You may also like

problem icon

Upsetting Pitagoras

Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2

problem icon

Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?

problem icon

Euler's Squares

Euler found four whole numbers such that the sum of any two of the numbers is a perfect square. Three of the numbers that he found are a = 18530, b=65570, c=45986. Find the fourth number, x. You could do this by trial and error, and a spreadsheet would be a good tool for such work. Write down a+x = P^2, b+x = Q^2, c+x = R^2, and then focus on Q^2-R^2=b-c which is known. Moreover you know that Q > sqrtb and R > sqrtc . Use this to show that Q-R is less than or equal to 41 . Use a spreadsheet to calculate values of Q+R , Q and x for values of Q-R from 1 to 41 , and hence to find the value of x for which a+x is a perfect square.

Phew I'm Factored

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Sue from Madras College first explored the factors of the numbers which are written as 10101 in different number bases, then proved that the numbers 10201, 11011 and 10101 are composite in any base, then she also proved the results that she had discovered. Here is her solution.

1. Written in base 10 we notice that the larger factor of the previous base becomes the smaller factor of the next base.

10101 2 = 7 x 3
10101 3 = 13 x 7
10101 4 = 21 x 13
10101 5 = 31 x 21
10101 6 = 43 x 31
10101 7 = 57 x 43
10101 8 = 73 x 57
10101 9 = 91 x 73
10101 = 111 x 91.

2. Here we see that one of the factors is always 111 (written in different bases) and the other factor has digits (the base number - 1) followed by 1.

In base 2: 10101 = 111 x 11
In base 3: 10101 = 111 x 21
In base 4: 10101 = 111 x 31
In base 5: 10101 = 111 x 41
In base 6: 10101 = 111 x 51
In base 7: 10101 = 111 x 61
In base 8: 10101 = 111 x 71
In base 9: 10101 = 111 x 81
In base 10: 10101 = 111 x 91.

3. Written in base 10, the factors form a number series 7, 13, 21, 31, 43, 57, 73, ...

4. The difference between the two factors is always twice the base number:

7-3 = 4 = 2x2
13-7 = 6 = 2x3
21-13 = 8 = 2x4 etc.

10201 is composite in any base.

Proof The number 10201 in base $a$ (for $a \geq3$) would be


10201 a = 1 + 2 a 2 + a 4
= (1 + a 2 ) 2 .

So the number 10201 in any base a would always be the square number 101 2 and hence it is always composite.

11011 is composite in any base.
Proof The number 11011 can be written in base a as:

11011 a = 1+ a + a 3 + a 4
= (1 + a 3 )+( a + a 4 )
= (1 + a 3 )+ a (1 + a 3 )
= (1 + a )(1 + a 3 )
= (1+ a ) 2 (1- a + a 2 ).

So 11011 always has the factors (1 + a ) 2 and (1 - a + a 2 ) so it is always composite.

10101 is composite in any base.
Proof To prove 10101 is composite in any base we must factorise it.

l0l0l a = a 4 + a 2 +1
= ( a 2 +1) 2 - a 2
= [( a 2 +1)+ a ][( a 2 +1)- a ].

10101 a therefore has the factors [ a 2 + a + 1] and [ a 2 - a + 1] and hence it is composite for all values of a .

Now we can explain what we found before:

1. In base ( a + 1) the smaller factor would be [( a + 1) 2 - ( a + 1) + 1] and this is equal to [ a 2 + a + 1] which is the larger factor in base a .

2. If 10101 a = [ a 2 + a + 1] [ a 2 - a + 1] it is clear that
[ a 2 + a + 1] = 111 a and
[ a 2 - a + 1] = a ( a - 1) + 1 = ( a - 1) l a

3. The series 3, 7, 13, 21, 31, ... has the general formula n 2 + n + 1 ( n = base number).

4. If the two factors in base a are [ a 2 + a + 1] and [ a 2 - a + 1] then their difference would be 2 a which is what we observed.