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Counting Factors

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

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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?

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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.

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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.

Venn diagram : A intersects with B.

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...