Categories

## 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

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= $\frac{34}{100}=\frac{17}{50}$

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.