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Algebra Math Olympiad PRMO USA Math Olympiad

Value of Sum | PRMO – 2018 | Question 16

Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

Value of Sum – PRMO 2018, Question 16


What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

  • $50$
  • $53$
  • $55$
  • $59$
  • $65$

Key Concepts


Odd-Even

Sum

integer

Check the Answer


Answer:$55$

PRMO-2018, Problem 16

Pre College Mathematics

Try with Hints


We have to find out the sum . Now substitite $i=1,2,3…9$ and observe the all odd-even cases……

Can you now finish the problem ……….

$i=1 \Rightarrow$$ 1+(2+4+6+8+10-3-5-7-9)$
$=1-4+10=7$
$i=2 \Rightarrow $$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$

$i=3 \Rightarrow$$ 1 \times 3+(4+6+8+10-5-7-9)$
$=3-3+10=10$
$i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10$
$=13$
$i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16$
$i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$ 1 \times 9+(10)=19$

Can you finish the problem……..

Therefore $ S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55 $



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Algebra Math Olympiad PRMO USA Math Olympiad

Digits Problem | PRMO – 2018 | Question 19

Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19.

Digits Problem – PRMO 2018, Question 19


Let $N=6+66+666+\ldots \ldots+666 \ldots .66,$ where there are hundred 6 ‘s in the last term in the sum. How many times does the digit 7 occur in the number $N ?$

  • $30$
  • $33$
  • $36$
  • $39$
  • $42$

Key Concepts


Number theorm

Digits Problem

integer

Check the Answer


Answer:$33$

PRMO-2018, Problem 19

Pre College Mathematics

Try with Hints


Given that $\mathrm{N}=6+66+666+……. \underbrace{6666 …..66}_{100 \text { times }}$

If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply \(\frac{6}{9}\) then it becomes $=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$

Can you now finish the problem ……….

$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+…….+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+…..+10^{100}\right)-100\right]$

Can you finish the problem……..

$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999….99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222…..200)}_{99 \mathrm{times}}-60$

$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times



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Algebra AMC 10 Math Olympiad USA Math Olympiad

Least Possible Value Problem | AMC-10A, 2019 | Quesstion19

Contents
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    Try this beautiful problem from Algebra based on Least Possible Value.

    Least Possible Value – AMC-10A, 2019- Problem 19


    What is the least possible value of \(((x+1)(x+2)(x+3)(x+4)+2019)\)

    where (x) is a real number?

    • \((2024)\)
    • \((2018)\)
    • \((2020)\)

    Key Concepts


    Algebra

    quadratic equation

    least value

    Check the Answer


    Answer: \((2018)\)

    AMC-10A (2019) Problem 19

    Pre College Mathematics

    Try with Hints


    To find out the least positive value of \((x+1)(x+2)(x+3)(x+4)+2019\), at first we have to expand the expression .\(((x+1)(x+2)(x+3)(x+4)+2019)\) \(\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)\)

    Let us take \(((x^2+5x+5=m))\)

    then the above expression becomes \(((m-1)(m+1)+2019)\) \(\Rightarrow m^2-1+2019\) \(\Rightarrow m^2+2018\)

    Can you now finish the problem ……….

    Clearly in \((m^2+2018)…….(m^2)\) is positive ( squares of any number is non-negative) and least value is 0

    can you finish the problem……..

    Therefore minimum value of \(m^2+2108\) is \(2018\) since \(m^2 \geq 0\) for all m belongs to real .

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    Algebra Arithmetic Math Olympiad PRMO

    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

    Check the Answer


    Answer: is 19.

    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.

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    Algebra Arithmetic Math Olympiad USA Math Olympiad

    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

    Check the Answer


    Answer: is 40.

    AIME, 2012, Question 1.

    Elementary Number Theory by David Burton .

    Try with Hints


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

    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

    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.

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    Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

    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

    Check the Answer


    Answer: is 195.

    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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    Algebra Arithmetic Math Olympiad PRMO

    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

    Check the Answer


    Answer: is 70.

    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.

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    Algebra Arithmetic Math Olympiad PRMO

    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

    Check the Answer


    Answer: is 21.

    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.

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    Geometry Math Olympiad USA Math Olympiad

    Length and Triangle | AIME I, 1987 | Question 9

    Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Length and Triangle.

    Length and Triangle – AIME I, 1987


    Triangle ABC has right angle at B, and contains a point P for which PA=10, PB=6, and \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA. Find PC.

    Length and Triangle
    • is 107
    • is 33
    • is 840
    • cannot be determined from the given information

    Key Concepts


    Angles

    Algebra

    Triangles

    Check the Answer


    Answer: is 33.

    AIME I, 1987, Question 9

    Geometry Vol I to Vol IV by Hall and Stevens

    Try with Hints


    First hint

    Let PC be x, \(\angle \)APB=\(\angle\)BPC=\(\angle\)CPA=120 (in degrees)

    Second Hint

    Applying cosine law \(\Delta\)APB, \(\Delta\)BPC, \(\Delta\)CPA with cos120=\(\frac{-1}{2}\) gives

    \(AB^{2}\)=36+100+60=196, \(BC^{2}\)=36+\(x^{2}\)+6x, \(CA^{2}\)=100+\(x^{2}\)+10x

    Final Step

    By Pathagorus Theorem, \(AB^{2}+BC^{2}=CA^{2}\)

    or, \(x^{2}\)+10x+100=\(x^{2}\)+6x+36+196

    or, 4x=132

    or, x=33.

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    Algebra Arithmetic Math Olympiad USA Math Olympiad

    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

    Check the Answer


    Answer: is 112.

    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.

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