Tag: divisibility

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

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Post date June 1, 2020
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Probability in Divisibility – AMC-10A, 2003- Problem 15
- Key Concepts
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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

But try the problem first…

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

Source

Suggested Reading

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints

First hint

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……..

Second Hint

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……..

Final Step

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

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

Natural Numbers Problem | PRMO 2019 | Question 30

Post author By Shirsendu Roy
Post date May 16, 2020
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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

But try the problem first…

Answer: is 64.

Source

Suggested Reading

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.

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

Post author By Shirsendu Roy
Post date May 13, 2020
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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

But try the problem first…

Answer: is 432.

Source

Suggested Reading

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.

Algebraic value | AIME I, 1990 | Question 2

Post author By Shirsendu Roy
Post date May 12, 2020
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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

But try the problem first…

Answer: is 828.

Source

Suggested Reading

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.

Positive solution | AIME I, 1990 | Question 4

Post author By Shirsendu Roy
Post date May 12, 2020
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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

But try the problem first…

Answer: is 13.

Source

Suggested Reading

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.

Right Rectangular Prism | AIME I, 1995 | Question 11

Post author By Shirsendu Roy
Post date May 11, 2020
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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

But try the problem first…

Answer: is 40.

Source

Suggested Reading

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.

Greatest Integer | PRMO 2019 | Question 22

Post author By Shirsendu Roy
Post date May 11, 2020
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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

But try the problem first…

Answer: is 78.

Source

Suggested Reading

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.

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

Post author By Shirsendu Roy
Post date May 10, 2020
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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

But try the problem first…

Answer: is 5.

Source

Suggested Reading

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.

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.

Problem on Largest Prime Factor | PRMO 2019 | Question 21

Post author By Shirsendu Roy
Post date May 10, 2020
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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

But try the problem first…

Answer: is 17.

Source

Suggested Reading

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.

Sum of digits | PRMO 2019 | Question 20

Post author By Shirsendu Roy
Post date May 9, 2020
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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

But try the problem first…

Answer: is No largest value.

Source

Suggested Reading

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.

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

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Natural numbers Problem – PRMO 2019

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Smallest positive Integer Problem – AIME I, 1990

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Algebraic value – AIME I, 1990

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

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Positive solution – AIME I, 1990

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Right Rectangular Prism – AIME I, 1995

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Greatest integer – PRMO 2019

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Pyramid with Squared base – AIME I, 1995

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

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Problem on Largest Prime Factor – PRMO 2019

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Sum of digits – PRMO 2019

Key Concepts

Check the Answer

Try with Hints

Other useful links

Related Program

Subscribe to Cheenta at Youtube