)+(100)(11)c

=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c

=7p+(c-a) where p is an integer

Second Hint

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

Final Step

since b is odd

number of such number=\(14 \times 5\)=70.

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Geometry Math Olympiad USA Math Olympiad

Length and Triangle | AIME I, 1987 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.

Length and Triangle – AIME I, 1987


Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA. Find PC.

Length and Triangle
  • is 107
  • is 33
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Algebra

Triangles

Check the Answer


Answer: is 33.

AIME I, 1987, Question 9

Geometry Vol I to Vol IV by Hall and Stevens

Try with Hints


First hint

Let PC be x, \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA=120 (in degrees)

Second Hint

Applying cosine law \(\Delta\)APB, \(\Delta\)BPC, \(\Delta\)CPA with cos120=\(\frac{-1}{2}\) gives

\(AB^{2}\)=36+100+60=196, \(BC^{2}\)=36+\(x^{2}\)+6x, \(CA^{2}\)=100+\(x^{2}\)+10x

Final Step

By Pathagorus Theorem, \(AB^{2}+BC^{2}=CA^{2}\)

or, \(x^{2}\)+10x+100=\(x^{2}\)+6x+36+196

or, 4x=132

or, x=33.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

Algebra and Positive Integer – AIME I, 1987


What is the largest positive integer n for which there is a unique integer k such that \(\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}\)?

  • is 107
  • is 112
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 112.

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

Try with Hints


First hint

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

Second Hint

for 91n-104k<n+k, K>\(\frac{6n}{7}\)

and 0<91n-104k gives k<\(\frac{7n}{8}\)

Final Step

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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Geometry Math Olympiad USA Math Olympiad

Distance and Spheres | AIME I, 1987 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Distance and Spheres.

Distance and Sphere – AIME I, 1987


What is the largest possible distance between two points, one on the sphere of radius 19 with center (-2,-10,5) and the other on the sphere of radius 87 with center (12,8,-16)?

  • is 107
  • is 137
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Algebra

Spheres

Check the Answer


Answer: is 137.

AIME I, 1987, Question 2

Geometry Vol I to Vol IV by Hall and Stevens

Try with Hints


First hint

The distance between the center of the spheres is \(\sqrt{(12-(-2)^{2}+(8-(-10))^{2}+(-16-5)^{2}}\)

Second Hint

=\(\sqrt{14^{2}+18^{2}+21^{2}}\)=31

Final Step

The largest possible distance=sum of the two radii+distance between the centers=19+87+31=137.

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