Continuing to explore four Consecutive Numbers
Problem
This problem follows on from, Starting to Explore Four Consecutive Numbers.
Take four consecutive numbers, $a$, $b$, $c$, $d$.
- Why can't $bd-ac$ be even?
- What is $bc-ad$ always equal to?
- Why must the sum $a+b+c+d$ have an odd factor?
- Why can't the sum $a+b+c+d$ be a multiple of $4$?
- Which consecutive numbers are such that $a+b+c+d$ divides exactly by $3$?
If you enjoyed working on this problem, you may now want to take a look at the follow-up problem, Is There More to Discover About Four Consecutive Numbers?.
With thanks to Don Steward, whose ideas formed the basis of this problem.
Getting Started
If the first number is $a$, can you write down the second number in terms of $a$? How about the third and fourth numbers?
Can you find an expression for the sum of the four consecutive numbers in terms of the first one?
What do all the numbers of the form $2k$ (where $k$ is an integer) have in common? What about numbers of the form $2k+1$?
Student Solutions
1. Why can't $bd-ac$ be even?
Ami from Garden International School in Malaysia and Ci Hui Minh Ngoc Ong from Kelvin Grove State College (Brisbane) in Australia thought about what happens when $a$ is even and when $a$ is odd. This is Ci Hui Minh Ngoc Ong's work:
Samuel from the UK, Eva, Vladimir, Arman, Ivan, Ivan, Ekaterina and Sama from Skolkovo in Russia, Yuk-Chiu from Harrow School in the UK and Rishik K used algebra. This is Rishik K's work:
$a,b,c,d$ are consecutive numbers so each number is $1$ more than the previous number.
So, we know that:
$b = a+1\\
c = a+2\\
d = a+3$
Now we can substitute $b, c, d$ using only one variable $a$.
So, the new [expression] is:
$((a+1)(a+3)) – ((a(a+2))$
When the inner brackets are expanded and simplified the equation is:
$(a^2 +4a+3) - (a^2 +2a)\\
a^22 +4a+3 - a^22 - 2a\\
2a+3$
So, we have found out that $bd – ac$ will always be equal to $2a+3$
I know that $(bd - ac)$ which is $(2a+3)$ will always be odd. If a number is multiplied by $2$ and $3$ is added it will always be odd. When any number, odd or even, is doubled it will always be even. After adding $3,$ an odd number is added to an even number so the answer must be odd as I know that even + odd = odd.
2. What is $bc-ad$ always equal to?
Samuel, Eva, Vladimir, Arman, Ivan, Ivan, Ekaterina and Sama, Ami, Yuk-Chiu and Rishik K used algebra. This is Samuel's work:
Ci Hui Minh Ngoc Ong also used algebra, but called the first number $n-1$:
3. Why must the sum $a+b+c+d$ have an odd factor?
Theekshi from NLCS in the UK used algebra and some examples to show that it usually does have an odd factor:
$a + b + c + d$ must always have an odd factor as $a + a+1 + a+2 + a+3 + a+4 = 4a+6$ meaning that we must see that when a multiple of $4$ is added to $6,$ if it will have an odd factor.
A chart investigating this:
Samuel, Eva, Vladimir, Arman, Ivan, Ivan, Ekaterina and Sama and Yuk Chui started with similar algebra to Theekshi, but used a different arguemt to explain why there is always an odd factor. This is Eva, Vladimir, Arman, Ivan, Ivan, Ekaterina and Sama's work:
Ci Hui Minh Ngoc Ong also used algebra, but called the first number $n-1$ again:
4. Why can't the sum $a+b+c+d$ be a multiple of $4$?
Theekshi used algebra and examples again:
The sum of $a, b, c$ and $d$ cannot equal a multiple of $4$ as the sum equals an [expression] of $4a+6.$ Meaning that, no matter what $a$ is equal to, $6$ will be added to it and as $6$ is not a multiple of $4,$ the sum also will not be a multiple of $4.$
Yuk-Chiu factorised the expression:
Eva, Vladimir, Arman, Ivan, Ivan, Ekaterina and Sama used a different argument:
Ci Hui Minh Ngoc Ong used a very similar method with different algebra:
5. Which consecutive numbers are such that $a+b+c+d$ divides exactly by $3$?
Samuel, Rishik K, Theekshi, Eva, Vladimir, Arman, Ivan, Ivan, Ekaterina and Sama and Ci Hui Minh Ngoc Ong made good progress on this question. Rishik K explained the answer:
We know that:
$a + b + c + d = 4a + 6$
So $4a + 6$ must be divisible by $3$ in this investigation. If we say $x$ is divisible $3,$ it is also the same as saying that $x$ is a multiple of $3.$ So, $4a + 6$ is a multiple of $3.$
$4a + 6 =$ a multiple of $3$
Now, let us delve into the expression $(4a + 6).$ $6$ is divisible by $3$ so we can ignore the $6$ as by adding the $6$ will not change anything.
$4a =$ a multiple of $3$
$a =$ (a multiple of 3) $\div 4$
So, the multiple of $3$ must also be a multiple of $12$ if it will be divided by $4.$
If we use the first multiple of $12,$ which is $12,$ then:
$a = 12 \div 4$
$a = 3$
Double checking:
If $a = 3$ then $b = 4, c = 5, d = 6.$
The total of $3, 4, 5, 6$ is $18$ and that is a multiple of $3.$
There is an infinite number of sequences of four consecutive numbers which have a sum that is a multiple of $3.$ If we continue using $a$ = (a multiple of 3) $\div 4$ and we substitute the (multiple of 3) part with different multiples of 12, we will realise that a can be any multiple of 3. So, we can have a sequence of four consecutive numbers which have a sum that is a multiple of 3 if our first consecutive number is a multiple of 3.
Other Examples include:
$6 + 7 + 8 + 9 = 30$
$321 + 322 + 323 + 324 = 1290$
$1002 + 1003 + 1004 + 1005 = 4014$
Yuk-Chiu used factorisation for a very neat result:
Teachers' Resources
This is the second problem in a set of three. The other two are:
Starting to Explore Four Consecutive Numbers and
Is There More to Discover About Four Consecutive Numbers?
This set of problems offers students the opportunity to explore, conjecture and use algebra to justify their findings.
This problem featured in the NRICH Secondary webinar in June 2022.