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  • Level: GCSE
  • Subject: Maths
  • Word count: 1402

222 and all that!

Extracts from this document...

Introduction

image00.pngimage01.png

222 and all that!

Maths Investigationimage02.png

image04.pngimage03.png

Sayan Dutta Chowdhury     9C

Contents

The Project………………………….Page 3

3 Digit Numbers…………………….Page 4

4 Digit Numbers…………………….Page 6

5 Digit Numbers…………………….Page 9

6 Digit Numbers............................Page 13

The Formula……………………….Page 14

Proving The Formula……………..Page 16

Bases………………………………Page 23

The Project

Everyone in year 9 was given the “222 and all that!” maths project to do. We had to write down all the three digit numbers that could be made by rearranging the digits 1, 2 and 7. The combinations were 127, 172, 217, 271, 712 and 721. Then by adding those digits, the answer came to 2220. Next we had to add the three digits, i.e. 1 + 2 + 3 = 10. In the end we divided 2220 by 10 giving us 222.

3 Digit Numbers

  1. Make all the combinations using the three   digits.
  2. Add the combinations together.
  3. Add the three numbers together.
  4. Divide ∑ the combinations divided by ∑ the three numbers and the answer will always be 222.

Examples

123 132 213 231 312 321          

Adding the above numbers gives 1332 and 1 + 2 + 3 = 6, so 1332 ÷ 6 = 222.

567 576 657 675 756 765

...read more.

Middle

5 Digit Numbers

  1. Make all the combinations using the five   digits.
  2. Add the combinations together.
  3. Add the five numbers together.
  4. Divide ∑ the combinations divided by ∑ the five numbers and the answer will always be 266664.

Examples

12345 12354 12435 12453 12534 12543 13245 13254 13425 13452 13524 13542 14235 14253 14325 14352 14523 14532 15234 15243 15324 15342 15423 15432 21345 21354 21435 21453 21534 21543 23145 23154 23415 23451 23514 23541 24135 24153 24315 24351 24513 24531 25134 25143 25314 25341 25413 25431 31245 31254 31425 31452 31524 31542 32145 32154 32415 32451 32514 32541 34125 34152 34215 34251 34512 34521 35124 35142 35214 35241 35412 35421 41235 41253 41325 41352 41523 41532 42135 42153 42315 42351 42513 42531 43125 43152 43215 43251 43512 43521 45123 45132 45213 45231 45312 45321 51234 51243 51324 51342 51423 51432 52134 52143 52314 52341 52413 52431 53124 53142 53214 53241 53412 53421 54123 54132 54213 54231 54312 54321

Adding the above numbers gives 2999960 and 1 + 2 + 3 + 4 + 5 = 15, so 2999960 ÷ 15 = 266664.

14678 14687 14768 14786 14867 14876 16478 16487 16748 16784 16847 16874 17468 17486 17648 17684 17846 17864 18467 18476 18647 18674 18746 18764 41678 41687 41768 41786 41867 41876 46178 46187 46718 46781 46817 46871 47168 47186 47618 47681 47816 47861 48167 48176 48617 48671 48716 48761 61478 61487 61748 61784 61847 61874 64178 64187 64718 64781 64817 64871 67148 67184 67418 67481 67814 67841 68147 68174 68417 68471 68714 68741 71468 71486 71648 71684 71846 71864 74168 74186 74618 74681 74816 74861 76148 76184 76418 76481 76814 76841 78146 78164 78416 78461 78614 78641 81467 81476 81647 81674 81746 81764 84167 84176 84617 84671 84716 84761 86147 86174 86417 86471 86714 86741 87146 87164 87416 87461 87614 87641

Adding the above numbers gives 6933264 and 1 + 4 + 6 + 7 + 8 = 26, so 6933264 ÷ 26 = 26664.

6 Digit Numbers

  1. Make all the combinations using the six   digits.
  2. Add the combinations together.
  3. Add the six numbers together.
  4. Divide ∑ the combinations divided by ∑ the five numbers and the answer will always be 13333320.

The Formula

(N-1)! x (10X – 1) ÷ 9

N = different amount of digits

X = number of digits

! = the factorial operation

N is how many different digits are there, e.g. 1, 2 or 3.

X is how many digits are there.

The exclamation mark the product of the integers from 1 to n, when the expression is n! From example 4!, read as four factorial is 4 x 3 x 2 x 1 = 24.

Proving The Formula

...read more.

Conclusion

112 + 121 + 211 = 444

1 + 1 + 2 = 4

444 ÷ 4 = 111

     If all the digits are the same, then the answer will be different again. If the digits are 1, 1 and 1, then N is replaced by 1 and X by 3.

(1-1)! x (103 -1) ÷ 9 = (0)! x (999) ÷ 9

                               = 1 x 999 ÷ 9

                               = 1 x 111

                               = 111

The long way is 111 + 111 + 111 = 333 and 1 + 1 + 1 = 3, so 333 ÷ 3 = 111.

0! Is defined as 1, which is a neutral element in multiplication, not multiplied by anything.

Bases

If you are given the number 13 in base ten, this means 1 ten and 3 ones. If you want to write it in base five, then it’s in number of fives and some number of ones. If the number was big enough, you might also need 25's, 125's and so on, just as in base ten you might need hundreds or thousands.

So here is 13:

     ___10___  1 1 1

    /        \

    oooooooooo o o o

    \________/ \___/

         1       3

        ten     ones

Let's group it by 5's rather than 10's:

     _5_   _5_  1 1 1

    /   \ /   \

    ooooo ooooo o o o

    \_________/ \___/

         2        3

       fives     ones

So we have 2 fives and 3 ones in base five, we write this as

23 (base 5).

...read more.

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