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AMC 8 Math Olympiad USA Math Olympiad

Probability in Divisibility | AMC-10A, 2003 | Problem 15

Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

Probability in Divisibility – AMC-10A, 2003- Problem 15


What is the probability that an integer in the set \({1,2,3,…,100}\) is divisible by \(2\) and not divisible by \(3\)?

  • \(\frac {33}{100}\)
  • \(\frac{1}{6}\)
  • \(\frac{17}{50}\)
  • \(\frac{1}{2}\)
  • \(\frac{18}{25}\)

Key Concepts


Number system

Probability

divisibility

Check the Answer


Answer: \(\frac{17}{50}\)

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints


There are total number of integers are \(100\).and numer of integers divisible by \(2\) is \(\frac{100}{2}\)=\(50\). Now we have to find out divisible by \(2\) and not divisible by \(3\). so at first we have to find out the numbers of integers which are divisible by \(2\) and \(3\) both…….

can you finish the problem……..

To be divisible by both \(2\) and \(3\), a number must be divisible by the lcm of \((2,3)=6\).

Therefore numbers of integers which are divisible by \(6\)=\(\frac{100}{6}=16\) (between \(1\) & \( 100\))

can you finish the problem……..

Therefore the number of integers which are divisible by \(2\) and not divisible by \(3\)= \(50 – 16=34\).

So require probability= \(\frac{34}{100}=\frac{17}{50}\)

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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


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 \(a_1a_2….=75\)

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

Minimizing n third factor =2

and factor 5 raised to least power

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

Algebraic value | AIME I, 1990 | Question 2

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

Algebraic value – AIME I, 1990


Find the value of \((52+6\sqrt{43})^\frac{3}{2}-(52-6\sqrt{43})^\frac{3}{2}\).

  • is 107
  • is 828
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 828.

AIME I, 1990, Question 2

Elementary Algebra by Hall and Knight

Try with Hints


here we consider \(S^{2}=[(52+6\sqrt{43})^\frac{3}{2}-(52-6\sqrt{43})^\frac{3}{2}]^{2}\)

or, \(S^{2}=(52+6\sqrt{43})^{3}+(52-6\sqrt{43})^{3}\)

\(-2[(52+6\sqrt{43})(52-6\sqrt{43})]^\frac{3}{2}\)

or, \(S^{2}\)=685584

or, S=828.

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AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

Positive solution | AIME I, 1990 | Question 4

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

Positive solution – AIME I, 1990


Find the positive solution to

\(\frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0\)

  • is 107
  • is 13
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 13.

AIME I, 1990, Question 4

Elementary Algebra by Hall and Knight

Try with Hints


here we put \(x^{2}-10x-29=p\)

\(\frac{1}{p}+\frac{1}{p-16}-\frac{2}{p-40}=0\)

or, (p-16)(p-40)+p(p-40)-2p(p-16)=0

or, -64p+(40)(16)=0

or, p=10

or, 10=\(x^{2}-10x-29\)

or, (x-13)(x+3)=0

or, x=13 positive solution.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Right Rectangular Prism | AIME I, 1995 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Right Rectangular Prism.

Right Rectangular Prism – AIME I, 1995


A right rectangular prism P (that is rectangular parallelopiped) has sides of integral length a,b,c with \(a\leq b \leq c\), a plane parallel to one of the faces of P cuts P into two prisms, one of which is similar to P, and both of which has non-zero volume, given that b=1995, find number of ordered tuples (a,b,c) does such a plane exist.

  • is 107
  • is 40
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 40.

AIME I, 1995, Question 11

Geometry Vol I to IV by Hall and Stevens

Try with Hints


Let Q be similar to P

Let sides of Q be x,y,z for \(x \leq y \leq z\)

then \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c} < 1\)

As one face of Q is face of P

or, P and Q has at least two side lengths in common

or, x <a, y<b, z<c

or, y=a, z=b=1995

or, \(\frac{x}{a}=\frac{a}{1995}=\frac{1995}{c}\)

or, \(ac=1995^{2}=(3)^{2}(5)^{2}(7)^{2}(19)^{2}\)

or, number of factors of \((3)^{2}(5)^{2}(7)^{2}(19)^{2}\)=(2+1)(2+1)(2+1)(2+1)=81

or, \([\frac{81}{2}]=40\) for a <c.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Pyramid with Square base | AIME I, 1995 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Pyramid with Square base.

Pyramid with Squared base – AIME I, 1995


Pyramid OABCD has square base ABCD, congruent edges OA,OB,OC,OD and Angle AOB=45, Let \(\theta\) be the measure of dihedral angle formed by faces OAB and OBC, given that cos\(\theta\)=m+\(\sqrt{n}\), find m+n.

  • is 107
  • is 5
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 5.

AIME I, 1995, Question 12

Geometry Vol I to IV by Hall and Stevens

Try with Hints


Let \(\theta\) be angle formed by two perpendiculars drawn to BO one from plane ABC and one from plane OBC.

Let AP=1 \(\Delta\) APO is a right angled isosceles triangle, OP=AP=1.

Pyramid with square base

then OB=OA=\(\sqrt{2}\), AB=\(\sqrt{4-2\sqrt{2}}\), AC=\(\sqrt{8-4\sqrt{2}}\)

taking cosine law

\(AC^{2}=AP^{2}+PC^{2}-2(AP)(PC)cos\theta\)

or, 8-4\(\sqrt{2}\)=1+1-\(2cos\theta\) or, cos\(\theta\)=-3+\(\sqrt{8}\)

or, m+n=8-3=5.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Smallest positive Integer | AIME I, 1993 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Smallest positive Integer.

Smallest positive Integer – AIME I, 1993


Find the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers.

  • is 107
  • is 495
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Algebra

Check the Answer


Answer: is 495.

AIME I, 1993, Question 6

Elementary Number Theory by David Burton

Try with Hints


Let us take the first of each of the series of consecutive integers as a,b,c

then n=a+(a+1)+…+(a+8)=9a+36=10b+45=11c+55

or, 9a=10b+9=11c+19

or, b is divisible by 9

10b-10=10(b-1)=11c

or, b-1 is divisible by 11

or, least integer b=45

or, 10b+45=10(45)+(45)=495.

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AMC 8 Math Olympiad USA Math Olympiad

Integer Problem | AMC 10A, 2020 | Problem 17

Try this beautiful problem from Number theory based on Integer.

Integer Problem – AMC-10A, 2020- Problem 17


Define \(P(x)=(x-1^2)(x-2^2)……(x-{100}^2)\)

How many integers \(n\) are there such that \(P(n) \geq 0\)?

  • \(4900\)
  • \(4950\)
  • \(5000\)
  • \(5050\)
  • \(5100\)

Key Concepts


Number system

Probability

divisibility

Check the Answer


Answer: \(5100\)

AMC-10A (2020) Problem 17

Pre College Mathematics

Try with Hints


Given \(P(x)=(x-1^2)(x-2^2)……(x-{100}^2)\). at first we notice that \(P(x)\) is a product of of \(100\) terms…..now clearly \(P(x)\) will be negetive ,for there to be an odd number of negetive factors ,n must be lie between an odd number squared and even number squared.

can you finish the problem……..

\(P(x)\) is nonpositive when \(x\) is between \(100^2\) and \(99^2\), \(98^2\) and \(97^2 \ldots\) , \(2^2\) and \(1^2\)

can you finish the problem……..

Therefore, \((100+99)(100-99)+((98+97)(98-97)+1)\)+….

.+\(((2+1)(2-1)+1)\)=\((200+196+192+…..+4)\) =\(4(1+2+…..+50)=4 \frac{50 \times 51}{2}=5100\)

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AMC 8 Math Olympiad USA Math Olympiad

Largest possible value | AMC-10A, 2004 | Problem 15

Try this beautiful problem from Number system: largest possible value

Largest Possible Value – AMC-10A, 2004- Problem 15


Given that \( -4 \leq x \leq -2\) and \(2 \leq y \leq 4\), what is the largest possible value of \(\frac{x+y}{2}\)

  • \(\frac {-1}{2}\)
  • \(\frac{1}{6}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)
  • \(\frac{1}{9}\)

Key Concepts


Number system

Inequality

divisibility

Check the Answer


Answer: \(\frac{1}{2}\)

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints


The given expression is \(\frac{x+y}{x}=1+\frac{y}{x}\)

Now \(-4 \leq x \leq -2\) and \(2 \leq y \leq 4\) so we can say that \(\frac{y}{x} \leq 0\)

can you finish the problem……..

Therefore, the expression \(1+\frac{y}x\) will be maximized when \(\frac{y}{x}\) is minimized, which occurs when \(|x|\) is the largest and \(|y|\) is the smallest.

can you finish the problem……..

Therefore in the region \((-4,2)\) , \(\frac{x+y}{x}=1-\frac{1}{2}=\frac{1}{2}\)

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Problem on Positive Integer | AIME I, 1995 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Problem on Positive Integer.

Problem on Positive Integer – AIME I, 1995


Let \(n=2^{31}3^{19}\),find number of positive integer divisors of \(n^{2}\) are less than n but do not divide n.

  • is 107
  • is 589
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Number of divisors

Check the Answer


Answer: is 589.

AIME I, 1995, Question 6

Elementary Number Theory by David Burton

Try with Hints


Let \(n=p_1^{k_1}p_2^{k_2}\) for some prime \(p_1,p_2\). The factors less than n of \(n^{2}\)

=\(\frac{(2k_1+1)(2k_2+1)-1}{2}\)=\(2k_1k_2+k_1+k_2\)

The number of factors of n less than n=\((k_1+1)(k_2+1)-1\)

=\(k_1k_2+k_1+k_2\)

Required number of factors =(\(2k_1k_2+k_1+k_2\))-(\(k_1k_2+k_1+k_2\))

=\(k_1k_2\)

=\(19 \times 31\)

=589.

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