Chaos, Challenge, Logic, Order Quiz

Answer: 7

In the interesting info sections you will find a step-by-step detail of the solution which utilizes the crevice purposely left within the questions and hints given.

In order to solve, you have to be able to think in terms of the J(Q) which is a composite term involving both the question and its answer while keeping in mind the uniqueness requirement on the answers. This is not easy!

In the question bodies and hints, I did my best to encourage you not to lose grip on this notion.

Answer: 2

Let's start with the key to the solution of the puzzle.

Look at the following:

1. There are exactly 4 values of J(Q) which are prime numbers (given in Q7).
2. The answers are unique (given in the intro and in Q10).

It stems from the above two facts that the 4 prime numbers MUST populate the answers 1,3,7,9.

There is no other option, as all prime numbers except for 2 are odd (2 itself is irrelevant, since by definition J(Q) has at least two digits).

Answer: 6

Armed with the above notion, give a second look to Q8 which states that there are exactly 2 values of J(Q) which are perfect squares.

The initial options are: 16,25,36,49,64 and 81.

Now, 49 and 81 are eliminated since they are not prime numbers.

25 is eliminated by the hint in Q2 which says that J(2) is even.

We are left to choose a pair out of 16, 36 and 64.

There are three pairs which should be considered: (16,36), (16,64) and (36,64).

Answer: 5

The pair (16,36) is disallowed because the answer 6 must be unique.

The pair (16,64) is also illegal: Q2 states that J(1)+J(2)+J(6)=J(10). J(10) is given as odd (by the hint in Q10). But J(1)+J(2)+J(6)=16+J(2)+64 would yield an even sum as J(2) is an even number (by the hint in Q2). This contradiction eliminates the pair (16,64).

Thus, the only possibility for a pair of perfect squares is 36 and 64.

Answer: 9

We already have: J(3)=36. J(6)=64.

The next step is to utilize the information given in Q6 and look for the 2 prime values of J(Q) which are mirror images of one another.

Find out which prime numbers are left relevant:

J(1) is greater than 13 (by the hint in Q1).
J(2) is not a prime number (by the hint in Q2).
J(3) is given as 36.
J(6) is given as 64.
J(7) is smaller than 77 (by the hint in Q7).
J(8) is not a prime number (by the hint in Q8).
J(9) is not a prime number (by the hint in Q9).

The above leaves the following prime numbers as valid options:

17,19,41,43,47,53,59,71,73,101,103,107,109.

And the only mirror image pair is 17 and 71.

Answer: 4

We already have: J(1)=17. J(3)=36. J(6)=64. J(7)=71.

At this stage, we have J(1) and J(6) in hand. Time to make use of Q2 stating that J(1)+J(2)+J(6)=J(10).

J(2)=24 is ruled out because we already have J(3)=36 as the only number where the answer equals double the question number (given in Q5).
J(2)=26 is ruled out as the answer 6 is already used by J(3)=36.
J(1)+J(2)+J(6)=J(10) means J(2) cannot be 210.

We are left with two options:

J(1)=17. J(2)=22. J(3)=64. J(10)=J(1)+J(2)+J(6)=103.
or
J(1)=17. J(2)=28. J(3)=64. J(10)=J(1)+J(2)+J(6)=109.

Answer: 1

We already have: J(1)=17. J(3)=36. J(6)=64. J(7)=71.

Let's explore the pair J(2)=28 and J(10)=109.

At this stage, the 4 remaining values of J(Q) can be easily derived:

Using Q4 and its hint, the only valid option left where the answer equals the question number is J(5)=55.

This leaves J(4)=43 as the only option for the fourth prime number, J(10)=109 being the third.

We assume that J(2)=28 as the only J(Q) evenly divisible by 7 (given in Q3) so J(9)=910 is not allowed (since 910/7=130). This necessitates J(8)=810.

The last remaining value to populate is J(9)=92.

We now have a set of answers which is consistent with all the puzzle's conditions and constraints, except Q9:

J(1)=17. J(2)=28. J(3)=36. J(4)=43. J(5)=55. J(6)=64. J(7)=71. J(8)=810. J(9)=92. J(10)=109.

Try to verify this set against Q9:

The sum of squares of differences between each question and answer above is: 36+36+9+1+0+4+36+4+49+1=176.

Since it is given that this sum should be 156 - this is not the correct solution.

By this we have established that J(2)=22 and J(10)=103.

Answer: 10

We already have: J(1)=17. J(2)=22. J(3)=36. J(6)=64. J(7)=71. J(10)=103.

At this stage, the 4 remaining values of J(Q) can be easily derived:

The only option remaining for a number evenly divisible by 7 is J(9)=98.

The last option remaining for a fourth prime number is J(5)=59, J(10)=103 being the third.

Using the hint in Q8, only J(8)=810 is available to satisfy an even J(8).

The last value to populate is J(4)=45.

This should be the solving set:

J(1)=17. J(2)=22. J(3)=36. J(4)=45. J(5)=59. J(6)=64. J(7)=71. J(8)=810. J(9)=98. J(10)=103.

Try to verify this set against Q9:

The sum of squares of differences between question and answer is: 36+0+9+1+16+4+36+4+1+49=156.

Bingo!

Answer: 8

We finally have:

J(1)=17. The answer to question 1 is 7.
J(2)=22. The answer to question 2 is 2.
J(3)=36. The answer to question 3 is 6.
J(4)=45. The answer to question 4 is 5.
J(5)=59. The answer to question 5 is 9.
J(6)=64. The answer to question 6 is 4.
J(7)=71. The answer to question 7 is 1.
J(8)=810. The answer to question 8 is 10.
J(9)=98. The answer to question 9 is 8.
J(10)=103. The answer to question 10 is 3.

In full accordance with the information brought in the question bodies and hints.

Answer: 3

The question body above reminded you to abandon the terminology of J(Q) while filling in the answers. I truly hope you paid attention to it. Thanks for your time. If you found a better way of solving the puzzle, please drop me a line or two.