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)

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

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.

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

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 ..........

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

Sum of Digits base 10 | PRMO II 2019 | Question 7

Try this beautiful problem from the PRMO II, 2019 based on Sum of Digits base 10.

Sum of Digits base 10 - PRMO II 2019

Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m)=20 and s(33m)=120, what is the value of s(3m)?

is 107
is 60
is 840
cannot be determined from the given information

Key Concepts

Real Numbers

Algebra

Integers

Check the Answer

Answer: is 60.

PRMO II, 2019, Question 7

Elementary Algebra by Hall and Knight

Try with Hints

taking sum of digit base 10 to (mod 9)

and s(ab)=s(a).s(b)(mod 9)

[ let x congruent r mod n, y congruent to s mod n,

$0 \leq r,s \leq n-1$,

x=in+r, y=jn+s, i,j are integers

xy=(in+r)(jn+s)=ij$n^2$+(is+jr)n+rs congruent to rs mod n

so, xy mod n =(x mod n)(y mod n) ]

given s(m)=20

s(33m)=120=$s(11) \times s(3m)$

or, 120=$2 \times s(3m)$ [ since s(11)=2(mod 9)]

or, 60=s(3m)

so, s(3m)=60.

Distance travelled | PRMO II 2019 | Question 26

Try this beautiful problem from the Pre-RMO II, 2019, Question 26, based on Distance travelled.

Distance travelled - Problem 26

A friction-less board has the shape of an equilateral triangle of side length 1 meter with bouncing walls along the sides. A tiny super bouncy ball is fired from vertex A towards the side BC. The ball bounces off the walls of the board nine times before it hits a vertex for the first time. The bounces are such that the angle of incidence equals the angle of reflection. The distance travelled by the ball in meters is of the form $\sqrt{N}$, where N is an integer

is 107
is 31
is 840
cannot be determined from the given information

Key Concepts

Equation

Algebra

Integers

Check the Answer

Answer: is 31.

PRMO II, 2019, Question 26

Higher Algebra by Hall and Knight

Try with Hints

x= length of line segment

and by cosine law on triangle of side x, 1, 5 and 120 (in degrees) as one angle gives

$x^2=5^2+1^2-2 \times 5 \times 1 cos 120^\circ$

$=25+1+5$

or, x=$\sqrt{31}$

or, N=31

Folding the triangle continuously each time of reflection creates the above diagram. 9 points of reflection can be seen in the diagram. Thus root (N) is the length of line which is root (31). Thus N=31 is the answer.

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

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 ..........

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 $m+n+k=75$

Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

Circle - PRMO 2017, Problem 27

Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. Suppose another circle $\Omega_{2}$ with centre P lies in the interior of $\Omega_{1}$. Tangents are drawn from A and B to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$, and $B_{1}$ are on the opposite sides of $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

$9$
$40$
$34$
$20$

Key Concepts

Geometry

Circle

Check the Answer

Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

Try with Hints

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$
From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$
[
\begin{array}{l}
\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \
\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}
\end{array}
]

Can you now finish the problem ..........

Again, $\Delta B P E \sim \Delta B A B_{1}$
Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$
$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem........

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

Acute angled Triangle | PRMO II 2019 | Question 29

Try this beautiful problem from the Pre-RMO II, 2019, Question 29, based on Acute angled triangle.

Acute angled triangle - Problem 29

Let ABC be a acute angled triangle with AB=15 and BC=8. Let D be a point on AB such that BD=BC. Consider points E on AC such that $\angle$DEB=$\angle$BEC. If $\alpha$ denotes the product of all possible val;ues of AE, find[$\alpha$] the integer part of $\alpha$.

is 107
is 68
is 840
cannot be determined from the given information

Key Concepts

Equation

Algebra

Integers

Check the Answer

Answer: is 68.

PRMO II, 2019, Question 29

Higher Algebra by Hall and Knight

Try with Hints

The pairs $E_1$,$E_2$ satisfies condition or $E_1$=intersection of CBO with AC and $E_2$=intersection of $\angle$bisector of B and AC

since that $\angle DE_2B$=$\angle CE_2B$ and for $E_1$$\angle BE_1C$=$\angle$BDC=$\angle$BCD=$\angle BE_1D$

or, $AE_1.AC$=$AD.AB$=$7 \times 15$

$\frac{AE_2}{AC}$=$\frac{XY}{XC}$

(for y is midpoint of OC and X is foot of altitude from A to CD)

$\frac{XD}{DY}=\frac{7}{8}$ and DY=YC

or, $\frac{XD+DY}{XC}$=$\frac{15}{7+8+8}$=$\frac{15}{23}$

or, $\frac{XY}{XC}=\frac{15}{23}$

or, $\frac{AE_2}{AC}$=$\frac{15}{23}$

or, $AE_1.AE_2$=$\frac{15}{23}(7.15)$=$\frac{225 \times 7}{23}$

$[\frac{225 \times 7}{23}]$=68.

Shortest Distance | PRMO II 2019 | Question 27

Try this beautiful problem from the Pre-RMO II, 2019, Question 27 based on Shortest Distance.

Shortest Distance - Pre-RMO II, Problem 27

A conical glass is in the form of a right circular cone. The slant height is 21 and the radius of the top rim of the glass is 14. An ant at the mid point of a slant line on the outside wall of the glass sees a honey drop diametrically opposite to it on the inside wall of the glass. If d the shortest distance it should crawl to reach the honey drop, what is the integer part of d?

is 107
is 36
is 840
cannot be determined from the given information

Key Concepts

Equation

Algebra

Integers

Check the Answer

Answer: is 36.

PRMO II, 2019, Question 27

Higher Algebra by Hall and Knight

Try with Hints

Rotate $\Delta$OAP by 120$^\circ$ in anticlockwise then A will be at B, P will be at P'

or, $\Delta$OAP is congruent to $\Delta$OBP'

or, PB+PA=P'B+PB $\geq$ P'P

Minimum PB+PA=P'P equality when P on the angle bisector of $\angle$AOB

or, P'P=2(21)sin60$^\circ$=21$\sqrt{3}$

[min(PB+PA)]=[21$\sqrt{3}$]=36 (Answer)

Length of side of Triangle | PRMO II 2019 | Question 28

Try this beautiful problem from the Pre-RMO II, 2019, Question 28, based on Length of side of triangle.

Length of side of triangle - Problem 28

In a triangle ABC, it is known that $\angle$A=100$^\circ$ and AB=AC. The internal angle bisector BD has length 20 units. Find the length of BC to the nearest integer, given that sin 10$^\circ$=0.174.

is 107
is 27
is 840
cannot be determined from the given information

Key Concepts

Equation

Algebra

Integers

Check the Answer

Answer: is 27.

PRMO II, 2019, Question 28

Higher Algebra by Hall and Knight

Try with Hints

given, BD=20 units

$\angle$A=100$^\circ$

AB=AC

In $\Delta$ABD

$\frac{BD}{sinA}=\frac{AD}{sin20^\circ}$

or, $\frac{BD}{sin100^\circ}=\frac{AD}{sin20^\circ}$

or, 20=$\frac{AD}{2sin10^\circ}$ or, AD=40sin10$^\circ$=6.96

finding the length of the side of triangle

In $\Delta$BDC

$\frac{BD}{sin40^\circ}=\frac{BC}{sin120^\circ}=\frac{CD}{sin20^\circ}$

or, CD=$\frac{20}{2cos20^\circ}$=$\frac{20}{2 \times 0.9394}$=10.65

So, AD+CD=AC=AB=17.6

since BD is angle bisector

$\frac{BC}{AB}=\frac{CD}{AD}$

or, BC=$\frac{AB \times CD}{AD}$=$\frac{17.6 \times 10.65}{6.96}$

=26.98=27.