# Find the number for the question mark

What is the number for the question mark? (My original puzzle)

3 $$\rightarrow$$ 39

1 $$\rightarrow$$ 3

4 $$\rightarrow$$ 84

5 $$\rightarrow$$ 155

2 $$\rightarrow$$ 14

10 $$\rightarrow\ \large?$$

By noting that each number maps to a multiple of itself:
$$3\rightarrow3*13$$
$$1\rightarrow1*3$$
$$4\rightarrow4*21$$
$$5\rightarrow5*31$$
$$2\rightarrow2*7$$
And that the quotient increases quadratically
$$1:3$$
$$2:7=3+4$$
$$3:13=7+6$$
$$4:21=13+8$$
$$5:31=21+10$$
The mapping must be a cubic function.
The above sequence is equal to $$n^2+n+1$$, where n is the input.
The original sequence is multiplied by $$n$$ again. So the function can be described by $$n^3+n^2+n$$.
So the final answer is $$10^3+10^2+10=1110$$.

• It looks like this is it!
– Duck
Sep 29 '19 at 23:57
• Congratulations Sep 30 '19 at 1:30
• This was the mathematical way to do what I described in words. Congratulations! Sep 30 '19 at 1:37
• @El-Guest Thanks, I saw that I started similar to you, but I didn't realise it was the same since you got a different result (but now I know you just miscounted). Sep 30 '19 at 3:14

Note that

1 x 3 = 3 (3 is the 2nd odd positive number)
2 x 7 = 14 (7 is the 4th positive odd number)
3 x 13 = 39 (13 is the 7th positive odd number)
4 x 21 = 84 (21 is the 11th positive odd number)
5 x 31 = 155 (31 is the 16th positive odd number)

Note that

The nth odd number is being used, where n forms a pattern of 2, 4, 7, 11, 16 — ie. differences increase by one. Then to continue this pattern, we have 22, 29, 37, 46, 56 — this sequence is the triangular numbers plus 1. (Note the triangular Numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55...)

So then the answer ought to be

10 = 1110, as 111 is the 56th positive odd number and 10 x 111 = 1110.

• You are probably on the right track, but maybe insisting on primes isn't necessary?
– Bass
Sep 29 '19 at 19:57
• Yes, it looks like you’re right — this fits better I think. Thanks for the heads up, @Bass Sep 29 '19 at 21:20
• No not the right answer Sep 29 '19 at 23:20
• As it turns out, my process IS right, I’m just a dumdum. Sep 30 '19 at 1:36
• Congrats on having the right process, but too bad for the miscount. Sep 30 '19 at 3:15

I came to the same conclusion as the others through a different method maybe, sorry if this is exactly what someone else said or I use some funky syntax:

I noted that

3/1 = 3
14/2 = 7
39/3 = 13
84/4 = 21
155/5 = 31
The difference between the RHS of each group increases by 2 each time.

Therefore:

Extrapolating the differences (up to 10) we get:
12, 14, 16, 18 and 20 Adding the differences to the RHs of the first set of equations starting from the last one (31), results in 111