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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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Algebra Arithmetic Math Olympiad PRMO

Natural Numbers Problem | PRMO 2019 | Question 30

Try this beautiful problem from the Pre-RMO, 2019 based on Natural Numbers.

Natural numbers Problem – PRMO 2019


Let E denote the set of all natural number n such that \(3< n<100\) and the set {1,2,3,…,n} can be partitioned in to 3 subsets with equal sums. Find the number of elements of E.

  • is 107
  • is 64
  • is 840
  • cannot be determined from the given information

Key Concepts


Divisibility

Equations

Integer

Check the Answer


Answer: is 64.

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

Try with Hints


First hint

{1,2,…,n}

This set can be partitioned into 3 subsets with equal sums so total sum is divisible by 3

\(\frac{n(n+1)}{2}\) is divisible by 3.

or, n of form 3k, 3k+2

or, n=6k,6k+2,6k+3, 6k+5

Second Hint

case I n=6k, we group numbers in bundles of 6 for each bundle 1,2,3,4,5,6(16,25,34)

case II n=6k+2 then we club last bundle of 8 numbers rest can be partitioned and those eight numbers can be done 1,2,3,4,5,6,7,8 (1236,48)

case III n=6k+3 we club last nine number and rest can be partitioned 1,2,3,4,5,6,7,8,9 (12345,69,78)

Final Step

case IV 6k+5 we take last five numbers, rest can be aprtitioned 1,2,3,4,,5(14,25,5)

Hence we select any number of form 6k(16), 6k+2(16), 6K+3(16), 6K+5(16)

or, total=64 numbers.

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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

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


First hint

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

Second Hint

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

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

Final Step

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


First hint

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

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

Second Hint

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

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

or, p=10

Final Step

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


First hint

Let Q be similar to P

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

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

Second Hint

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

Final Step

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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Algebra Arithmetic Math Olympiad PRMO

Greatest Integer | PRMO 2019 | Question 22

Try this beautiful problem from the Pre-RMO, 2019 based on Greatest Integer.

Greatest integer – PRMO 2019


Find the greatest integer not exceeding the sum \(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)

  • is 107
  • is 78
  • is 840
  • cannot be determined from the given information

Key Concepts


Largest integer

Divisibility

Integer

Check the Answer


Answer: is 78.

PRMO, 2019, Question 22

Elementary Number Theory by David Burton

Try with Hints


First hint

\(\int\limits_1^{1600}\frac{1}{x}{d}x < \sum_{x=1}^{1599}\frac{1}{\sqrt{n}}\)

\(< 1+\sum_{n=1}^{1599}\frac{1}{\sqrt{x}}{d}x\)

Second Hint

or, \([2\sqrt{x}]_{1}^{1600}< \sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)

\(< 1+|2{\sqrt{x}}|_1^{1599}\)

Final Step

or, 78<\(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}} <2\sqrt{1599}-1\)

or, 78 < \(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)<78.97

or,\([\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}]\)=78.

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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


First hint

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

Second Hint

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

Final Step

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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Algebra Arithmetic Math Olympiad PRMO

Problem on Largest Prime Factor | PRMO 2019 | Question 21

Try this beautiful problem from the Pre-RMO, 2019 based on Largest prime factor.

Problem on Largest Prime Factor – PRMO 2019


Consider the set E={5,6,7,8,9}, for any partition {A,B} of E, with both A and B non empty. Consider the number obtained by adding the product of elements of A to the product of elements of A to the product of elements of B. Let N be the largest prime number among these numbers, find the sum of the digits of N.

  • is 107
  • is 17
  • is 840
  • cannot be determined from the given information

Key Concepts


Largest prime

Divisibility

Integer

Check the Answer


Answer: is 17.

PRMO, 2019, Question 21

Elementary Number Theory by David Burton

Try with Hints


First hint

here one of the set A or set B contains odd number only

Second Hint

set A set B

I 5 6,7,8,9 5+(6)(7)(8)(9)=3029 not prime

II 7 5,6,8,9 7+(5)(6)(7)(8)(9)=2167 not prime

III 9 5,6,7,8 9+(5)(6)(7)(8)=not prime

Final Step

IV 5,7 6,8,9 (5)(7)+(6)(8)(9)=467 prime

V 5,9 6,7,8 (5)(9)+(6)(7)(8)=not prime

VI 7,9 5,6,8 (7)(9)+(5)(6)(8)=not prime

VII 5,7,9 6,8 (5)(7)(9)+(6)(8)not prime

N=467

or, 4+6+7=17.

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Algebra Arithmetic Math Olympiad PRMO

Sum of digits | PRMO 2019 | Question 20

Try this beautiful problem from the Pre-RMO, 2019 based on Sum of digits.

Sum of digits – PRMO 2019


Consider the set E of all natural numbers n such that when divided by 11,12,13 respectively, the remainders, in that order, are distinct prime numbers in an arithmetic progression. If N is the largest number in E, find the sum of digits of N.

  • is 107
  • is No largest value
  • is 840
  • cannot be determined from the given information

Key Concepts


Largest Number

Divisibility

Integer

Check the Answer


Answer: is No largest value.

PRMO, 2019, Question 20

Elementary Number Theory by David Burton

Try with Hints


First hint

here N can be of the form (13)(12)(11)(k)+29

Second Hint

where k belongs to an integer

Final Step

then no largest value.

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