Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### Advanced mathematics

### For younger learners

# Continuing to Explore Four Consecutive Numbers

Or search by topic

Age 14 to 16

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

**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: