# The mathematical quiz

There are 4 equations. What should I put in place of the ? in the last one?

1 + 4 = 5
5 + 3 = 11
7 * 7 = 61
10 - 1 = ?

• Is this an original question? If not, you should put the source in your puzzle text just to give due credit to its original author. – Xenocacia Mar 14 '18 at 6:54
• My mother-in-law is author. How should I credit her? – asiniy Mar 14 '18 at 7:01
• @asiniy Are you sure the second equation is not 5+3=10? – rhsquared Mar 14 '18 at 7:34
• @asiniy You already did. Just by stating who's the author is crediting him/her. – Paul Karam Mar 14 '18 at 7:49
• Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful. – Rubio Mar 17 '18 at 1:42

It looks to me like a simple octal calculation (i.e. base-8), assuming that there was an error on the second line. If this is the case then the answer is 7.

• so you mean calculate in decimal then convert to Octal/Base 8? – ALinuxLover Mar 14 '18 at 10:59
• @ALinuxLover No, I mean all the numbers are Base 8. – rhsquared Mar 14 '18 at 11:31
• oh ok makes more sense heh – ALinuxLover Mar 14 '18 at 11:50

I guess the answer provided by @rhsquared is correct provided the OP corrects the question. But a more far fetched answer by assumption is

9 = 10-1 = 49 - (8*5) - From Normal Maths 10-1 = (7*7) - [(1+4)*(5+3)] - Rearrangement 10-1 = 61 - [5*8] -Substitution from above rules 10-1 = 61 - 55= 6. -Substitution from above rules & From Normal Maths

I think that one possible answer to the equations can be:

(Given data) + (3 * n * n) in which the n represents the iteration number, starting by 0.
(1 + 4) + (3 * 0 * 0) = 5 + 0 = 5 -> Iteration 0
(5 + 3) + (3 * 1 * 1) = 8 + 3 = 11 -> Iteration 1
(7 * 7) + (3 * 2 * 2) = 49 + 12 = 61 -> Iteration 2
(10 - 1) + (3 * 3 * 3) = 9 + 27 = 36 -> Iteration 3

One possible answer using the previous answer and 2 to the power of the line index:

1 + 4 = 5
5 + 3 = 11 = 8 + (5-2^1)
7 * 7 = 61 = 49 + (11+5-2^2)
10 - 1 = 78 = 9 + (61+11+5-2^3)

• But what if the last sum is alternating, where $10 - 1 = 38 = 9 + (61 - 11 + 5 - 2^3)\,?$ – Mr Pie Mar 21 '18 at 5:49
• @user477343 an infinite number of solutions exist, and therefore an infinite number of "what ifs". The simplest solutions are preferred. – SwiftPanda Mar 21 '18 at 19:46
• Yeah, I guess you might be right then. Perhaps the pattern was subtracting by $2^n$ instead of $2^n$ being the last summand of an alternating sum. My answer above was an entirely new pattern that I found, so maybe there are infinitely many patterns... – Mr Pie Mar 21 '18 at 20:50

$10 - 1 = 23$
The first prime after $(1 + 4) - 1 = 4$ is $5$. Notice that first corresponds to $1 = 2^0$. $\quad\qquad$ The second prime after $(5 + 3) - 3 = 5$ is $11$. Notice that second corresponds to $2 = 2^1$. $\quad$ The fourth prime after $(7\times 7) - 5 = 44$ is $61$. Notice that fourth corresponds to $4 = 2^2$.
We look at the fourth expression, namely ($10 - 1$), and then we subtract this by the next odd number from $5$, namely $7$, to make $$(10 - 1) - 7 = 2.$$ And now since $2^3 = 8$, we find the $8^\text{th}$ prime number after $2$, which is $23$. $$\therefore 10 - 1 = 23$$