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

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.

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## Problem – Number Series ( B.Stat Objective Problem )

We are going to discuss about Number Series from B.Stat Objective Problem .

A student studying the weather for d days observed that(i) it rained on 7 days morning or afternoon, (ii) when it rained in the afternoon it was clear in the morning, (iii) there were five clear afternoons (iv) there were six clear mornings. Then d equals

• 8
• 9
• 11
• 10

### Key Concepts

Series

Algebra

Integers

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Here A1, A2, A3, A4, A5, A6, A7, A8, A9 are mornings of day 1, day 2, day 3, day 4, day 5, day 6, day 7, day 8 and day 9 again B1, B2, B3, B4, B5, B6, B7, B8, B9 are afternoons of day 1, day 2, day 3, day 4, day 5, day 6 and day 7, day 8 and day 9

A1 A2 clear morning B1 B2 rainy afternoon from first two conditions

Second Hint

A3 rainy A4 clear A5 rainy A6 clear A7 rainy and B3 clear B4 rainy B5 clear B6 rainy B7 clear from first two conditions

Final Step

So, from first 7 days, 4 clear mornings and 3 clear afternoons. Then A8 clear A9 clear B8 clear B9 clear from last two conditions .Then all condition satisfied that is 4+2=6 mornings clear 3+2 =5 afternoons clear. d=9 that is in 9 days.

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## Series Problem | PRMO 2017 | Question 6

Try this beautiful problem from the Pre-RMO, 2017 based on Series.

## Series Problem – PRMO 2017

Let the sum $\displaystyle\sum_{n=1}^{9}\frac{1}{n(n+1)(n+2)}$ written in its lowest terms be $\frac{p}{q}$, find the value of q-p.

• is 107
• is 83
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Integers

Rearrangement of terms in series

PRMO, 2017, Question 6

Calculus Vol 1 and 2 by Apostle

## Try with Hints

First hint

here $\displaystyle\sum_{n=1}^{9}\frac{n+2-n}{2n(n+1)(n+2)}$

=$\frac{1}{2}\displaystyle\sum_{n=1}^{9}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$

Second Hint

=$\frac{1}{2}(\frac{1}{(1)(2)}-\frac{1}{(2)(3)}$

+$\frac{1}{(2)(3)}-\frac{1}{(3)(4)}+….+\frac{1}{(9)(10)}-\frac{1}{(10)(11)})$

=$\frac{1}{2}(\frac{1}{2}-\frac{1}{110})$

Final Step

$\Rightarrow \frac{27}{110}$

$\Rightarrow q-p=110-27$

=83.

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## Series and Integers | B.Stat Objective | TOMATO 81

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Series and Integers.

## Series and Integers (B.Stat Objective)

The value of $\sum ij$, where the summation is over all i and j such that $1 \leq i \lt j \leq 10$

• 2640
• 1320
• 3025
• none of these

### Key Concepts

Series

Integers

Inequality

B.Stat Objective Question 81

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

required sum=1(2+3+4+….+10)+2(3+4+….+10)+3(4+5+….+10)+4(5+6+….+10)+5(6+7+….+10)+6(7+8+..+10)+7(8+9+10)+8(9+10)+9(10)

Second Hint

=54+104+147+180+200+204+189+152+90

Final Step

=1320

Categories

## Theory of Equations | AIME I, 2015 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Theory of Equations.

## Theory of Equations – AIME I, 2015

The expressions A=$1\times2+3\times4+5\times6+…+37\times38+39$and B=$1+2\times3+4\times5+…+36\times37+38\times39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.Find the positive difference between integers A and B.

• is 722
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Equations

Number Theory

AIME I, 2015, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

A = $(1\times2)+(3\times4)$

$+(5\times6)+…+(35\times36)+(37\times38)+39$

Second Hint

B=$1+(2\times3)+(4\times5)$

$+(6\times7)+…+(36\times37)+(38\times39)$

Final Step

B-A=$-38+(2\times2)+(2\times4)$

$+(2\times6)+…+(2\times36)+(2\times38)$

=722.

Categories

## Probability of divisors | AIME I, 2010 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

## Probability of divisors – AIME I, 2010

Ramesh lists all the positive divisors of $2010^{2}$, she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

$2010^{2}=2^{2}3^{2}5^{2}67^{2}$

Second Hint

$(2+1)^{4}$ divisors, $2^{4}$ are squares

Final Step

probability is $\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}$ implies m+n=107

Categories

## Equations with number of variables | AIME I, 2009 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2009 based on Equations with a number of variables.

## Equations with number of variables – AIME 2009

For t=1,2,3,4, define $S^{t}=a^{t}_1+a^{t}_2+…+a^{t}_{350}$, where $a_{i}\in${1,2,3,4}. If $S_{1}=513, S_{4}=4745$, find the minimum possible value for $S_{2}$.

• is 905
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2009, Question 14

Polynomials by Barbeau

## Try with Hints

First hint

j=1,2,3,4, let $m_{j}$ number of $a_{i}$ s = j then $m_{1}+m{2}+m{3}+m{4}=350$, $S_{1}=m_{1}+2m_{2}+3m_{3}+4m_{4}=513$ $S_{4}=m_{1}+2^{4}m_{2}+3^{4}m_{3}+4^{4}m_{4}=4745$

Second Hint

Subtracting first from second, then first from third yields $m_{2}+2m_{3}+3m_{4}=163,$ and $15m_{2}+80m_{3}+255m_{4}=4395$ Now subtracting 15 times first from second gives $50m_{3}+210m_{4}=1950$ or $5m_{3}+21m_{4}=195$ Then $m_{4}$ multiple of 5, $m_{4}$ either 0 or 5

Final Step

If $m_{4}=0$ then $m_{j}$ s (226,85,39,0) and if $m_{4}$=5 then $m_{j}$ s (215,112,18,5) Then $S_{2}=1^{2}(226)+2^{2}(85)+3^{2}(39)+4^{2}(0)=917$ and $S_{2}=1^{2}(215)+2^{2}(112)+3^{2}(18)+4^{2}(5)=905$ Then min 905.

Categories

## Coordinate Geometry Problem | AIME I, 2009 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Coordinate Geometry.

## Coordinate Geometry Problem – AIME 2009

Consider the set of all triangles OPQ where O  is the origin and P and Q are distinct points in the plane with non negative integer coordinates (x,y) such that 41x+y=2009 . Find the number of such distinct triangles whose area is a positive integer.

• is 107
• is 600
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Geometry

AIME, 2009, Question 11

Geometry Revisited by Coxeter

## Try with Hints

First hint

let P and Q be defined with coordinates; P=($x_1,y_1)$ and Q($x_2,y_2)$. Let the line 41x+y=2009 intersect the x-axis at X and the y-axis at Y . X (49,0) , and Y(0,2009). such that there are 50 points.

here [OPQ]=[OYX]-[OXQ] OY=2009 OX=49 such that [OYX]=$\frac{1}{2}$OY.OX=$\frac{1}{2}$2009.49 And [OYP]=$\frac{1}{2}$$2009x_1$  and [OXQ]=$\frac{1}{2}$(49)$y_2$.

Second Hint

2009.49 is odd, area OYX not integer of form k+$\frac{1}{2}$ where k is an integer

Final Step

41x+y=2009 taking both 25  $\frac{25!}{2!23!}+\frac{25!}{2!23!}$=300+300=600.

.

Categories

## Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

## Geometric Sequence Problem – AIME 2009

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

• is 500
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Series

Real Analysis

AIME, 2009

Introduction to Real Analysis, 4th Edition  by Robert G. Bartle, Donald R. Sherbert

## Try with Hints

First hint

3-digit sequence a, ar, $ar^{2}$. The largest geometric number must have a<=9.

Second Hint

ar $ar^{2}$ less than 9 r fraction less than 1 For a=9 is $\frac{2}{3}$ then number 964.

Final Step

a>=1 ar and $ar^{2}$ greater than 1 r is 2 and number is 124. Then difference 964-124=840.

Categories

## Problem on Series | SMO, 2009 | Problem No. 25

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.

## Problem on Series | SMO Test

Given that $x+(1+x)^2+(1+x)^3+………(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$ where each $a_r$ is an integer , $r = 0,1,2,…,n$

Find the value of n such that $a_0 +a_1 +a_2+a_3 +………..+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}$?

• 2
• 5
• 6
• 0

### Key Concepts

Series Problem

Algebra

Challenges and Thrills – Pre – college Mathematics

## Try with Hints

If you got stuck in this sum then we can try by understanding the pattern we are using here.If we assume x = 1 thenn the expression will be

$x+(1+x)^2+(1+x)^3+………(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$

The right hand side of this equation will be

$a_0 +a_1 + a_2 +a_3+………………+ a_n$

and the left hand side of the given equation be like

$1 + 2^2 + 2^ 3 + ……+2^n$

$1 + 2^2 + 2^ 3 + ……+2^n$ = $a_0 +a_1 + a_2 +a_3+………………+ a_n$

Try the rest of the sum…………..

In the next hint we continue from the previous hint:

so the expression , $1 + 2^2 + 2^ 3 + ……+2^n$ = $2^{n+1} – 3$

Again , $a_1 = 1+2+3+……….+n = \frac {n(n+1)}{2}$

so $a_n = 1$

Now we have almost reach the final step.I am not showing it now.Try first………

Now coming back to the last step :

$60 – \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} – 3$