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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998 based on LCM and Integers.

Lcm and Integer – AIME I, 1998


Find the number of values of k in \(12^{12}\) the lcm of the positive integers \(6^{6}\), \(8^{8}\) and k.

  • is 107
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Lcm

Algebra

Integers

Check the Answer


Answer: is 25.

AIME I, 1998, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

here \(k=2^{a}3^{b}\) for integers a and b

\(6^{6}=2^{6}3^{6}\)

\(8^{8}=2^{24}\)

\(12^{12}=2^{24}3^{12}\)

Second Hint

lcm\((6^{6},8^{8})\)=\(2^{24}3^{6}\)

\(12^{12}=2^{24}3^{12}\)=lcm of \((6^{6},8^{6})\) and k

=\((2^{24}3^{6},2^{a}3^{b})\)

=\(2^{max(24,a)}3^{max(6,b)}\)

Final Step

\(\Rightarrow b=12, 0 \leq a \leq 24\)

\(\Rightarrow\) number of values of k=25.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

Integer – AIME I, 1993


Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

  • is 107
  • is 870
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Algebra

Check the Answer


Answer: is 870.

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

Second Hint

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

Final Step

taking first two solutions a<b<c<d<500

or,\(1 \leq c-93, c+1 \leq 499\)

or, \(94 \leq c \leq 498 \) gives 405 solutions

and \(1 \leq c-31, c+3 \leq 499\)

or, \(32 \leq c \leq 496\) gives 465 solutions

or, 405+465=870 solutions.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

Positive Integer – AIME I, 1996


For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that \(n \lt 1000\) and that \([log_{2}n]\) is a positive even integer.

  • is 107
  • is 340
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequality

Greatest integer

Integers

Check the Answer


Answer: is 340.

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

here Let \([log_{2}n]\)=2k for k is an integer

\(\Rightarrow 2k \leq log_{2}n \lt 2k+1\)

\(\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}\) and \(n \lt 1000\)

Second Hint

\(\Rightarrow 4 \leq n \lt 8\)

\(16 \leq n \lt 32\)

\(64 \leq n \lt 128\)

\(256 \leq n \lt 512\)

Final Step

\(\Rightarrow 4+16+64+256\)=340.

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Algebra Arithmetic Math Olympiad PRMO

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


First hint

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) ]

Second Hint

given s(m)=20

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

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

Final Step

or, 60=s(3m)

so, s(3m)=60.

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Algebra Arithmetic Geometry Math Olympiad PRMO

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


First hint

x= length of line segment

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

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

\(=25+1+5\)

Distance travelled graph

Second Hint

or, x=\(\sqrt{31}\)

or, N=31

Final Step

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.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Trapezoid Problem | AIME I, 1992 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on trapezoid.

Trapezoid – AIME I, 1992


Trapezoid ABCD has sides AB=92, BC=50,CD=19,AD=70 with AB parallel to CD. A circle with centre P on AB is drawn tangent to BC and AD. Given that AP=\(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 164
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Trapezoid

Angle Bisectors

Check the Answer


Answer: is 164.

AIME I, 1992, Question 9

Coordinate Geometry by Loney

Try with Hints


First hint

Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Trapezoid Problem

Second Hint

Let F be point on CD where it meets

Taking angle bisector theorem,

let YB=z(92-y), YA=zy for some z

YD=zy-70, YC=z(92-y)-50

\(\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}\)

Final Step

solving we get 120y=(70)(92)

or, AP=y=\(\frac{161}{3}\)

or, 161+3=164.

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Algebra Arithmetic Geometry Math Olympiad PRMO

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?

Shortest Distance
  • 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’

Shortest Distance figure

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)

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Algebra Arithmetic Geometry Math Olympiad PRMO

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


First hint

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

Second Hint

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

Final Step

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.

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Algebra Arithmetic Geometry Math Olympiad PRMO

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


First hint

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)

Acute angled Triangle problem

Second Hint

\(\frac{XD}{DY}=\frac{7}{8}\) 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, \(AE_1.AE_2\)=\(\frac{15}{23}(7.15)\)=\(\frac{225 \times 7}{23}\)

Final Step

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

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Algebra Arithmetic Math Olympiad PRMO

Missing Integers | PRMO II 2019 | Question 1

Try this beautiful problem from the PRMO II, 2019 based on Missing Integers.

Missing Integers – PRMO II 2019


Consider the sequence of numbers [n+\(\sqrt{2n}+\frac{1}{2}\)] for \(n \geq 1\), where [x] denotes the greatest integer not exceeding x. If the missing integers in the sequence are \(n_1<n_2<n_3<…\) then find \(n_{12}\).

  • is 107
  • is 78
  • is 840
  • cannot be determined from the given information

Key Concepts


Real Numbers

Algebra

Integers

Check the Answer


Answer: is 78.

PRMO II, 2019, Question 1

Elementary Algebra by Hall and Knight

Try with Hints


First hint

\([n+\sqrt{2n}+\frac{1}{2}]\)=[\((\sqrt{n}+\frac{1}{\sqrt{2}})^2\)]

Let P=[\((\sqrt{n}+0.7)^2\)]

Second Hint

given \(n \geq 1\), put n=1 gives P=2

n=2 gives P=4

n=3 gives P=5

n=4 gives P=7

n-5 gives P=8

n=6 gives P=9

n=7 gives P=11

Final Step

here missing number are

1,3,6,10,… which is following a certain pattern

1, 1+2, 3+3, 6+4, 10+5, 15+6, 21+7, 28+8, 36+9, 45+10, 55+11, 66+12.

so, \(n_{12}\)=78.

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