Math Kangaroo (Benjamin) 2016 Problem 24 | Play With Numbers

Try this Problem based on Playing With Numbers from Math Kangaroo (Benjamin) 2016 Problem 24

Playing With Numbers | Math Kangaroo (Benjamin) 2016 | Problem 24

Two three-digit numbers are made up of six different digits. The first digit of the second number is twice as big as the last digit of the first number. (Note: 0 is also a digit but cannot be the first digit of a number!) How big is the smallest possible sum of the two numbers?

Key Concepts

Numbers

Arithmetic

Counting

Suggested Book | Source | Answer

Mathematical Circle

Math Kangaroo (Benjamin) 2016 | Problem 24

537

Try with Hints

Let us assume these three digit numbers are $ABC$, $DEF$.

According to the question $D=2C$.

Let's follow the given condition and try to construct the smallest numbers.

So here $ABC=102$.

And if I follow the given condition then $DEF= 435$.

We did this keeping in mind that repetitions are not allowed.

Now calculate the answer.

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Math Kangaroo (Benjamin) 2016 | Problem 20 | Algebra

Try this Problem based on Algebra from Math Kangaroo (Benjamin) 2016 Problem 20

Equation Solving | Math Kangaroo (Benjamin) 2016 | Problem 20

Luigi owns a few square tables and some chairs for his little restaurant. If he sets out his tables individually with 4 chairs each, then he is 6 chairs short. If he always puts two tables together to create a bigger table with 6 chairs, then he has 4 chairs left over.
How many tables does Luigi have?

Key Concepts

Algebra

Arithmetic

Equation Solving

Suggested Book | Source | Answer

Mathematical Circle

Math Kangaroo (Benjamin) 2016 | Problem 20

10

Try with Hints

Let us assume the number of tables and chairs are $x, y$ respectively.

Let's follow the given condition and construct the equations.

For the first case if he set table individually with 4 chairs each then he is 6 chairs short.

So, $4x-6=y$.

Now if he put two tables together with 6 chairs each, then he has 4 chairs left over.

So, $6\frac{x}{2}+4=y$.

Comparing the equation of hint 3 and hint 4 we get,

$4x-6= 3x+4$

So, $ x=10$

Hence the answer is $10$.

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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

\(\frac{y}{x}\)=\(\frac{cos2cos6cos10.....cos86}{cos1cos2cos3....cos89}\)

=\(2^{44}\times\sqrt{2}\frac{cos2cos6cos10...cos86}{sin2sin4...sin88}\)

[ since cos\(\theta\)=sin(90-\(\theta\)) from cos 46 upto cos 89 and 2sin\(\theta\)cos\(\theta\)=sin2\(\theta\)]

=\(\frac{2^{\frac{89}{2}}sin4sin8sin12...sin88}{sin2sin4sin6...sin88}\)

[ since sin\(\theta\)=cos(90-\(\theta\))]

=\(\frac{2^{\frac{89}{2}}}{cos4cos8cos12..cos88}\)

[ since cos\(\theta\)=sin(90-\(\theta\))]

=\(\frac{2^\frac{89}{2}}{\frac{1}{2}^{22}}\)

[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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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

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

Now if c-a is a multiple of 7

c-a=7,0,-7

hence number of ordered pairs of (a,c) is 14

since b is odd

number of such number=\(14 \times 5\)=70.

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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

2a-b=\(k_1^2\) is equation 1

a-2b=\(k_2^2\) is equation 2

a+b=\(k_3^2\) is equation 3

adding 2 and 3 we get

2a-b=\(k_2^2+k_3^2\)

or, \(k_2^2+k_3^2\)=\(k_1^2\) \((k_2<k_3)\)

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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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

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

for 91n-104k<n+k, K>\(\frac{6n}{7}\)

and 0<91n-104k gives k<\(\frac{7n}{8}\)

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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Digits and Numbers | AIME I, 2012 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on digits and numbers.

Digits and numbers - AIME I, 2012

Let S be set of all perfect squares whose rightmost three digits in base 10 are 256. T be set of numbers of form \(\frac{x-256}{1000}\) where x is in S, find remainder when 10th smallest element of T is divided by 1000.

  • is 107
  • is 170
  • is 840
  • cannot be determined from the given information

Key Concepts

Digits

Algebra

Numbers

Check the Answer

Answer: is 170.

AIME I, 2012, Question 10

Elementary Number Theory by David Burton

Try with Hints

x belongs to S so perfect square, Let x=\(y^{2}\), here \(y^{2}\)=1000a+256 \(y^{2}\) element in S then RHS being even y=2\(y_1\) then \(y_1^{2}=250a+64\) again RHS being even \(y_1=2y_2\) then \(y_2^{2}\)=125\(\frac{a}{2}\)+16 then both sides being integer a=2\(a_1\) then \(y_2^{2}=125a_1+16\)

\(y_2^{2}-16=125a_1\) then \((y_2-4)(y_2+4)=125a_1\)

or, one of \((y_2+4)\) and \((y_2-4)\) contains a non negative multiple of 125 then listing smallest possible values of \(y_2\)

or, \(y_2+4=125\) gives \(y_2=121\) or, \(y_2-4=125\) gives \(y_2=129\) and so on

or, \(y_2=4,121,129,upto ,621\) tenth term 621

\(y=4y_2\)=2484 then \(\frac{2483^{2}-256}{1000}\)=170.

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Arbitrary Arrangement | TOMATO B.Stat Objective 119

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Arbitrary Arrangement.

Arbitrary Arrangement ( B.Stat Objective Question )

Let \(a_1, a_2, ....,a_{11}\) be an arbitrary arrangement (ie permutation) of the integers 1,2,....,11. Then the numbers \((a_1-1)(a_2-2)....(a_{11}-{11})\) is

  • necessarily \(\leq\) 0
  • necessarily even
  • necessarily 0
  • none of these

Key Concepts

Permutation

Numbers

Even and Odd

Check the Answer

Answer: necessarily even.

B.Stat Objective Problem 119

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

necessarily \(\leq\) 0 case

taking values (2-1)(3-2)(4-3).....(10-9)(1-10)(10-11)

here all the terms except last two terms are positive and there are 2 negetive terms whose product will be even

then product > 0

then not necessarily < 0 or = 0

necessarily even case

by contradiction

we assume that the product is not necessarily even

that is each of the factor have to be odd

then the arrangement look like

(even-1)(odd-2)(even-3)(odd-4)....(even-9)(odd-10)

but only one odd number left which will pair with 11 that a contradiction

\(\Rightarrow\) product is even

\(\Rightarrow\) necessarily even.

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Relations and Numbers | B.Stat Objective | TOMATO 63

Try this TOMATO problem from I.S.I. B.Stat Objective based on Relations and Numbers.

Relations and Numbers (B.Stat Objective problems)

We consider the relation , "a person x shakes hand with a person y".Obviously if x shakes hand with y, then y shakes hand with x. In a gathering of 99 persons , one of the following statements is always true, considering 0 to be an even number, find which one is it.

  • there is at least one person who shakes hand with an odd number of persons
  • there is at least one person who shakes hand with an even number of persons
  • there are even number of persons who shake hand exactly with an even number of persons
  • none of these

Key Concepts

Logic

Relations

Numbers

Check the Answer

Answer: there is at least one person who shakes hand with an even number of persons

B.Stat Objective Question 63

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints

Let R be handshakes among 99 persons holds

first person may handshake with at most 98(even) other persons, for second person similar arguments hold and this holds with similar arguments for all persons

then there exists at least one person who shakes hand with an even number of persons.

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Percentage Problem | AIME I, 2008 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Percentage.

Percentage Problem - AIME I, 2008

Of the students attending a party, 60% of the students are girls, and 40% of the students like to dance. After these students are joined by 20 more boy students, all of whom like to dance, the party is now 58% girls, find number of students now at the party like to dance.

  • is 107
  • is 252
  • is 840
  • cannot be determined from the given information

Key Concepts

Ratios

Percentage

Numbers

Check the Answer

Answer: is 252.

AIME I, 2008, Question 1

Elementary Number Theory by David Burton

Try with Hints

Let number of girls and boys be 3k and 2k, out of 3k girls, 2k likes to dance and 2k+20(boys) like to dance

here given that \(\frac{3k}{5k+20}\)=\(\frac{58}{100}\) then k=116

then 2k+20=252.

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