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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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Math Olympiad PRMO USA Math Olympiad

Ordered Pairs | PRMO-2019 | Problem 18

Try this beautiful problem from PRMO, 2019, Problem 18 based on Ordered Pairs.

Orderd Pairs | PRMO | Problem-18


How many ordered pairs \((a, b)\) of positive integers with \(a < b\) and \(100 \leq a\), \(b \leq 1000\) satisfy \(gcd (a, b) : lcm (a, b) = 1 : 495\) ?

  • $20$
  • $91$
  • $13$
  • \(23\)

Key Concepts


Number theory

Orderd Pair

LCM

Check the Answer


Answer:\(20\)

PRMO-2019, Problem 18

Pre College Mathematics

Try with Hints


At first we assume that \( a = xp\)
\(b = xq\)
where \(p\) & \(q\) are co-prime

Therefore ,

\(\frac{gcd(a,b)}{LCM(a ,b)} =\frac{495}{1}\)

\(\Rightarrow pq=495\)
Can you now finish the problem ……….

Therefore we can say that

\(pq = 5 \times 9 \times 11\)
\(p < q\)

when \( 5 < 99\) (for \(x = 20, a = 100, b = 1980 > 100\)),No solution
when \(9 < 55\) \((x = 12\) to \(x = 18)\),7 solution
when,\(11 < 45\) \((x = 10\) to \(x = 22)\),13 solution
Can you finish the problem……..

Therefore Total solutions = \(13 + 7=20\)

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Geometry Math Olympiad PRMO USA Math Olympiad

Maximum area | PRMO-2019 | Problem 23

Try this beautiful problem from PRMO, 2019 based on Maximum area

Maximum area | PRMO-2019 | Problem-23


Let $\mathrm{ABCD}$ be a convex cyclic quadrilateral. Suppose $\mathrm{P}$ is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least. If ${\mathrm{PA}, \mathrm{PB}, \mathrm{PC}, \mathrm{PD}}={3,4,6,8} .$ What is the maximum possible area of ABCD?

  • $20$
  • $55$
  • $13$
  • \(23\)

Key Concepts


Geometry

Triangle

Area

Check the Answer


Answer:\(55\)

PRMO-2019, Problem 23

Pre College Mathematics

Try with Hints


problem figure

Given that $\mathrm{PA}=\mathrm{a}, \mathrm{PB}=\mathrm{b}, \mathrm{PC}=\mathrm{c}, \mathrm{PD}=\mathrm{d}$
Now from the above picture area of quadrilateral ABCD
Area=$[\mathrm{APB}]+[\mathrm{BPC}]+[\mathrm{CPD}]+[\mathrm{DPA}]$

finding the maximum area

Therefore area $\Delta=\frac{1}{2} \mathrm{ab} \sin \mathrm{x}+\frac{1}{2} \mathrm{bc} \sin \mathrm{y}+\frac{1}{2} \mathrm{cd} \sin \mathrm{z}+\frac{1}{2}$ da $\sin \mathrm{w}$
$\Delta_{\max }$ when $x=y=z=w=90^{\circ}$
$\Delta_{\max }=\frac{1}{2}(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{d})$
Now ac $=$ bd (cyclic quadrilateral) As $(a, b, c, d)=(3,4,6,8)$
$\Rightarrow{(a, c)(b, d)}={(3,8)(4,6)}$

So $\Delta_{\max }=\frac{1}{2} \times 11 \times 10=55$

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Geometry Math Olympiad PRMO USA Math Olympiad

Ratio of the areas | PRMO-2019 | Problem 19

Try this beautiful problem from PRMO, 2019 based on Ratio of the areas.

Ratio of the areas | PRMO | Problem-19


Let $\mathrm{AB}$ be a diameter of a circle and let $\mathrm{C}$ be a point on the segment $\mathrm{AB}$ such that $\mathrm{AC}: \mathrm{CB}=6: 7 .$ Let $\mathrm{D}$ be a point on the circle such that $\mathrm{DC}$ is perpendicular to $\mathrm{AB}$. Let DE be the diameter through $\mathrm{D}$. If $[\mathrm{XYZ}]$ denotes the area of the triangle XYZ. Find [ABD] / $[\mathrm{CDE}]$ to the nearest integer.

  • $20$
  • $91$
  • $13$
  • \(23\)

Key Concepts


Geometry

Triangle

Area

Check the Answer


Answer:\(13\)

PRMO-2019, Problem 19

Pre College Mathematics

Try with Hints


ratio of the areas problem figure

\(\angle \mathrm{AOC} \quad=\frac{6 \pi}{13}, \angle \mathrm{BOC}=\frac{7 \pi}{13}\)

$\mathrm{Ar} \Delta \mathrm{ABD}=\mathrm{Ar} \Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{OC} \sin \frac{6 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{CDE}=\frac{1}{2} \mathrm{DE} \times \mathrm{OC} \sin \left(\frac{7 \pi}{13}-\frac{6 \pi}{13}\right)$

figure

$\frac{[\mathrm{ABD}]}{[\mathrm{CDE}]}=\frac{\sin \frac{6 \pi}{13}}{\sin \frac{\pi}{13}}=\frac{1}{2 \sin \frac{\pi}{26}}=\mathrm{p}$

because $\sin \theta \cong \theta$ if $\theta$ is small
$\Rightarrow \sin \frac{\pi}{26} \cong \frac{\pi}{26}$

$\mathrm{p}=\frac{13}{\pi} \Rightarrow$ Nearest integer to $\mathrm{p}$ is 4

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Algebra AMC 8 Arithmetic India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Problem on Positive Integers | PRMO-2019 | Problem 26

Try this beautiful problem from Algebra PRMO 2019 based on Positive Integers.

Positive Integers | PRMO | Problem 26


Positive integers x,y,z satisfy xy+z=160 compute smallest possible value of x+yz.

  • 24
  • 50
  • 29
  • 34

Key Concepts


Algebra

Integer

sum

Check the Answer


Answer:50

PRMO-2019, Problem 26

Higher Algebra by Hall and Knight

Try with Hints


x+yz=\(\frac{160-z}{y}\)+yz

=\(\frac{160}{y}+\frac{z(y^{2}-1)}{y}=\frac{160-z}{y}+\frac{zy^{2}}{y}=\frac{160-z}{y}+zy\)

for particular value of z, \(x+yz \geq 2\sqrt{z(160-z)}\)

or, least value=\(2\sqrt{z(160-z)}\) but an integer also

for least value z is also

case I z=1, \(x+yz=\frac{159}{y}+y\) or, min value at y=3 which is 56

case II z=2, \(x+yz=\frac{158}{y}+2y\) or, min value at y =2 which is 83 (not taken)

case III z=3, \(x+yz=\frac{157}{y}+3y\) or, min value at y=1 which is 160 (not taken)

case IV z=4, \(x+yz=\frac{156}{y}+4y\) or, min at y=6 which is 50 (taken)

case V z=5, \(x+yz=\frac{155}{y}+5y\) or, minimum value at y=5 which is 56 (not taken)

case VI z=6, \(x+yz=\frac{154}{y}+6y\) \( \geq 2\sqrt{924}\)>50

smallest possible value =50.

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AMC 8 Math Olympiad USA Math Olympiad

Largest Possible Value | PRMO-2019 | Problem 17

Try this beautiful problem from PRMO, 2019 based on Largest Possible Value.

Largest Possible Value | PRMO | Problem-17


Let a, b, c be distinct positive integers such that \(b + c – a\),\( c + a – b\) and \(a + b – c\) are all perfect squares.
What is the largest possible value of \(a + b + c\) smaller than \(100\)?

  • $20$
  • $91$
  • $13$

Key Concepts


Number theory

Perfect square

Integer

Check the Answer


Answer:\(91\)

PRMO-2019, Problem 17

Pre College Mathematics

Try with Hints


Let \(b + c – a = x^2\) … (i)
\(c + a – b = y^2\) … (ii)
\(a + b – c = z^2\) … (iii)

Now since \(a\),\( b\), \(c\) are distinct positive integers,
Therefore, \(x\), \(y\), \(z\) will also be positive integers,
add (i), (ii) and (iii)
\(a + b + c = x^2 + y^2 + z^2\)
Now, we need to find largest value of \(a + b + c or x^2 + y^2 + z^2\) less than \(100\)
Now, to get a, b, c all integers \(x\),\( y\), \(z\) all must be of same parity, i.e. either all three are even or all three
are odd.

Can you now finish the problem ……….

Let us maximize\(x^2 + y^2 + z^2\), for both cases.
If \(x\), \(y\), \(z \)are all even.
Therefore,

\(b + c – a = 8^2 = 64\)
\(c + a – b = 42 = 16\)
\(a + b – c = 22 = 4\)
Which on solving, give\( a = 10\),\( b = 34\), \(c = 40\) and \(a + b + c = 84\)
If x, y, z are all odd
\(\Rightarrow b + c – a = 92 = 81\)
\(c + a – b = 32 = 9\)
\(a + b – c = 12 = 1\)
Which on solving, give \(a = 5\) ,\(b = 41\), \(c = 45\) and\( a + b + c = 91\)

Can you finish the problem……..

Therefore Maximum value of \(a + b + c < 100 = 91\)

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Smallest positive value | Algebra | PRMO-2019 | Problem 13

Try this beautiful problem from Algebra PRMO 2019 based on smallest positive value

Smallest positive value| PRMO | Problem 13


Each of the numbers \(x_1, x_2,……….x_{101}\) is \(±1\). What is the smallest positive value of \(\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) ?

  • $24$
  • $10$
  • $34$

Key Concepts


Algebra

Integer

sum

Check the Answer


Answer:\(10\)

PRMO-2019, Problem 13

Pre College Mathematics

Try with Hints


\(S=\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j\) .

we have \((x_1+x_2+x_3+….+x_{101})^2={x_1}^2+{x_2}^2+…..+{x_{101}}^2+2S\)

\(\Rightarrow 2S\)=\((\displaystyle\sum_{i=1}^{101} {x_i})^2-\displaystyle\sum_{j=1}^{101} {x_j}^2\)

Can you now finish the problem ……….

Since we have \(x_i=\pm 1\) so \({x_i}^2=1\)

so \(2S=(\displaystyle\sum_{i=1}^{101} {x_i})^2-{101}\)

Since \(\displaystyle\sum_{i=1}^{101} {x_i}\) will be an integer

so \((\displaystyle\sum_{i=1}^{101} {x_i})^2\) will be a perfect square .

For smalll positive \(S\), \((\displaystyle\sum_{i=1}^{101} {x_i})^2\)must be smallest perfect square greater than \({101}\)

So \((\displaystyle\sum_{i=1}^{101} {x_i})^2={121}\)

\(\Rightarrow (\displaystyle\sum_{i=1}^{101} {x_i})\)=\({11}\) or \({-11}\)

Can you finish the problem……..

We can verify that the desired sum can be achieved by putting \(45\) \(x_i\)’s to be –1 and \(56\) \(x_i\)’s to be \(1\) So, \(2S = 121 – 101 = 20\)

\(\Rightarrow s=10\)

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AMC 8 India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Regular polygon | Combinatorics | PRMO-2019 | Problem 15

Try this beautiful problem from combinatorics PRMO 2019 based on Regular polygon

Regular polygon| PRMO | Problem 15


In how many ways can a pair of parallel diagonals of a regular polygon of \(10\) sides be selected

  • $24$
  • $45$
  • $34$

Key Concepts


Combinatorics

Regular polygon

geometry

Check the Answer


Answer:\(45\)

PRMO-2019, Problem 15

Pre College Mathematics

Try with Hints


regular polygon

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ……….

Polygon 2
Fig.2
Polygon 1
Fig. 1

If we joined the diagonals (shown in Fig. 1), i.e \((P_3 \to P_10)\),\((P_4\to P_9)\),\((P_5 \to P_8)\) then then we have 3 diagonals.so we have 5\(4 \choose 2\) ways=\(15\) ways.

If we joined the diagonals (shown in Fig.2), i.e \((P_1 \to P_3)\),\((P_10\to P_4)\),\((P_9\to P_5)\),\((P_8\to P_6)\)then we have \(4\)diagonals.so we have 5\(3 \choose 2\) ways=\(30\) ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of \(10\) sides be selected is \(15+30=45\)

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AMC 8 Math Olympiad USA Math Olympiad

Good numbers Problem | PRMO-2019 | Problem 12

Try this beautiful problem from PRMO, 2019 based on Good numbers.

Good numbers Problem | PRMO | Problem-12


A natural number \(k > \) is called good if there exist natural numbers
\(a_1 < a_2 < ………. < a_k\)

\(\frac{1}{\sqrt a_1} +\frac{1}{\sqrt a_2}+………………. +\frac{1}{\sqrt a_k}=1\)

Let \(f(n)\) be the sum of the first \(n\) good numbers, \(n \geq 1\). Find the sum of all values of \(n\) for which
\(f(n + 5)/f(n)\) is an integer.

  • $20$
  • $18$
  • $13$

Key Concepts


Number theory

Good number

Integer

Check the Answer


Answer:\(18\)

PRMO-2019, Problem 12

Pre College Mathematics

Try with Hints


A number n is called a good number if It is a square free number.

Let \(a_1 ={A_1}^2\),\(a_2={A_2}^2\),………………\(a_k={A_k}^2\)
we have to check if it is possible for distinct natural number \(A_1, A_2………….A_k\) to satisfy,
\(\frac{1}{A_1}+\frac{1}{A_2}+………..+\frac{1}{A_k}=1\)

Can you now finish the problem ……….

For \(k = 2\); it is obvious that there do not exist distinct\( A_1, A_2\), such that \(\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2\) is not a good number

For \(k = 3\); we have \(\frac{1}{2} +\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3\) is a good number.

\(\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\) \(\Rightarrow 4\) is a good number

Let \(k\) wil be a good numbers for all \(k \geq 3\)

\(f(n) = 3 + 4 +… n\) terms =\(\frac{n(n + 5)}{2}\)
\(f(n + 5) =\frac{(n + 5)(n +10)}{2}\)

\(\frac{f(n+5}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n}\)

Can you finish the problem……..

Therefore the integer for n = \(1\), \(2\), \(5\) and \(10\). so sum=\(1 + 2 + 5 + 10 = 18\).

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AMC 8 Math Olympiad USA Math Olympiad

Time & Work Problem | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 from Arithmetic based on Time & Work.

Time & Work Problem | PRMO | Problem 3


A contractor has two teams of workers: team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins A after four days. Team A withdraws after two more days. For how many more days should team B work to complete the job?

  • \(24\)
  • \(16\)
  • \(22\)
  • \(18\)

Key Concepts


Arithmetic

Unitary process

Work done

Check the Answer


Answer:\(16\)

PRMO-2017, Problem 3

Pre College Mathematics

Try with Hints


At first we have to find out A’s 1 days work and B’s 1 days work.next find out A and B both together 1 day’s work .

Can you now finish the problem ……….

Team A completes job in 12 days and Team B completes job in 36 days

1 day work of team A =\(\frac{1}{12}\)

1 day work of team B=\(\frac{1}{36}\)

1 day work of team A and team B (when they both work together \(\frac{1}{12} +\frac{1}{36}\)=\(\frac{1}{9}\)

Now according to question,
Let more number of days should team B works to complete the job be x days

\(4 \times \frac{1}{12} +2 \times \frac{1}{9} + x \times \frac{1}{36}=1\)

\(\Rightarrow x=16\)

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