Afreen says, "I am thinking of 3 consecutive numbers. The first is a multiple of 4, the second is a multiple of 5 and the third is a multiple of 6." What could the numbers be? Can you find 3 possible sets of numbers?

The 3 ses that we found are:

{4, 5, 6}

{64, 65, 66}

{124, 125, 126}

Step-by-step explanation:

3 consecutive numbers are written as:

n, (n + 1), (n + 2)

We know that the first one is a multiple of 4.

Then n is a multiple of 4:

n = k*4

where k is an integer.

The second one is a multiple of 5, then:

(n + 1) = j*5

where j is an integer.

The last one is a multiple of 6.

(n + 2) = p*6

where p is an integer.

We want to find 3 sets.

The first one is trivial:

n = 4

n + 1 = 5

n + 2 = 6

4 is a multiple of 4

5 is a multiple of 5

6 is a multiple of 6.

Now let's find a set that is not trivial.

First, remember that all the multiples of 5 end with a 0 or a 5,

Then we can look for a value n, that is multiple of 4, and that has a last units digit equal to 4 (then the next one will have a units digit and will be multiple of 5)

For example, 4*6 = 24

24 is a multiple of 4.

The next number is 24 + 1 = 25, which is multiple of 5.

The next number is 25 + 1 = 26, which is not multiple of 6.

So let's try again.

4*11 = 44, is a multiple of 4.

The next number is 44 + 1 = 45

45 is a multiple of 5.

The next number is 45 + 1 = 46, which is not multiple of 6

So we need to try with another set.

4*16 = 64, is a multiple of 4.

The next number is 64 + 1 = 65, which is multiple of 5.

The next number is 66, which is multiple of 6:

6*11 = 66

Then the set:

n = 64

n + 1 = 65

n + 2 = 66

Is a possible set.

Now we can keep trying this, we can see that the next set  has the numbers:

n = 4*31  = 124 is a multiple of 4.

The next number, is 124 + 1 = 125, which is a multiple of 5.

The next number is 125 + 1 = 126, which is multiple of 6

6*21 = 126

Then the set:

n = 124

n + 1 = 125

n + 2 = 126

Is another possible set.

The 3 ses that we found are:

{4, 5, 6}

{64, 65, 66}

{124, 125, 126}

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺