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Algebra Math Olympiad PRMO USA Math Olympiad

Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

Positive Integer – PRMO 2017, Question 1


How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$

  • $9$
  • $7$
  • $28$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$28$

PRMO-2017, Problem 1

Pre College Mathematics

Try with Hints


Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$

Can you now finish the problem ……….

Also $n<1000 \Rightarrow s \leq 27$
therefore $\mathrm{s}=21$
Clearly, n must be a 3 digit number Let $x_{1}, x_{2}, x_{3}$ be the digits, then $x_{1}+x_{2}+x_{3}=21$
where $1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9$
$\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18$
$\Rightarrow x_{1} \geq 3$

Can you finish the problem……..

For $x_{1}=3,4, \ldots ., 9,$ the equation (1) has $1,2,3, \ldots ., 7$ solutions
therefore total possible solution of equation (1)

=$1+2+\ldots+7=\frac{7 \times 8}{2}=28$

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Algebra Math Olympiad PRMO USA Math Olympiad

Pen & Note Books Problem| PRMO-2017 | Question 8

Try this beautiful Pen & Note Books Problem from Algebra, from PRMO 2017.

Pen & Note Books – PRMO 2017, Question 8


A pen costs $Rs 11$ and a notebook costs $Rs. 13 .$ Find the number of ways in which a person can spend exactly Rs.1000 to buy pens and notebooks.

  • $9$
  • $7$
  • $11$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$7$

PRMO-2017, Problem 8

Pre College Mathematics

Try with Hints


Given A pen costs Rs.\(11\) and a note book costs Rs.\(13\)

$11 x+13 y=1000$………………….(1)

Can you now finish the problem ……….

Now $11 x+13 y=1000$
$\Rightarrow 11 x=1000-13 y=(1001-11 y)-(2 y+1)$

=$11(91-y)-(2 y+1)$
$\Rightarrow 11 | 2 y+1$
Let $2 y+1=11(2 k-1), k \in I^{+}$
$\Rightarrow y=11 \mathrm{k}-6$
therefore $11 x=11(97-11 k)-11(2 k-1)$
$\Rightarrow x=98-13 k$
But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$

Can you finish the problem……..

But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$
therefore for each $\mathrm{k} \in{1,2, \ldots ., 7},$ we get a unique pair $(\mathrm{x}, \mathrm{y})=(98-13 \mathrm{k}, 11 \mathrm{k}-6)$ satisfying equation
Hence 7 ways are possible.

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Algebra Math Olympiad PRMO USA Math Olympiad

Rectangle Problem | Geometry | PRMO-2017 | Question 13

Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

Rectangle Problem – Geometry – PRMO 2017, Question 13


In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

  • $9$
  • $24$
  • $11$

Key Concepts


Geometry

Triangle

Trigonometry

Check the Answer


Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

Try with Hints


Rectangle Problem

We have to find out the value of \(AB\). Join \(BD\). \(BF\) is perpendicular on \(AC\).

Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Rectangle Problem

Therefore

Therefore$ \angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

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Algebra Math Olympiad PRMO USA Math Olympiad

Chords in a Circle | PRMO-2017 | Question 26

Try this beautiful Problem based on Chords in a Circle, Geometry, from PRMO 2017.

Chords in a Circle – PRMO 2017, Question 26


Let $A B$ and $C D$ be two parallel chords in a circle with radius 5 such that the centre $O$ lies between these chords. Suppose $A B=6, C D=8 .$ Suppose further that the area of the part of the circle lying between the chords $A B$ and $C D$ is $(m \pi+n) / k,$ where $m, n, k$ are positive integers with gcd$(m, n, k)=1$ . What is the value of $m+n+k ?$

  • $9$
  • $75$
  • $11$

Key Concepts


Geometry

Triangle

Circle

Check the Answer


Answer:$75$

PRMO-2017, Problem 26

Pre College Mathematics

Try with Hints


Chords in a Circle

Draw OE $\perp A B$ and $O F \perp C D$

Clearly $\mathrm{EB}=\frac{\mathrm{AB}}{2}=3, \mathrm{FD}=\frac{\mathrm{CU}}{2}=4$

$\mathrm{OE}=\sqrt{5^{2}-3^{2}}=4$ and $\mathrm{OF}=\sqrt{5^{2}-4^{2}}=3$

Therefore $\Delta O E B \sim \Delta D F O$

Can you now finish the problem ……….

Chords in a Circle

Let $\angle \mathrm{EOB}=\angle \mathrm{ODF}=\theta,$ then

$\angle B O D=\angle A O C=180^{\circ}-\left(\theta+90^{\circ}-\theta\right)=90^{\circ}$

Now area of portion between the chords

= \(2 \times\) (area of minor sector BOD)+2 \times ar\((\triangle AOB)\)
$=2 \times \frac{\pi \times 5^{2}}{4}+2 \times \frac{1}{2} \times 6 \times 4=\frac{25 \pi}{2}+24=\frac{25 \pi+48}{2}$

Therefore $m=25, n=48$ and $k=2$

Can you finish the problem……..

Therefore $m+n+k=75$

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Algebra Math Olympiad PRMO USA Math Olympiad

Pen & Note Books Problem | PRMO-2019 | Question 16

Try this beautiful Problem from Algebra based on Pen & Note Books from PRMO 2019.

Pen & Note Books – PRMO 2019, Problem 16


A pen costs Rs.13 and a notebook costs Rs.17. A school spends exactly Rs.10000 in the year 2017 – 18 to buy x pens and y notebooks such that $x$ and $y$ are as close as possible (i.e. $|x-y|$ is minimum). Next year, in $2018-19,$ the school spends a little more than Rs.10000 and buys y pens and x notebooks. How much more did the school pay?

  • $9$
  • $40$
  • $34$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$40$

PRMO-2019, Problem 16

Pre College Mathematics

Try with Hints


Given A pen costs Rs.\(13\) and a note book costs Rs.\(17\)

Nunber of pens \(x\) and numbers of notebooks \(y\)

According to question

\(13 x+17 y=10,000\) ……………………..(1)
\(17 x+13 y=10000+P\)………………….(2)

Where $\mathrm{P}$ is little more amount spend by school in $2018-19$.we have to find out the value of \(P\).Can you find out?

Can you now finish the problem ……….

Adding (1) & (2)

$x+y=\frac{20,000+P}{30}$

Subtract (1) form (2)

$x-y=\frac{P}{4}$

Let $P=4 a$
$\begin{aligned} \text { So, } x+y=& \frac{20,000+4 a}{30} \ &=666+\frac{10+2 a}{15} \end{aligned}$

Therefore \(a\)=\(10\)

Can you finish the problem……..

Since $P=4 a$ \(\Rightarrow P=40\)

So school had to pay Rs. 40 extra in 2018 – 2019

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Math Olympiad Number Theory PRMO USA Math Olympiad

Integer based Problem | PRMO-2018 | Question 20

Try this beautiful Integer based Problem from Algebra, from PRMO 2018.

Integer based Problem – PRMO 2018, Question 20


Determine the sum of all possible positive integers n, the product of whose digits equals \(n^2 -15n – 27 \)

  • $9$
  • $17$
  • $34$

Key Concepts


Algebra

Integer

multiplication

Check the Answer


Answer:$17$

PRMO-2018, Problem 17

Pre College Mathematics

Try with Hints


Product of digits = \(n^2 – 15n – 27 = n(n – 15) – 27\)

so at first we observe when n=one digit ,two digit and 3 digit numbers…..

If n is a more than 2-digit number, say 3-digit number, then product has to be\(\leq 9 × 9 × 9 = 729\) but \((n(n – 15) – 27)\) is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n). Hence, n can be either one-digit or 2-digit.

If n is 1-digit then \(n^2 – 15n – 27 = n\) \(\Rightarrow n\)= not an integer , so n is a two digit number

now we will observe for 2-digit numbers…..

Can you now finish the problem ……….

For Two-digit numbers:

As product is positive so n(n-15)-27>0\(\Rightarrow n\geq 17\)

Now two digit product is less than equal to 81

so \(n(n-15)-27\leq 1\)\(\Rightarrow n(n-15)\leq 108\) \(\Rightarrow n\leq 20\)

Therefore n can be \(17\),\(18\),\(19\) or \(20\)

Can you finish the problem……..

For \(n\)= \(17\),\(18\),\(19\) or \(20\)

when n=17,then \(n(n-15)-27=7=1 \times 7\)

when n=18,then \(n(n-15)-27=27\neq 1\times 8\)

when n=19,then \(n(n-15)-27=49=1 \neq 9\)

when n=20,then \(n(n-15)-27=73=1 \neq 0\)

Therefore \(n\)=17

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AMC 8 Math Olympiad

Integer Problem |ISI-B.stat | Objective Problem 156

Try this beautiful problem Based on Integer, useful for ISI B.Stat Entrance.

Integer | ISI B.Stat Entrance | Problem 156


Let n be any integer. Then \(n(n + 1)(2n + 1)\)

  • (a) is a perfect square
  • (b) is an odd number
  • (c) is an integral multiple of 6
  • (d) does not necessarily have any foregoing properties.

Key Concepts


Integer

Perfect square numbers

Odd number

Check the Answer


Answer: (c) is an integral multiple of 6

TOMATO, Problem 156

Challenges and Thrills in Pre College Mathematics

Try with Hints


\(n(n + 1)\) is divisible by \(2\) as they are consecutive integers.

If \(n\not\equiv 0\) (mod 3) then there arise two casess……..
Case 1,,

Let \(n \equiv 1\) (mod 3)
Then \(2n + 1\) is divisible by 3.

Let \(n \equiv2\) (mod 3)
Then\( n + 1\) is divisible by \(3\)

Can you now finish the problem ……….


Now, if \(n\) is divisible by \(3\), then we can say that \(n(n + 1)(2n + 1)\) is always
divisible by \(2*3 = 6\)

Therefore option (c) is the correct

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AMC 8 Math Olympiad USA Math Olympiad

Problem based on LCM | AMC 8, 2016 | Problem 20

Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.

Problem based on LCM – AMC 8, 2016


The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

  • \(26\)
  • \(20\)
  • \(28\)

Key Concepts


Algebra

Divisor

multiplication

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


We have to find out the least common multiple of $a$ and $c$.if you know the value of \(a\) and \(c\) then you can easily find out the required LCM. Can you find out the value of \(a\) and \(c\)?

Can you now finish the problem ……….

Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore \(a\)=\(\frac{12}{3}=4\)

can you finish the problem……..

so\(b\)=3. Given that LCM of \(b\) and \(c\) is 15. Therefore c=5

Now lcm of \(a\) and \(c\) that is lcm of 4 and 5=20

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AMC 8 Math Olympiad

Linear Equations | AMC 8, 2007 | Problem 20

Try this beautiful problem from Algebra based on Linear equations from AMC-8, 2007.

Linear equations – AMC 8, 2007


Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

  • \(40\)
  • \(48\)
  • \(58\)

Key Concepts


Algebra

linear equation

multiplication

Check the Answer


Answer:48

AMC-8, 2007 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


At first, we have to Calculate the number of won games and lost games. Unicorns had won $45$% of their basketball game.so we may assume that out of 20 unicorns woned 9.

Can you now finish the problem ……….

Next unicorns won six more games and lost two.so find out the total numbers of won game and total numbers of games i.e won=\(9x+6\) and the total number of games become \(20x+8\)

can you finish the problem……..

Given that Unicorns had won \(45\)% of their basketball games i.e \(\frac{45}{100}=\frac{9}{20}\)

During district play, they won six more games and lost two,

Therefore they won\(9x+6\) and the total number of games becomes \(20x+8\)

According to the question, Unicorns finish the season having won half their games. …

Therefore,\(\frac{9x+6}{20x+8}=\frac{1}{2}\)

\(\Rightarrow 18x+12=20x+8\)

\(\Rightarrow 2x=4\)

\(\Rightarrow x=2\)

Total number of games becomes \(20x+8\) =\((20 \times 2) +8=48\)

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AMC 8 Math Olympiad

Integer | ISI-B.stat Entrance(Objective from TOMATO) | Problem 72

Try this beautiful problem Based on Integer useful for ISI B.Stat Entrance.

Integer| ISI B.Stat Entrance | Problem 72


The number of integer (positive ,negative or zero)solutions of \(xy-6(x+y)=0\) with \(x\leq y\) is

  • 5
  • 10
  • 12
  • 9

Key Concepts


Integer

Algebra

Divisor

Check the Answer


Answer: 10

TOMATO, Problem 72

Challenges and Thrills in Pre College Mathematics

Try with Hints


Factorize the given equation

Can you now finish the problem ……….

Find the divisor

can you finish the problem……..

Given equation is \(xy-6(x+y)=0\)

\(\Rightarrow xy-6x-6y=0\)

\(\Rightarrow xy-6x-6y+36=36\)

\(\Rightarrow (x-6)(y-6)=3^2 \times 2^2\)

Now the numbers of factpr of \(36=9\) i.e \(\{1,2,3,4,6,9,12,18,36\}\)

Thus we may say that 36 has 9 positive divisors, and 9 negative. and x=0 and y=0 is also a solution

the given condition \(x\leq y\) ,so there are 10 non-negetive solution

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