## Number Investigations

Investigations:

1.   Look at the table which shows the powers of the first few counting numbers. Look at the last digits in each row and column and write down what you can see. Are some digits missing? You might write as many as ten different observations.

 Number Square Cube 4th power 5th power 6th power 7th power 8th power 9th power 1 1 1 1 1 1 1 1 1 2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 2187 6561 19683 4 16 64 256 1024 4096 16384 65536 262144 5 25 125 625 3125 15625 78125 390625 1953125 6 36 216 1296 7776 46656 279936 1679616 10077696 7 49 343 2401 16807 117649 823543 5764801 40353607 8 64 512 4096 32768 262144 2097152 16777216 134217728 9 81 729 6561 59049 531441 4782969 43046721 387420489 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000

2  Draw a 10x10 table of the first 100 counting numbers. Using a ruler, cross out multiples of each prime number. The gaps left are the primes. You shouldn’t really need to cross out the multiples of eleven and thirteen, but you might check that you did cross them out.

Write down how many primes there are under 100. Use your table to predict how many there are between 100 and 110.

Write down the squares of eleven and thirteen.

I think 139 might be a prime.  Which numbers do you need to test as factors before you are sure it is prime? Why can I claim that you do not need to test whether 13 is a factor? Is there a rule you can (perhaps guess and) write down?

Use your answer to predict which numbers you would test as factors to show that 173 is a prime. I tested only one division on my calculator. I briefly considered only six possible factors; which numbers did I think about?

Determine whether each these is prime: 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047,

3  The factors of a number add to a total. Exclude from your total the number itself, but include  the number one. So the total for six is 1+2+3=6.    Six is said to be ‘perfect’. The total for 12 is 1+2+3+4+6=16 and because 16 > 12 (16 is bigger than 12) 12 is said to be super-perfect. 17 is prime so its factor total is one; it is ‘sub-perfect’ as well as being prime. Is 15 sub-perfect?

a)  make a table showing the factor total and classify the numbers as perfect, sub-perfect or super-perfect. You might use the first thirty numbers, and maybe Excel.

b)  Predict the case for these numbers and then test your answers

i) 48, ii) 63  iii) 28  iv) 56  v) 144 vi) 12vii) 16  viii) 496

c)  Perfect numbers are pretty rare; the questions of the previous exercise should have given you the first four. There is no 5-digit one. Find the 6-digit one and use your answer to predict the next one.  The last part of Investigation 2 may help. The rule was found first by Euler (him again: he discovered a lot of the more interesting maths we use up to GCSE).

1. repeating patterns of last digit in each row, especially 0,1,4,5,6
missing digits in columns (e.g. squares can’t end in 2,3,7,8)
all digits in cube, only 1 and 6 in 4th power
patterns repeat every 4th power increase, e.g. no missing digits in 7th power.

2.  2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97….
101, 103, 107, 109  i.e. 4 (1st time that happens since 20)
for 139 test 2,3,5,7,11.    139<12x13 so if factor existed it would be found before 13 needed.
173 needs 2,3,5,7,11 and 13 because 173>169. Obvious not even, not 0/5, 1+7+3=10 so not
factor of 3, test div by 7 but 7x25=175, 11x16=176, 173 too close to 169 for 13 to be factor.
prime: 3, 7, 31, 127, 2047 not prime 15 (3.5), 63 (3.3.7), 255 (3.5.17), 511 (7.73), 1023 (3.11.31).

the primes fit 2^p -1. It can’t be that easy: test to see if a subsequent is not prime.

3.  perfect numbers: 6, 28, 496, 8028… these fit (22p)(22p+1-1)   and/or (2p-1)(2p-1), though the first suggestion fails at the start. If p is a rpime number, this suggests that the next perfect is 1024.2047  or is 512.1023. 1023 is not prime and 2047 is; does this help? 511 is prime, so 256.511 might well be perfect – and invalidate my suggested formula, unless it is just that p is a power, not prime, with 2p-1 to be prime; can p be even once p>2?