### Counting Factors

Is there an efficient way to work out how many factors a large number has?

### Summing Consecutive Numbers

Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?

### Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

# One or Both

##### Age 11 to 14 Challenge Level:

Joseph's solution involved trial and improvement:

First I decided that the number of pupils in the class had to be a mulitple of 10 so that I could work out the percentages with no fractions of people.

I started with 20 pupils in the class. This meant that 14 passed one question (70%) and 12 the other (60%). As 9 pupils passed both questions this meant that 5 got only the first one right (14-9) and 3 the second one, making a total of 9 + 5 + 3 = 17 pupils. This is not enough as I started my working with 20 pupils.

Next I tried 30 pupils in the class. This time 21 passed one question (70% of 30) and 18 passed the other question (60%). Taking out the 9 pupils who passed both gave:
21-9 = 12 passing one question,
18-9 = 9 passing the other
9 passing both.
This makes a total of 30 (12+9+9), which is right

There were 30 pupils in the class.

***

This is Zi Heng's solution:

 70% + 60% = 130% 130% - 100% = 30% 30% = 9 pupils 100% = 9/30 * 100 = 30 pupils.

30 pupils took the exam.

Andrei, School 205, Bucharest, Romania solved this problem using a Venn diagram.

Let A be the set of solvers of the first problem, and B the set of solvers of the second problem and the number in set A be written $n(A)$ etc. Their intersection has 9 elements: $$n(A\cap B)=9$$ Their union contains all students. It is evident that: $$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$ If x is the number of students participating in the exam, then A has 70 per cent of x elements, B has 60 per cent of x elements, and relation (2) can be re-written as $x=0.7x+0.6x-9$ or $x=30$. So, 30 pupils came to the exam, 21 solved the first problem and 18 the second one.

Prateek , James , Alan , Jenny and Robert also sent in good solutions. Joseph's solution to the second part of the problem stated:

As all the pupils solved at least one problem 44% solved both and this is 11/25 in its lowest form.
72% - 44% = 28% = 7/25

This means that the number of pupils in the group must be a multiple of 25.

So, if 25 pupils took the exam14 solved one problem (7+7) and 11 both.
If 50 pupils took the exam 28 solved one problem and 22 both and so on...