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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Consecutive positive Integers | AIME I, 1990| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.

Consecutive positive integer – AIME I, 1990


Someone observed that 6!=(8)(9)(10). Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers.

  • is 107
  • is 23
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Inequality

Algebra

Check the Answer


Answer: is 23.

AIME I, 1990, Question 11

Elementary Number Theory by David Burton

Try with Hints


First hint

The product of (n-3) consecutive integers=\(\frac{(n-3+a)!}{a!}\) for a is an integer

Second Hint

\(n!=\frac{(n-3+a)!}{a!}\) for \(a \geq 3\) \((n-3+a)! \geq n!\)

or, \(n!=\frac{n!(n+1)(n+2)….(n-3+a)}{a!}\)

Final Step

for a=4, n+1=4! or, n=23 which is greatest here

n=23.

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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


First hint

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

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

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

Second Hint

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

Final Step

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


First hint

5 tails flipped, any less,

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

Second Hint

(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

Final Step

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


First hint

for no carrying required

Second Hint

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

Final Step

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


First hint

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

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

Second Hint

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

Final Step

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


First hint

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.

Second Hint

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

Final Step

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


First hint

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.

Second Hint

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

Final Step

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

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

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Smallest positive Integer Problem | AIME I, 1990 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on smallest positive integer.

Smallest positive Integer Problem – AIME I, 1990


Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find \(\frac{n}{75}\).

  • is 107
  • is 432
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 432.

AIME I, 1990, Question 5

Elementary Number Theory by David Burton

Try with Hints


First hint

75=\(3 \times 5^{2}\)=(2+1)(4+1)(4=1)

or, \(n=p_1^{a_1-1}p_2^{a_2-1}…..\) such that \(a_1a_2….=75\)

Second Hint

or, 75|n with two prime factors 3 and 5

Minimizing n third factor =2

and factor 5 raised to least power

Final Step

or, \(n=(2)^{4}(3)^{4}(5)^{2}\)

and \(\frac{n}{75}=(2)^{4}(3)^{4}(5)^{2}\)=(16)(27)=432.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Proper divisors | AIME I, 1986 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1986 based on Proper divisors.

Proper Divisor – AIME I, 1986


Let S be the sum of the base 10 logarithms of all the proper divisors (all divisors of a number excluding itself) of 1000000. What is the integer nearest to S?

  • is 107
  • is 141
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisors

Algebra

Check the Answer


Answer: is 141.

AIME I, 1986, Question 8

Elementary Number Theory by David Burton

Try with Hints


First hint

1000000=\(2^{6}5^{6}\) or, (6+1)(6+1)=49 divisors of which 48 are proper

\(log1+log2+log4+….+log1000000\)

\(=log(2^{0}5^{0})(2^{1}5^{0})(2^{2}5^{0})….(2^{6}5^{6})\)

Second Hint

power of 2 shows 7 times, power of 5 shows 7 times

total power of 2 and 5 shows=7(1+2+3+4+5+6)

=(7)(21)=147

Final Step

for proper divisor taking out \(2^{6}5^{6}\)=147-6=141

or, \(S=log2^{141}5^{141}=log10^{141}=141\).

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Combinatorics in Tournament | AIME I, 1985 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on Combinatorics in Tournament.

Combinatorics in Tournament- AIME I, 1985


In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two player earned \(\frac{1}{2}\) point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

  • is 107
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Combinatorics

Algebra

Check the Answer


Answer: is 25.

AIME I, 1985, Question 14

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let there be n+10 players

Case I from n players not in weakest 10, \({n \choose 2}\) games played and \({n \choose 2}\) points earned

Case II n players also earned \({n \choose 2}\) points against weakest 10

Second Hint

Case III now weakest 10 played among themselves \({10 \choose 2}\)=45 games and 45 points earned

Case IV 10 players also earned 45 points against stronger n

So total points earned= 2[\({n \choose 2}\)+45]=\(n^{2}-n+90\)

Final Step

case V 1 point earned per game \({n+10 \choose 2}\)=\(\frac{(n+10)(n+9)}{2}\) games and \(\frac{(n+10)(n+9)}{2}\) points earned

So \(n^{2}-n+90=\frac{(n+10)(n+9)}{2}\)

or, \(n^{2}-21n+90=0\)

or, n=6, n=15 here taking n>10,

or, n=15 or, n+10=25.

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