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Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

Logarithms and Equations – AIME I 2000


\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).

  • is 905
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Logarithms

Theory of Equations

Number Theory

Check the Answer


Answer: is 25.

AIME I, 2000, Question 9

Polynomials by Barbeau

Try with Hints


Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)

taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).

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Probability in Games | AIME I, 1999 | Question 13

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.

Probability in Games – AIME I, 1999 Question 13


Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find \(log_{2}n\)

  • 10
  • 742
  • 30
  • 11

Key Concepts


Probability

Theory of equations

Combinations

Check the Answer


Answer: 742.

AIME, 1999 Q13

Course in Probability Theory by Kai Lai Chung .

Try with Hints


\({40 \choose 2}\)=780 pairings with \(2^{780}\) outcomes

no two teams win the same number of games=40! required probability =\(\frac{40!}{2^{780}}\)

the number of powers of 2 in 40!=[\(\frac{40}{2}\)]+[\(\frac{40}{4}\)]+[\(\frac{40}{8}\)]+[\(\frac{40}{16}\)]+[\(\frac{40}{32}\)]=20+10+5+2+1=38 then 780-38=742.

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Theory of Equations | AIME I, 2015 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Theory of Equations.

Theory of Equations – AIME I, 2015


The expressions A=\(1\times2+3\times4+5\times6+…+37\times38+39\)and B=\(1+2\times3+4\times5+…+36\times37+38\times39\) are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.Find the positive difference between integers A and B.

  • is 722
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Equations

Number Theory

Check the Answer


Answer: is 722.

AIME I, 2015, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


A = \((1\times2)+(3\times4)\)

\(+(5\times6)+…+(35\times36)+(37\times38)+39\)

B=\(1+(2\times3)+(4\times5)\)

\(+(6\times7)+…+(36\times37)+(38\times39)\)

B-A=\(-38+(2\times2)+(2\times4)\)

\(+(2\times6)+…+(2\times36)+(2\times38)\)

=722.

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Area of Equilateral Triangle | AIME I, 2015 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry, based on Area of Equilateral Triangle (Question 4).

Area of Triangle – AIME I, 2015


Point B lies on line segment AC with AB =16 and BC =4. Points D and E lie on the same side of line AC forming equilateral triangle ABD and traingle BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of triangle BMN is x. Find \(x^{2}\).

Area of Triangle Problem
  • is 107
  • is 507
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Theory of Equations

Geometry

Check the Answer


Answer: is 507.

AIME, 2015, Question 4

Geometry Revisited by Coxeter

Try with Hints


Let A(0,0), B(16,0),C(20,0). let D and E be in first quadrant. then D =\((8,8\sqrt3)\), E=\((18,2\sqrt3\)).

M=\((9,\sqrt3)\), N=(\(14,4\sqrt3\)), where M and N are midpoints

since BM, BN, MN are all distance, BM=BN=MN=\(2\sqrt13\). Then, by area of equilateral triangle, x=\(13\sqrt3\) then\(x^{2}\)=507.

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Probability Problem | Combinatorics | AIME I, 2015 – Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Probability.

Probability Problem – AIME I, 2015


In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 341
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Theory of Equations

Probability

Check the Answer


Answer: is 341.

AIME, 2015, Question 5

Geometry Revisited by Coxeter

Try with Hints


Wednesday case – with restriction , select the pair on wednesday in \(5 \choose 1 \) ways

Tuesday case – four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is \(8 \choose 2\) – 4

Monday case – a total of 6 socks and a pair not picked \(6 \choose 2\) -2

by multiplication and principle of combinatorics \(\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}\)=\(\frac{26}{315}\). That is 341.

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Probability of tossing a coin | AIME I, 2009 | Question 3

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

Probability of tossing a coin – AIME I, 2009 Question 3


A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to \(\frac{1}{25}\) the probability of five heads and three tails. Let p=\(\frac{m}{n}\) where m and n are relatively prime positive integers. Find m+n.

  • 10
  • 20
  • 30
  • 11

Key Concepts


Probability

Theory of equations

Polynomials

Check the Answer


Answer: 11.

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

Try with Hints


here \(\frac{8!}{3!5!}p^{3}(1-p)^{5}\)=\(\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}\)

then \((1-p)^{2}\)=\(\frac{1}{25}p^{2}\) then 1-p=\(\frac{1}{5}p\)

then p=\(\frac{5}{6}\) then m+n=11

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Equations with number of variables | AIME I, 2009 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2009 based on Equations with a number of variables.

Equations with number of variables – AIME 2009


For t=1,2,3,4, define \(S^{t}=a^{t}_1+a^{t}_2+…+a^{t}_{350}\), where \(a_{i}\in\){1,2,3,4}. If \(S_{1}=513, S_{4}=4745\), find the minimum possible value for \(S_{2}\).

  • is 905
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Theory of Equations

Number Theory

Check the Answer


Answer: is 905.

AIME, 2009, Question 14

Polynomials by Barbeau

Try with Hints


j=1,2,3,4, let \(m_{j}\) number of \(a_{i}\) s = j then \(m_{1}+m{2}+m{3}+m{4}=350\), \(S_{1}=m_{1}+2m_{2}+3m_{3}+4m_{4}=513\) \(S_{4}=m_{1}+2^{4}m_{2}+3^{4}m_{3}+4^{4}m_{4}=4745\)

Subtracting first from second, then first from third yields \(m_{2}+2m_{3}+3m_{4}=163,\) and \(15m_{2}+80m_{3}+255m_{4}=4395\) Now subtracting 15 times first from second gives \(50m_{3}+210m_{4}=1950\) or \(5m_{3}+21m_{4}=195\) Then \(m_{4}\) multiple of 5, \(m_{4}\) either 0 or 5

If \(m_{4}=0\) then \(m_{j}\) s (226,85,39,0) and if \(m_{4}\)=5 then \(m_{j}\) s (215,112,18,5) Then \(S_{2}=1^{2}(226)+2^{2}(85)+3^{2}(39)+4^{2}(0)=917\) and \(S_{2}=1^{2}(215)+2^{2}(112)+3^{2}(18)+4^{2}(5)=905\) Then min 905.

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Exponents and Equations | AIME I, 2010 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Exponents and Equations.

Exponents and Equations – AIME 2010


Suppose that y=\(\frac{3x}{4}\) and \(x^{y}=y^{x}\). The quantity x+y can be expressed as a rational number \(\frac{r}{s}\) , where r and s are relatively prime positive integers. Find r+s.

.

  • is 107
  • is 529
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Number Theory

Check the Answer


Answer: is 529.

AIME, 2010, Question 3.

Elementary Number Theory by Sierpinsky

Try with Hints


y=\(\frac{3x}{4}\) into  \(x^{y}=y^{x}\)  and \(x^{\frac{3x}{4}}\)=\((\frac{3x}{4})^{x}\) implies \(x^{\frac{3x}{4}}\)=\((\frac{3}{4})^{x}x^{x}\) implies \(x^{-x}{4}\)=\((\frac{3}{4})^{x}\) implies \(x^{\frac{-1}{4}}=\frac{3}{4}\) implies \(x=\frac{256}{81}\).

y=\(\frac{3x}{4}=\frac{192}{81}\).

x+y=\(\frac{448}{81}\) then 448+81=529.

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Theory of Equations | AIME I, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.

Theory of Equations – AIME 2015


Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

  • is 107
  • is 72
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Theory of Equations

Algebra

Check the Answer


Answer: is 72.

AIME, 2015, Question 10.

Polynomials by Barbeau.

Try with Hints


Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

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Cube of Positive Integer | Number Theory | AIME I, 2015 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Cube of Positive Integer.

Cube of Positive Numbers – AIME I, 2015


There is a prime number p such that 12p+1 is the cube of positive integer.Find p..

  • is 107
  • is 183
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Theory of Equations

Number Theory

Check the Answer


Answer: is 183.

AIME, 2015, Question 3

Elementary Number Theory by David Burton

Try with Hints


\(a^{3}=12p+1\) implies that \(a^{3}-1=12p\) that is (a-1)(\(a^{2}\)+a+1)=12p

a is odd, a-1 even, \(a^{2} +a+1 odd implies a-1 multiple of 12 that is here =12 then a=12+1 =13

\(a^{2}+a+1=p implies p= 169+13+1=183.

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