The value increases by 3 when you go to the next number.

First number is 12.
Last is 99.

99 - 12 = 87 (total increase from first number to last)

Number of increases = 87 ÷ 3 = 29

1st number → increase 29 times → 30th
number
(1 + 29 = 30)

So there are 30 numbers altogether.


Next, pair them up like this :


12 + 99 = 111
15 + 96 = 111
18 + 93 = 111
21 + 89 = 111
...
...
54 + 57 = 111



There are 30 numbers , so 15 pairs can be obtained. (30 ÷ 2 = 15)
Each pair has a total of 111.

Sum = 15 × 111 = 1665

Alternatively, do this :

12 + 15 + 18 + ... + 93 + 96 + 99

= 3 × (4 + 5 + 6 + ... 31 + 32 + 33)

Do the same pairing method.

Sum of 1 pair = 4 + 33 = 37

There are 15 pairs, so their sum = 37 × 15
= 555


Then, 3 × (4 + 5 + 6 + ... 31 + 32 + 33)
= 3 × 555
= 1665

For b) ,

It is easier to count those numbers that contain 3. Then subtract from the total.

Numbers that contain 3 :


30 33 36 39 63 93
(Total 6)


So, 30 - 6 = 24

24 numbers don't contain 3

Thank you so much !!!

A question : numbers containing 3 are the following
13 23 30 33 43 53 63 73 83 and 93 right ? So 9 no ?

Welcome .


For b) , there is an observation

The last digits of the numbers repeat every 10 numbers. (3 6 9 2 5 8 1 4 7 0)


3 6 9 12 15 18 21 24 27 30
There are two numbers with '3'

For the subsequent sets of 10 numbers,

33 36 39 42 45 48 51 54 57 60
This line has three numbers with '3'

63 66 69 72 75 78 81 84 87 90
Only one number has '3'

93 96 99 102 105 108 111 114 117 120
Also only one number with '3'

So most lines have only 1 number with 3 except those like the first line and others like :

123 126 129 133 136 139 142 145 147 150

333 336 339 342 345 348 351 354 357 360

You'll just need to be careful for such lines to not miss out

Oh yes !

Answer to your question : nope.

Your sequence does not contain many of those numbers.


Remember your sequence goes like this :


12 15 18 21 24 27 30 33 36....


There is no 13,23,43,53,73,83. These are not multiples of 3.