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Algebra AMC 10 Math Olympiad USA Math Olympiad

Sum of the numbers | AMC-10A, 2001 | Problem 16

Try this beautiful problem from Algebra based on Sum of the numbers….

Sum of the numbers – AMC-10A, 2001- Problem 16


The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?

  • \(5\)
  • \(20\)
  • \(30\)
  • \(25\)
  • \(36\)

Key Concepts


algebra

Mean

Median

Check the Answer


Answer: \(30\)

AMC-10A (2001) Problem 16

Pre College Mathematics

Try with Hints


 Mean of three numbers means average of three numbers……

Let \(x\) be the mean of three numbers then we can say that the least of the numbers is $m-10$ and the greatest is $m + 15$

Can you now finish the problem ……….

Given that The median of the three numbers is $5$. Now “median” means the middle of the three numbers

so we can write $\frac{1}{3}[(m-10) + 5 + (m + 15)] = m$,

\(\Rightarrow m=10\)

Therefore The sum of three numbers are \(3(10)=30\)

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Convex polyhedron Problem | AIME I, 1988 | Question 10

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

Convex polyhedron Problem – AIME I, 1988


A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

  • is 107
  • is 840
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Edges

Algebra

Check the Answer


Answer: is 840.

AIME I, 1988, Question 10

Geometry Revisited by Coxeter

Try with Hints


\({48 \choose 2}\)=1128

Every vertex lies on exactly one vertex of a square/hexagon/octagon

V=(12)(4)=(8)(6)=(6)(8)=48

each vertex is formed by the trisection of three edges and every edge is counted twice, once at each of its endpoints, the number of edges E=\(\frac{3V}{2}\)=72

each of the segment on face of polyhedron is diagonal of that face, so each square gives \(\frac{n(n-3)}{2}=2\) diagonals, each hexagon=9,each octagon=20. The number of diagonals is \((2)(12)+(9)(8)+(20)(6)\)=216

or, number of space diagonals =1128-72-216=840.

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Fair coin Problem | AIME I, 1990 | Question 9

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

Fair Coin Problem – AIME I, 1990


A fair coin is to be tossed 10 times. Let i|j, in lowest terms, be the probability that heads never occur on consecutive tosses, find i+j.

  • is 107
  • is 73
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Combinatorics

Algebra

Check the Answer


Answer: is 73.

AIME I, 1990, Question 9

Elementary Algebra by Hall and Knight

Try with Hints


5 tails flipped, any less,

by Pigeonhole principle there will be heads that appear on consecutive tosses

(H)T(H)T(H)T(H)T(H)T(H) 5 tails occur there are 6 slots for the heads to be placed but only 5H remaining, \({6 \choose 5}\) possible combination of 6 heads there are

\(\sum_{i=6}^{11}{i \choose 11-i}\)=\({6 \choose 5} +{7 \choose 4}+{8 \choose 3}+{9 \choose 2} +{10 \choose 1} +{11 \choose 0}\)=144

there are \(2^{10}\) possible flips of 10 coins

or, probability=\(\frac{144}{1024}=\frac{9}{64}\) or, 9+64=73.

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AMC 10 Math Olympiad USA Math Olympiad

Algebraic Equation | AMC-10A, 2001 | Problem 10

Try this beautiful problem from Algebra based on Algebraic Equation.

Algebraic Equation – AMC-10A, 2001- Problem 10


If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is

  • \(5\)
  • \(20\)
  • \(22\)
  • \(25\)
  • \(36\)

Key Concepts


algebra

Equation

sum

Check the Answer


Answer: \(22\)

AMC-10A (2001) Problem 10

Pre College Mathematics

Try with Hints


The given equations are $xy=24$ and $xz=48$.we have to find out \(x+y+z\)

Now using two relations we can write \(\frac{xy}{xz}=\frac{24}{48}\)\(\Rightarrow 2y=z\)

Can you now finish the problem ……….

Now if we substitute the value $z = 2y$ into the equation $yz = 72$ then we will get \(2y^2=72\) \( \Rightarrow y=6\) and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$.

Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)

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AIME I Algebra Arithmetic Coordinate Geometry Math Olympiad USA Math Olympiad

Ordered pair Problem | AIME I, 1987 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Ordered pair.

Ordered pair Problem – AIME I, 1987


An ordered pair (m,n) of non-negative integers is called simple if the additive m+n in base 10 requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to 1492.

  • is 107
  • is 300
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Ordered pair

Algebra

Check the Answer


Answer: is 300.

AIME I, 1987, Question 1

Elementary Algebra by Hall and Knight

Try with Hints


for no carrying required

the range of possible values of any digit m is from 0 to 1492 where the value of n is fixed

Number of ordered pair (1+1)(4+1)(9+1)(2+1)

=(2)(5)(10)(3)

=300.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Positive divisor | AIME I, 1988 | Question 5

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

Positive Divisors- AIME I, 1988


Let \(\frac{m}{n}\) in lowest term, be the probability that a randomly chosen positive divisor of \(10^{99}\) is an integer multiple of \(10^{88}\). Find m+n.

  • is 107
  • is 634
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

DIvisors

Algebra

Check the Answer


Answer: is 634.

AIME I, 1988, Question 5

Elementary Number Theory by David Burton

Try with Hints


\(10^{99}=2^{99}5^{99}\)

or, (99+1)(99+1)=10000 factors

those factors divisible by \(10^{88}\)

are bijection to number of factors \(10{11}=2^{11}5^{11}\) has, which is (11+1)(11+1)=144

one required probability =\(\frac{m}{n}=\frac{144}{10000}=\frac{9}{625}\)

m+n=634.

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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 Combinatorics Math Olympiad USA Math Olympiad

Head Tail Problem | AIME I, 1986 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1986 based on Head Tail Problem.

Head Tail Problem – AIME I, 1986


In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head , a head is immediately followed by ahead and etc. We denote these by TH, HH, and etc. For example in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH, three HT, four TH and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH, three HT, four TH, and five TT subsequences?

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

Key Concepts


Integers

Combinatorics

Algebra

Check the Answer


Answer: is 560.

AIME I, 1986, Question 13

Elementary Algebra by Hall and Knight

Try with Hints


Let us observe the sequences.

H switches to T three times, T switches to H four times.

There are 5 TT subsequences.

We are to add 5 T’s into, the string. There are already 4 T’s in the sequence.

We are to add 5 balls in 4 urns which is same as 3 dividers \({5+3 \choose 3}\)=56

We do the same with 2H’s to get \({2+3 \choose 3}\)=10

so, \(56 \times 10\)=560.

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Algebra AMC 10 Math Olympiad USA Math Olympiad

Sum of the digits | AMC-10A, 2007 | Problem 25

Try this beautiful problem from Algebra based on Sum of the digits.

Sum of the digits – AMC-10A, 2007- Problem 25


For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is \(n+s(n)+s(s(n))=2007\)

  • \(1\)
  • \(2\)
  • \(3\)
  • \(4\)

Key Concepts


algebra

function

multiplication

Check the Answer


Answer: \(4\)

AMC-10A (2007) Problem 25

Pre College Mathematics

Try with Hints


Let \(P(n)=(n+s(n)+s(s(n))=2007)\) tnen obviously \(n<2007\)

For \(n\)=\(1999\).the sum becoms \(28+10)=38\)

so we may say that the minimum bound is \(1969\)

Now we want to break it in 3 parts …..

Case 1:\(n \geq 2000\),

Case 2:\(n \leq 2000\) (\(n = 19xy,x+y<10\)

Case 3:\(n \leq 2000\) (\(n = 19xy,x+y \geq 10\)

Can you now finish the problem ……….

Case 1:\(n \geq 2000\),

Then \(P(n)=(n+s(n)+s(s(n))=2007)\) gives \(n=2001\)

Case 2:\(n \leq 2000\) (\(n = 19xy,x+y<10\)

Then \(P(n)=(n+s(n)+s(s(n))=2007)\) gives \(4x+y=32\) which satisfying the constraints \(x = 8\), \(y = 0\).

Case 3:\(n \leq 2000) ((n = 19xy,x+y \geq 10\) gives \(4x+y=35\) which satisfying the constraints \(x = 7\), \(y = 7\) and \(x = 8\), \(y = 3\).

can you finish the problem……..

Therefore The solutions are thus \(1977, 1980, 1983, 2001\)

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Algebra AMC 8 Arithmetic India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Problem on Positive Integers | PRMO-2019 | Problem 26

Try this beautiful problem from Algebra PRMO 2019 based on Positive Integers.

Positive Integers | PRMO | Problem 26


Positive integers x,y,z satisfy xy+z=160 compute smallest possible value of x+yz.

  • 24
  • 50
  • 29
  • 34

Key Concepts


Algebra

Integer

sum

Check the Answer


Answer:50

PRMO-2019, Problem 26

Higher Algebra by Hall and Knight

Try with Hints


x+yz=\(\frac{160-z}{y}\)+yz

=\(\frac{160}{y}+\frac{z(y^{2}-1)}{y}=\frac{160-z}{y}+\frac{zy^{2}}{y}=\frac{160-z}{y}+zy\)

for particular value of z, \(x+yz \geq 2\sqrt{z(160-z)}\)

or, least value=\(2\sqrt{z(160-z)}\) but an integer also

for least value z is also

case I z=1, \(x+yz=\frac{159}{y}+y\) or, min value at y=3 which is 56

case II z=2, \(x+yz=\frac{158}{y}+2y\) or, min value at y =2 which is 83 (not taken)

case III z=3, \(x+yz=\frac{157}{y}+3y\) or, min value at y=1 which is 160 (not taken)

case IV z=4, \(x+yz=\frac{156}{y}+4y\) or, min at y=6 which is 50 (taken)

case V z=5, \(x+yz=\frac{155}{y}+5y\) or, minimum value at y=5 which is 56 (not taken)

case VI z=6, \(x+yz=\frac{154}{y}+6y\) \( \geq 2\sqrt{924}\)>50

smallest possible value =50.

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