Categories

## Nearest value | PRMO 2018 | Question 14

Try this beautiful problem from the PRMO, 2018 based on Nearest value.

## Nearest Value – PRMO 2018

If x=cos1cos2cos3…..cos89 and y=cos2cos6cos10….cos86, then what is the integer nearest to $\frac{2}{7}log_2{\frac{y}{x}}$?

• is 107
• is 19
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Numbers

Multiples

But try the problem first…

Source

PRMO, 2018, Question 14

Higher Algebra by Hall and Knight

## Try with Hints

First hint

$\frac{y}{x}$=$\frac{cos2cos6cos10…..cos86}{cos1cos2cos3….cos89}$

=$2^{44}\times\sqrt{2}\frac{cos2cos6cos10…cos86}{sin2sin4…sin88}$

[ since cos$\theta$=sin(90-$\theta$) from cos 46 upto cos 89 and 2sin$\theta$cos$\theta$=sin2$\theta$]

Second Hint

=$\frac{2^{\frac{89}{2}}sin4sin8sin12…sin88}{sin2sin4sin6…sin88}$

[ since sin$\theta$=cos(90-$\theta$)]

=$\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}$

[ since cos$\theta$=sin(90-$\theta$)]

Final Step

=$\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}$

[since $cos4cos8cos12…cos88$

$=(cos4cos56cos64)(cos8cos52cos68)(cos12cos48cos72)(cos16cos44cos76)(cos20cos40cos80)(cos24cos36cos84)(cos28cos32cos88)cos60$

$=(1/2)^{15}(cos12cos24cos36cos48cos60cos72cos84)$

$=(1/2)^{16}(cos12cos48cos72)(cos24cos36cos84)$

$=(1/2)^{20}(cos36cos72)$

$=(1/2)^{20}(cos36sin18)$

$=(1/2)^{22}(4sin18cos18cos36/cos18)$

$=(1/2)^{22}(sin72/cos18)$

$=(1/2)^{22}$]

=$2^\frac{133}{2}$

$\frac{2}{7}log_2{\frac{y}{x}}$=$\frac{2}{7} \times \frac{133}{2}$=19.

Categories

## Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

## Sequence and permutations – AIME II, 2015

Call a permutation $a_1,a_2,….,a_n$ of the integers 1,2,…,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,….,7.

• is 107
• is 486
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Permutations

Integers

But try the problem first…

Source

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

## Try with Hints

First hint

While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Second Hint

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements

or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements

……

or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 $\times$ $3^{n-2}$

forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$

for n=3 all six permutations taken and go up 18, 54, 162, 486

Final Step

for n=7, here $2 \times 3^{5} =486.$

as

sds

Categories

## Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

## Numbers of positive integers – AIME 2012

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

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

### Key Concepts

Integers

Number Theory

Algebra

But try the problem first…

Source

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

## Try with Hints

First hint

Here a number divisible by 4 if a units with tens place digit is divisible by 4

Second Hint

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

Final Step

So for 10 b’s with 2 a’s and 2 c’s for every b gives $10 \times 2 \times 2$

Then number of ways $10 \times 2 \times 2$ = 40 ways.

Categories

## Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

## Number of points and planes – AIME I, 1999

Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is $\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.

• is 107
• is 489
• is 840
• cannot be determined from the given information

### Key Concepts

Number of points

Plane

Probability

But try the problem first…

Source

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

$10 \choose 3$ sets of 3 points which form triangles,

Second Hint

fourth distinct segment excluding 3 segments of triangles=45-3=42

Final Step

Required probability=$\frac{{10 \choose 3} \times 42}{45 \choose 4}$

where ${45 \choose 4}$ is choosing 4 segments from 45 segments

=$\frac{16}{473}$ then m+n=16+473=489.

Categories

## Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

## Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

• is 107
• is 195
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Number Theory

Algebra

But try the problem first…

Source

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

First hint

After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.

Second Hint

Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.

Final Step

Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$

Hence option B correct.

Categories

## Digits of number | PRMO 2018 | Question 3

Try this beautiful problem from the PRMO, 2018 based on Digits of number.

## Digits of number – PRMO 2018

Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.

• is 107
• is 70
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Numbers

Multiples

But try the problem first…

Source

PRMO, 2018, Question 3

Higher Algebra by Hall and Knight

## Try with Hints

First hint

abccba (b is odd)

=a($10^5$+1)+b($10^4$+10)+c($10^3$+$10^2$)

=a(1001-1)100+a+10b(1001)+(100)(11)c

=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c

=7p+(c-a) where p is an integer

Second Hint

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

Final Step

since b is odd

number of such number=$14 \times 5$=70.

Categories

## Smallest value | PRMO 2018 | Question 15

Try this beautiful problem from the PRMO, 2018 based on Smallest value.

## Smallest Value – PRMO 2018

Let a and b natural numbers such that 2a-b, a-2b and a+b are all distinct squares. What is the smallest possible value of b?

• is 107
• is 21
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Numbers

Multiples

But try the problem first…

Source

PRMO, 2018, Question 15

Higher Algebra by Hall and Knight

## Try with Hints

First hint

2a-b=$k_1^2$ is equation 1

a-2b=$k_2^2$ is equation 2

a+b=$k_3^2$ is equation 3

Second Hint

adding 2 and 3 we get

2a-b=$k_2^2+k_3^2$

or, $k_2^2+k_3^2$=$k_1^2$ $(k_2<k_3)$

Final Step

For least ‘b’ difference of $k_3^2$ and $k_2^2$ is also least and must be multiple of 3

or, $k_2^2$=a-2b=$9^2$ and $k_3^2$=a+b=$12^2$

or, $k_3^2-k_2^2$=3b=144-81=63

or, b=21

or, least b is 21.

Categories

## Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

## Algebra and Positive Integer – AIME I, 1987

What is the largest positive integer n for which there is a unique integer k such that $\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}$?

• is 107
• is 112
• is 840
• cannot be determined from the given information

### Key Concepts

Digits

Algebra

Numbers

But try the problem first…

Source

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

## Try with Hints

First hint

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

Second Hint

for 91n-104k<n+k, K>$\frac{6n}{7}$

and 0<91n-104k gives k<$\frac{7n}{8}$

Final Step

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

Categories

## Arithmetic Mean | AIME I, 2015 | Question 12

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

## Arithmetic Mean of Number Theory – AIME 2015

Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

• is 107
• is 431
• is 840
• cannot be determined from the given information

### Key Concepts

Inequalities

Algebra

Number Theory

But try the problem first…

Source

AIME, 2015, Question 12

Elementary Number Theory by David Burton

## Try with Hints

First hint

Each 1000-element subset ${ a_1, a_2,a_3,…,a_{1000}}$ of ${1,2,3,…,2015}$ with $a_1<a_2<a_3<…<a_{1000}$ contributes $a_1$ to sum of least element of each subset and set ${a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$. $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive

Second Hint

Thus, the number of ways to choose the set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is equal to the sum. But choosing a set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is same as choosing a 1001-element subset from ${1,2,3,…,2016}$!

Final Step

average =$\frac{2016}{1001}$=$\frac{288}{143}$. Then $p+q=288+143={431}$

Categories

## Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

## Distance Time – AIME 2012

When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile.

• is 107
• is 279
• is 840
• cannot be determined from the given information

### Key Concepts

Time

Distance

Speed

But try the problem first…

Source

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

## Try with Hints

First hint

After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Second Hint

Then
$\frac{a}{6} + \frac{n-a}{4}$ = $\frac{n-a}{6} + \frac{2a}{5}$ implies that $a = \frac{5}{19}n$

Final Step

Then integral value of n is 19 and a = 5 and $t = \frac{13}{3}$ hours that is 260 minutes. Then $19 + 260 = {279}$.