1. Smallest (First) or Largest (Last) number in the set

2. The increment

3. The number of items in the set

1. Arithmetic Mean (Ave.) = Median ... you can find out the ave. by figuring out the Median (i.e. MIDDLE number)

2. Mean & Median = (First + Last terms) / 2... i.e. the average of the First and Last terms

3. Sum(Elements in Set) = Ave. x #Elements

n!

According to the Factor Foundation Rule, every number is divisible by all the factors of its factors. The product of any set of n consecutive integers is divisible by n.

e.g. the product of 3 consecutive integers will always be a multiple of 3, and a multiple of 2, and a multiple of 1, therefore 3 x 2 x 1 = 6

e.g. the product of 5 consecutive integers will always be a multiple of 5 (as one of those integers will always be a multiple of 5), and a multiple of 4 (i.e. two 2's), and a multiple of 3, and a multiple of 2, therefore 5 x 4 x 3 x 2 x 1 = 6

k!

e.g. 4x3x2x1

Because sum = ave. x # items... the average for an ODD # items is an integer, so the SUM is a MULTIPLE of the number of items.

e.g. The average of {13,14,15,16,17} is 15, so 15 x 5 = 13 + 14 + 15 +16 + 17

i.e. 13 + 14 + 15 + 16 + 17 = 15 x 5

For a set of consecutive integers with an EVEN number of items, the sum of ALL the integers is NEVER a multiple of the number of items... Why is this so?

Because the sum equals the average x the # items... the average for an EVEN # items is NEVER an integer, so the SUM of all the items is NEVER a MULTIPLE of the number of items.

the average of {8,9,10,11} is 9.5, so 9.5 x 4 = 8 + 9 + 10 + 11... That is, 8 + 9 + 10 + 11 is NOT a multiple of 4...

4 x 9.5 = 38

x + (x+1) + (x+2) + (x+3) = 38...

4x + 6 = 38

4x = 32

x = 8...

Therefore, the set is {8,9,10,11}

The average of a set of consecutive integers with 5 elements is 15 . What is the set?

5 x 15 = 75

x + (x+1) + (x+2) + (x+3) + (x+4) = 75...

5x + 10 = 75

5x = 65

x = 13 ...

Therefore, the set is {13,14,15,16}

The average of a set of 2 consecutive integers is EVEN...

TRUE or FALSE?

FALSE.

The average of a set of 2 consecutive integers is NEVER an integer because you are averaging an ODD and an EVEN i.e. (ODD+EVEN)/2 (...2 is EVEN) --> (O + E) / E  = O / E = NON-INT so it can't possibly be EVEN.

The average of a set of 2 consecutive integers is ODD...

TRUE or FALSE?

FALSE.

The average of a set of 2 consecutive integers is NEVER an integer because you are averaging an ODD and an EVEN i.e. (ODD+EVEN)/2 (...2 is EVEN) --> (O + E) / E  = O / E = NON-INT so it can't possibly be ODD.

4! = 4x3x2x1 = 24

The product of any set of 4 consecutive integers will have at least one multiple of 4, one multiple of 3, and an even number (a multiple of 2), and of course the product is also a multiple of 1.

According to the Factor Foundation Rule, every number is divisible by all the factors of its factors... so --> FILL THIS IN 

What does this tell us about k ? -->

The sum of k consecutive integers is divisible by k

* The sum divided by k results in an integer. 

* The sum divided by k is also the average.

So, the average is an integer... Therefore k MUST be ODD.

What is the relationship between evenly spaced sets, consecutive multiples and consecutive integers?

For an evenly-spaced set to be fully defined, what 3 parameters are required ?

1.  The smallest (first) OR largest (last) # in the set

2. The increment (always 1 for consecutive integers)

3. The number of items in the set

What is an evenly-spaced set?

A set of numbers whose values go up or down by the same amount (the increment) from one item in the sequence to the next

The product of k consecutive integers is always divisible by?

Why? (use the example of 4! ...)

k!

e.g. The product of any set of 4 consecutive integers will be divisible by 4! = 4 x 3 x 2 x 1 = 24, since that set will always contain one multiple of 4, one multiple of 3, and another even number (a multiple of 2).

In an evenly-spaced set, what do you get when you subtract the average from the median, and why?

0, because in evenly-spaced sets the average and the median will always be the same

Are the consecutive multiples of an integer evenly spaced?

Yes

What's an easy way of finding the average of an evenly-spaced set?

Simply find the average of the first and last terms in the set

The average of an ODD number of consecutive integers will always be an integer... Why?

Because the median('middle number') will always be an integer.

a) Consecutive integer Set A has an integer mean... Does this set have an EVEN or ODD number of elements? Why?

b) Consecutive integer Set B has an integer mean, + 1/2.     Does this set have an EVEN or ODD number of

    elements? Why?

a) ODD. If the number of elements is odd, the set of consecutive integers will have an integer mean. This is because there is only one 'middle term' in a set with an odd number of elements.

b) EVEN. If the number of elements is EVEN, the set of consecutive integers will have an integer mean + 1/2. This is because there is are 2 'middle terms' in a set with an even number of elements, and thus the median is the average of these 2 middle terms, which is a number/2 therefore resulting in an integer, + 1/2

any set of consecutive integers must have either an integer mean (if the number of integers is odd) or a mean that is an integer + 1/2 (if the number of integers is even). 

Because there is no true 'middle number'/median

Is x(x + 1)(x + 2) divisible by 4?

Use prime boxes (put them next to each other for each integer) to keep track of factors of consecutive integers...

x  |  (x + 1) | (x + 2)

------------------------ 

2  |             |     2    

A set of 3 consecutive integers must always contain ...?

One multiple of 3. Therefore the product of the set of consecutive integers is divisible by 3.

How many multiples of 7 between 100 and 150?

First integer in set = 105

Last integer in set = 147

Therefore, (147 - 105)/7 + 1 = 6 + 1 = 7 multiples of 7 between 100 and 150.

Note: the answer is not (150 - 100)/7 + 1 = 49/7 + 1 (i.e. 8) because 100 and 150 are not multiples of 7 therefore can't be included in the set in question.

What's the sum of all the integers from 20 to 100, inclusive?

1.) Average the first and last term to find the median of the set (which equals the average) = (100 + 20)/2 = 60

2) Count the number of terms ( 100 - 20 + 1 = 81)

3. Sum = Ave. x Number of terms = 60 x 81 = 4860

Answer = 4860

The SUM of n consecutive integers is divisible by n. What does this tell us about n, and why?

It means that n is ODD.  This is because the sum of n consecutive integers divided by n is the average/mean of that set of integers. Because the average is itself an integer, n can only be odd.

This is because the average of an odd number of consecutive integers will always be an integer, because the median/ave. (i.e. "middle number") will be a single integer.