# 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 at 23:57
• Congratulations – Deepthinker101 Sep 30 at 1:30
• This was the mathematical way to do what I described in words. Congratulations! – El-Guest Sep 30 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). – Matthew Jensen Sep 30 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 at 19:57
• Yes, it looks like you’re right — this fits better I think. Thanks for the heads up, @Bass – El-Guest Sep 29 at 21:20
• No not the right answer – Deepthinker101 Sep 29 at 23:20
• As it turns out, my process IS right, I’m just a dumdum. – El-Guest Sep 30 at 1:36
• Congrats on having the right process, but too bad for the miscount. – Matthew Jensen Sep 30 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