Fun with Mathematics

A problem 

Three boys randomly drew 4 digits from a bag having a large collection of numbers from 0 to 9. Boy A got 4 different digits without zero, boy B had 3 different digits with one zero while boy C was given with 3 different digits and one identical digit. How many different numbers each one of them can display with their digits?

Boy A got 4 different digits [1,2,3,4]

Single digit display  4 [1,2,3,4] = ⁴C₁

Double digits display. Fix a digit as first digit, then choose any one out of three. Repeat this for other digits. 4 [³C₁] =12  { 1(,2,3,4)-3;2(1,3,4)-3; 3(1,2,4)-3 ; 4(1,2,3)-3}

Triple digits display. Fix first digit. then choose any one out of three and then choose any one out of two.

4[³C₁ x ²C₁] = 4 x 3 x 2 = 24   e.g.,    with 1 as first digit there will be 6 numbers 1   {23,32,34,43,42,24}

For four digit display  ⁴C₁ [³C₁ x²C₁ x²C₁ x¹C₁ ] = 24  for example with 1 as first digit there will be 6 numbers 1 { 234,243,324,342,423,432}

Hece Boy A can show 4+12+24+24 = 64 numbers

Boy B has  3 different digits [1,2,3] and one zero [0]. Zero cannot exist in the beginning of any number.

Single digit display 3 (1,2,3]= ³C₁

Double digits display. Choose one digit out of three and then choose any one out of the remaining three

3 [³C₁] – 9  e.g., with one as first digit 1 [0,2,3]- 3

Triple digits numbers  3 [ ³C₁ x ²C₁ ] =3x3x2 = 18

Four digits numbers 3 [³C₁x²C₁x¹C₁ ] = 3x3x2 = 18

B can display 3 +9+18+18 = 48 numbers

Boy C has 3 different numbers [1,2,3] and one identical number [1].

Single digit number  ³C₁ = 3  [1,2,3]

Double digits numbers  for 1 ³C₁ and for 2,3 – 2 x²C₁ = 3 +2x2 = 7

Triple digits numbers  3 [³C₁ x ²C₁ ] = 18 numbers

Four digit numbers 3 [³C₁x²C₁x¹C₁ ]= 18 numbers

Boy C can display 3+7+18+18 = 46 numbers

What is the smallest integer under certain conditions ?

What is the smallest integer greater than 2 that leaves a remainder of 2 when divided by 3,4,5 or 6?

Let x be the required smallest integer. According to the conditions given,

3q₁ + 2 = 4q₂ + 2 = 5q₃ + 2 = 6q₄ + 2= x

Or 3q₁ = 4q₂= 5q₃ = 6q₄ = x-2

It means that x-2 is divisible by the divisors 3,4,5 and 6.. The smallest number to which 3,4,5 and 6 are all perfect divisors can be determined by multiplying all the common factors of the divisors.

3[ 3,4,5,6

2[ 1,4,5,2

   1,2,5,1

The smallest integer x -2 to which 3,4,5 and 6 are perfect divisors is given by 3x2x2x5 = 60;or x = 62.

Variant-1

Condition                                        Answer                              explanation

Divisors 3,4                                      14                               3x4+2

               3,4,5                                   62                               3x4x5+2

              3,4,5,6                                62                                2x5x6+2

              3,4,5,6,7                           422                               3x4x5x7+2

              3,4,5,6,7,8                       842                               3x5x7x8+2

              3,4,5,6,7,8,9                  2522                              5x7x8x9 +2

Variant-2

Find the smallest integer (x) greater than 3 that leaves a remainder when divided by 4,5,6,7.

4q₁ +3 = 5q₂ + 3 = 6q₃ + 3 = 7q₄ + 3 =  x

 Or 4q₁ = 5q₂ = 6q₃ = 7q₄ = x – 3

The smallest x-3 is given by the product of all common factors to the divisors 4,5,6,7 that is 3x4x5x7 = 420 or x = 423

 Condition               solution         explanation

 Divisors 4,5               23                   4x5+3

                4,5,6            63                  3x4x5+3

               4,5,6,7       423                  3x4x5x7+3

              4x5x6,7,8   843                 2x3x4x5x7+3

             4,5,6,7,8,9  2523              2x4x5x7x8 + 3

 

Fun with Mathematics

Let n be the number n =(9999….9)²₂₀₀₆ – (666666….6)²₂₀₀₆ Find the remainder when n is divided by (a)11 (b)8 and (c)7.

 

As there are even number of  9s and 6s in the base numbers(2006 9s and 2006 6s) ,each factor will be divisible by 11. So the remainder will be zero when n is divided by 11.

Instead if there are odd number of 9s and 6s in the base number, It can be expressed as (y+9)² – (y’+6)² where y and y’ are divisible by 11. Y²+18y+81 –(y’²+12y’+36) = the remainder is determined by 81-36=45 44 is divisible by 11, so the remainder is 1. This is true for all odd number of 9s and 6s in the base numbers.

It is noted that when n is divided by 3 (like 11) ,both the factors are divisible by 3 and hence the remainder will be zero.

 The number n will always end with  45

n-5 is divisible by 8 (5x8 = 40) hence the remainder is 5 whatever may be the number of 9s and 6s.

Since n ends with 5 , the remainder will be zero when it is divided by 5.

9²- 6² = 45 ; the remainder when divided by 7  is 3

99² – 66² = 5445 the remainder when divided by 7 is 6 ;(y+1)² – (y’ +3)² ;y² +2y +1 – y’² -6y’ -9; 14+1 -9 = 6

164 x 33 = (y+4)(y’+5) ,the remainder is determined from 20(=4x5) , the remainder 20-14 = 6

999² – 666² = 554445,the remainder when divided by 7 is 3

9999²- 6666² = 55544445, the remainder is 5

99999²- 66666² = 5555444445, the remainder is 5
999999²- 666666² = 555554444445, the remainder is 0

It is observed that the remainder of each square varies cyclically, in the case of all 9s,when divided by 7

No.of 9s      remainder        no.of 6s   remainder

      1                   2                        1                 6

      2                   1                        2                 3

     3                    5                        3                  1

      4                   3                        4                   2

      5                   4                         5                  5

      6                   0                         6                   0              

 Since there are 2006 9s and 2006 6 s, when divided by 6  it gives a remainder 2.

Hence (999999….9)₂₀₀₆ is divided by 7 ,it will give a remainder 1

(666666…6)₂₀₀₆ is divided by 7 , it will give a remainder 3. The resultant remainder can be determined from (y+1)² + (y’+3)² All terms containing y and y’ will be divisible by 7 , the remaining term is 1+9 = 10. So the ultimate remainder is 3

PROBLEM WITH RECTANGLE TRIANGLE

  

                                                                                  

                                                   

 

                                                                   

In a rectangle triangle with sides AB=a,AC=b and BC=c, a perpendicular is drawn from the vertex A to BC, the hypotenuse. Express the length of the perpendicular AD in terms of a,b,c. What is the ratio of BD:DC ?

According to Pythagoras theorem a² + b² = c²

Considering the inner rectangle triangles

AD² + BD² = a² and AD² + CD² = b² . If BD = x and CD = c - x Adding these two relations

c² + 2x² – 2cx + 2AD² = a² + b² = c² or x²- cx + AD² = o Solving this quadratic equation, we get x = [c – (c²-4AD²]^1/2 / 2.

 On subtraction, CD² – BD² = c²- 2cx =a² + b² -2cx = b² – a². Or a² = cx or x = BD = a²/c and c-x = CD =

 b² /c. Substituting this value of x in the above equation we get,

AD = ab/c and BD:CD = a²: b²

Fun with Mathematics

 

                                                                             (2010)²⁰⁰⁹

What would be the last two digits in (2011)

If the root number ends with 1, whatever may its power, it will always end with 1. So the problem is to determine the last but one digit.

11² = 121, when the power ends with 2, the last but one digit is 2

11³ = 1331

11⁴= 14641
11⁵ = 161051
11⁶ = 1771561
11⁷ = 19487171
11⁸= 214358881
11⁹ = 2357947691
11¹⁰ = 25937424601

Since the power number of 2011 is 2010 ²⁰⁰⁹   [=201²⁰⁰⁹x 10²⁰⁰⁹]  ends with zero(zeros), the last but one digit will be 0.

If the power number of 2010 is changed from 1 to any real number , our prediction will remain unchanged.

                                                    (2009)²⁰¹⁰

If the given number is  (2011)                      the result will be quite different. It is noted that the power of any numbers ending with 9 will always end either with 9 (for odd power) and 1 for even power

[for example 9² = 81;9³ = 729;9⁴ = 6561; 9⁵ =59049 and so on.].Since the power of 2009 is even, the power of 2011 will end with 1. Hence the, last two digits will be 11 and this is true for all even powers of 2009. In the case of odd power ,the power of 2011 will ends with 9, and hence the last two digits will be 91.

Fun with Mathematics

Problem

                                                                             (2010)²⁰⁰⁹

What would be the last two digits in (2011)

 

If the root number ends with 1, whatever may its power, it will always end with 1. So the problem is to determine the last but one digit.

11² = 121, when the power ends with 2, the last but one digit is 2

11³ = 1331

11⁴= 14641

11⁵ = 161051

11⁶ = 1771561

11⁷ = 19487171

11⁸= 214358881

11⁹ = 2357947691

11¹⁰ = 25937424601

Since the power number of 2011 is 2010 ²⁰⁰⁹   [=201²⁰⁰⁹x 10²⁰⁰⁹]  ends with zero(zeros), the last but one digit will be 0.

If the power number of 2010 is changed from 1 to any real number , our prediction will remain unchanged.

                                                    (2009)²⁰¹⁰

If the given number is  (2011)                      the result will be quite different. It is noted that the power of any numbers ending with 9 will always end either with 9 (for odd power) and 1 for even power

[for example 9² = 81;9³ = 729;9⁴ = 6561; 9⁵ =59049 and so on.].Since the power of 2009 is even, the power of 2011 will end with 1. Hence the, last two digits will be 11 and this is true for all even powers of 2009. In the case of odd power ,the power of 2011 will ends with 9, and hence the last two digits will be 91.

Problem

What is the remainder when 2⁵⁵  + 1 is divided by 33.?

The simple way to solve this problem is to reduce the number

conveniently smaller to workout.When 1000 is divided by 33, the remainder will be 10 ,If it is 1024, the remainder will be 10+24=34, and the final remainder will be 1

2⁵⁰ = (2¹⁰)⁵ = (1024)⁵ = (y+1)⁵ , where y is perfectly divisible by 33.On expanding it into a power series,all the terms will contain y except the last term which will be equal to 1

2⁵⁵ = (2¹⁰)⁵ x 2⁵ = (y+1)⁵ x 32 .The ultimate remainder will be 32 times the remainder in (y+l)⁵.i.e., 32 x 1 = 32.

Alternatively it can be worked out by considering 2⁵⁵ = (2¹¹)⁵ = (2048)⁵

 The remainder for 2000 is 20 and for 2048 it is 68 and final remainder is 2

2⁵⁵ = (y+2)⁵, where y is perfectly divisible by 33. The polynomial technique predicts  its last term (2⁵) as remainder .It is 33 when + 1 is also taken into account.

Problem

What is the remainder when (918273)³ when divided by 7 ?

One may know that all cube numbers when divided by 7, the remainder will always be 1 or 6 or 0.

When the base number is divisible by 7, the remainder will be 0, It the base number is in the form (1+7n) or(2+7n) or (4+7n), the remainder will be 1 ,If it is in the form (3+7n) or (5+7n) or (6+7n), the remainder will be 6.

918273/7 =131181(7) +6, so the remainder is 6.

It shows that the remainder of any base number is equal to the remainder of cube of the number when divided by 7.