• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month   # 3 Digit Number - Maths Investigations

Extracts from this document...

Introduction

## 3 Digit Number

Take any 3 digit number, write down all possible numbers that can be made with the three digits, add them up, divide the total by the sum of the 3 digits. Investigate.

If All Digits Are Different : -

123                        456                        789

132                        465                        798

213                        546                        879

231                        564                        897

321                        654                        987

+312                          +645                          +978

1332÷ 24=2223330 ÷ 15=2225328 ÷ 24=222

147                        258                        369

174                        285                        396

471                        528                        639

417                        582                        693

741                        825                        936

+714 _                          +852                          +963

2664÷ 12=2223330 ÷ 15=2223996 ÷ 18=222

It seems that when all 3 digits are different the answer to the problem is 222.

Can I use Algebra to explain this?

abc=100a+10b+c

acb=100a+10c+b

bac=100b+10a+c

bca=100b+10c+a

cba=100c+10b+a

+cab=100c+10a+b

222a+222b+222c = 222(a+b+c)

a+b+c

=222

What if 2 of the 3 digits are the same?

If 2 digits are the same : -

223                        334                        566

322                        343                        656

+232                          +433                          +665

777÷ 7=1111110 ÷ 10=1111887 ÷ 17=111

224                        559                        772

242                        595                        727

+422                          +955                          +277

888÷ 8=1112109 ÷19=111        1776 ÷ 16=111

Middle

=111

### What if all 3 digits are the same number?

If All Digits Are The Same : -

333    = 37444    = 37666        = 37999        = 37777 _        = 37

9                 12                 18                   27                 21

It seems that when all the 3 digits are the same number the answer to the problem is 37.

Can I use Algebra to explain this?

aaa=100a+10a+a

=111a             = 37

3a

### What if

Conclusion

+cbaa=1000c+100b+10a+a

6666a+3333b+3333c

2a+b+c

=3333(2a+b+c) =3333

2a+b+c

What if 3 of the 4 digits are the same?

If 3 Of The 4 Digits Are The Same : -

4447

4474

4744

+7444

21109÷19=1111

### Is this the same answer every time 3 of the 4 digits are the same?

Lets usr Algebraic terms to find out.

aaab=1000a+100a+10a+b

aaba=1000a+100a+10b+a

abaa=1000a+100b+10a+a

baaa=1000b+100a+10a+a

3333a+1111b

3a+b

=1111(3a+b)        =1111

3a+b

What if all 4 digits are the same?

If All 4 Digits Are The Same : -

9999        =277.75

36

Is this the same answer every time all 4 digits are the same?

Let’s use Algebraic terms to find out.

aaaa=1000a+100a+10a+a

=1111a        =277.75

4a

What if I do the same problem but with 5 digits instead of 4 or 3?

From the data I have gathered maybe I can guess the outcomes for 5 digits.

## 5 Digit Number

If All 5 Digits Are Different : -

This student written piece of work is one of many that can be found in our GCSE Pythagorean Triples section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Pythagorean Triples essays

1. ## Investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean ...

= 4 4 = 4 My formula works for the first term, so, I will now check it on the 2nd term. 2 x 22 + 2 x 2 = 12 2 x 4 + 4 = 12 8 + 4 = 12 12 = 12 My formula also works for the 2nd term.

2. ## Investigating families of Pythagorean triples.

When factorised, it is: 4 ( 2n + 3 ) The second difference between the numbers in column b is 4, meaning the formula starts with: 4n2 By adding: 12n + 5 To the formula, b is proved correct: 4n2 + 12n + 5 Or ( 2n + 5 )( 2n + 1 )

1. ## Maths GCSE coursework: Beyond Pythagoras

18 + 6 = 24 It is correct! We will now look at the long side and look for a pattern: 5 13 25 41 61 85 8 12 16 20 24 1st difference 4 4 4 4 2nd difference The 2nd difference is 4 so we must find the formula for the 'largest number'.

2. ## Maths Number Patterns Investigation

Shortest Side = Length of Shortest Side. Let�s see if this works for a triangle that I already know. 72 = Middle number + Largest number 49 = Middle number + Largest number 49 = 24.5 2 Lower bound = 24, Upper bound = 25. Middle Side = 24, Largest Side =25.

1. ## Maths Investigation: Number of Sides

Shortest Side = Length of Shortest Side. Let�s see if this works for a triangle that I already know. 72 = Middle number + Largest number 49 = Middle number + Largest number 49 = 24.5 2 Lower bound = 24, Upper bound = 25. Middle Side = 24, Largest Side =25.

2. ## Beyond Pythagoras - Pythagorean Triples

32 50 R2 - R3 2 4 6 8 10 Difference 2 2 2 2 I have found the nth term formula to be 2n2 + 2n This can also be expressed in the form of triangle numbers because: (b) • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 