# Missing number in a circle? [closed]

Here in this circle there are seven segments. 6 numbers have been written in 6 segments following a particular pattern. Find the number which will replace ??? in the seventh segment and also describe the pattern that you have derived. While I think MisterEman22's answer is better, I have an alternate solution.

The missing number should be 51.

How I arrived at this number,

Starting with the number 15, we add twice of the next number clockwise to get the number in the next sector. Then we choose that sector as the first sector and repeat. So,

15 + 2 x 3 = 21

21 + 2 x 6 = 33

And following the formula again gives
33 + 2 x 9 = 51

• Correct Answer. – prog_SAHIL Aug 23 '15 at 5:10

39

Explanation:

Starting with the part that has 3 in it, every other square clockwise is $3*n$ where $n$ is the next prime number. So:
$3*1=3$
$3*2=6$
$3*3=9$
$3*5=15$
$3*7=21$
$3*11=33$
$3*13=39$

• This seems to be the intended answer, but 1 is not a prime – Rohcana Aug 22 '15 at 3:19
• @Anachor Wow I completely overlooked that. Thanks for pointing it out – MisterEman22 Aug 22 '15 at 3:23
• Try Again.Good try though – prog_SAHIL Aug 22 '15 at 3:25
• Of course, there is always this – Rohcana Aug 22 '15 at 3:32
• You could just say n are non-composite natural numbers – CodeNewbie Aug 22 '15 at 4:52

Another possibility:

45

Explanation:

3 + 3 = 6
6 + 3 = 9
9 + 6 = 15
15 + 6 = 21
21 + 12 = 33
33 + 12 = 45

13

Because

15 + 3 + 21 + 6 + 33 + 9 + n = 100

This is likely not the true answer, but it's worth a shot.

I'm almost certain that this is the correct answer, because of the simple explanation:

39

For each pair of adjacent numbers, caluclate $\text{largest number} - 2*\text{smallest number}$

$15-2*3 = 9$, 9 is already in the circle
$21-2*3 = 15$, 15 is already in the circle
$21-2*6 = 9$, 9 is already in the circle
$33-2*6 = 21$, 21 is already in the circle
$33-2*9 = 15$, 15 is already in the circle

The only values for the missing number that follow this rule are: $3$, $33$, and $39$. The circle already contains $3$ and $33$, so we'll use $39$.