Logical Maths is not the only maths

The statements of A, B, C and D can be summarized as in the table below.

Look at the left table. First let us assume that statement D1: The number is less than 20 $(n < 20)$ is true (green). Then A1: $n < 12$ and A3: $n < 14$ are true and A2: $7 \not{|} \space n$ must be false (red), so that at least one of the three statements of A is false. B1: $n > 83$ and B3: $n > 100$ are false and B2: $10 | n$ must be true, so that at least one of the three statements of B is true. But there is no $n < 12$ which is divisible by $7$ and $10$ . Therefore, the assumption of D1 is true is wrong that is D1 is false.

Now look at the right table. Since D1 is false then A1 and A2 are false and A2: $7 \not{|} \space n$ is true. This means that D3: $7 | n$ is false. That means D2: $n$ is a prime is true. Then C1: $4 | n$ and C3: $9 | n$ are false and C2: $n < 90$ is true. B2: $10 | n$ and B3: $n > 100$ are false and B1: $n > 83$ is true. This means that $n$ is a prime larger than $83$ and less than $90$ which is $\boxed{89}$ .

Sorry that I don't understand how to write solutions properly using LaTex. So here is the only way I could describe how I found solution for this case.

I'll try to describe this picture into words.

Generally all 12 of Andy-Becky-Chloe-Danny's statements consist of two main conditions:

1. How big the number is

2. Divisibility rules that apply to the number

Based on such conditions, I created a horizontal line which consists of six numbers. 12 derived from A1 statement 14 derived from A3 statement 20 derived from D1 statement 84 derived from B1 statement 91 derived from C2 statement and 100 derived from B3 statement

So there I was, a horizontal lines with 12, 14, 20, 84, 91, and 100 as the "limits of analysis".

I went back to Andy's other statements and Danny's other statements. Seems apparent that both of them are contradictory about divisibility of the number by 7. So based on them, the number should be prime (D2 is true). If D2 is true, we don't need all other Danny's statements to be true. And we also see that Andy's statements "do not require the number to be anything but multiply of 7". So Andy's statements and Danny's were OK. But from here afterwards, I used D2 statement as my key statement to analyze all other possibilities.

Then I went to the girls. They provided information about 84, 91, and 100. If the number is less than 84 (B1 is false), then this number must be able to be divided by 10. Based on D2, we know this is impossible. Therefore I concluded that this number must not be lower than 80. It is higher than 84 and it is a prime.

Then, I checked Chloe's statements. As long as the number is less than 91 (C2 is true), any number may be correct as long as it is not 36 or 92 (C1 and C3 must be wrong). But if it is higher than 91, then it is impossible to be prime (remember our D2 as cornerstone statement here).

And I come to conclusion that this number is higher than 84, it is a prime, and it is less than 91.

My final answer: The only number to fulfill all criteria is 89.

If A1 is true, A2 will be definitely false because A3 is true as A1 is true. It is also possible that only A3 and A2 are true. Under this assumption, the only possible value are 7(for A1 and A3 true), 12 and 13(for A2 and A3 true ). Put this three values to a test for other people. It is clearly that three values fails on B1,2,3. Meaning, only A2 is true because it allows higher positive integer. As A2 true , D3 will be false. D1 is also a false statement because 15,16,17,18,19 are definitely false for B1,2,3. Therefore, only D2 is true. As D2 is true, B2,C1,C3 are all false under the definition of prime number. Since only C2 is true among C1,2,3, we can conclude that B3 is false. In the end, we just need to fine a prime number that fulfilling B1 and C2, which is 89.

We can observe that $A2$ and $D3$ are contrast, so to make the problem easier, we first focus on these two statements. $x$ is the number we need to find. If $A2$ is false then $x\vdots7$ So $A1$ and $A3$ can be both true or one and only one of them is false. Assume $A1$ is true then $A3$ is also true. We have then $x<12$ and $x\vdots7$ , so $x=7$ , therefore $B1, B2, B3$ are all false. It can't be the answer. $A1$ is false, so $A3$ is true. Then we have $x=12$ or $x=13$ , then $B1, B2, B3$ are all false too. After all of this we can see that $A2$ is true, so $7$ does not divide $x$ Then $A1$ and $A3$ can be both false or one and only one of them is true. If $A1$ and $A3$ are all false we have $x>14$ ( $x$ can't be $14$ because $7$ does not divide $x$ , and otherwise we have $x=12$ or $x=13$ . We eliminate the latter because if $x=12$ or $x=13$ , then $B1, B2, B3$ are all false too. This time, we come to D's statements. Because $D3$ is false, $D1$ and $D2$ can be both true or one and only one of them is false. If both of them are true we have $14<x<20$ and $x$ is a prime, so $x=17$ or $x=19$ . Then, all B's statements are false. If $D1$ is true and $D2$ is false then $B1$ and $B3$ are false, so $x\vdots10$ . We have $14<x<20$ and $x\vdots10$ , therefore we can't find $x$ So, $x>14$ and $x$ is a prime. Therefore $B2$ , $C1$ and $C3$ are all false. $C2$ has to be true, so $x<90.9090$ . Then $B3$ is false, so $B1$ has to be true, which means $x>83.3333$ Finally, we have $83.3333<x<90.9090$ and $x$ is a prime $x=89$

The conditions are:

A1: x<12
A2: x!=7m
A3: 5x<70

B1: 12x>1000
B2: x=10n
B3: x>100

C1: x=4p
C2: 11x<1000
C3: x=9q

D1: x<20
D2: x prime
D3: x=7r

here, m, n, p, q, r are constants.

1. B1,B3 are not compatible with A1,A3. If A1,A3 are true then B2 must be true and x=10 (20 or more goes against A1,A3). But if x=10, all A becomes true, which is against the rule. So A1, A3 are false and A2 is true.

2. That makes D3 false. So either D1 or D2 or both are true.

3. If D1 is true then B1 and B3 are false, so B2 must be true, that is x=10. But that is against the rule proved before. So, D1 is false so D2 is only true in D.

4. If D2 is true then B2, C1 and C3 are false so C2 is true. If C2 is true then B3 is false so B1 is only true.

So after deduction we see the true statements are A2, B1, C2, D2. If C2 and B2 are both true, 83<x<90. And if D2 is true then only prime in that range is 89. Which satisfies all the conditions.

Let's begin with Becky. If the number is less than 84, her 1st and 3rd statements are false, so in that case, her second statement must be true. So if the number is less than 84, it must be divisible by 10. Therefore, by Becky's statements, we can limit the number to:

• 10, 20, 30, 40, 50, 60, 70, 80,
• or something > 83.

Now let's add Andy's constraints to Becky's. His 1st and 3rd statements want to constraint the number below 12 and 14, respectively. Then anything below 12 must be divisible by 7 so that his 2nd statement will be false. So 10 is eliminated. Similarly, anything greater than 14 must not be divisible by 7. (As the 1st and 3rd statements are both false, then the 2nd must be true.) Because everything less than 14 has already been eliminated, this effectively now demands that the number is not divisible by 7. So, this is our new list of possibilities:

• 20, 30, 40, 50, 60, 80,
• or a number > 84 and not divisible by 7.

Now we add Danny's constraints. We now know that his 1st and 3rd statements must be false. Therefore, his 2nd statement must be true. The number is prime. This also eliminates every multiple of 10. So now our constraints are:

• n > 84
• n is prime
• (This also implies that it is not divisible by 7.)

Finally let's add Chloe's input. Her 1st and 3rd statements must also be false, because we know that the number is prime. So her 2nd statement must be true. this means that the number is < 91. So these are our final constraints:

• 84 < n < 91
• n is prime.

Fortunately, there is only one prime number between 84 and 91: 89.

...

Now, I know what you're thinking. Four school children independently took 89 to be their favorite number?! Look at the picture. Those kids are probably about 10 years old! And four of them favor 89 more than any other number? Statistically, people rarely favor a number greater than 10. But these kids all love a prime number larger than the age of their parents? Not likely. However, once you have eliminated the impossible, whatever is left, however improbable, must be true.

Maybe there is some external influence at work that was not stated in the problem. Perhaps some local sports hero has the jersey number 89. (Yeah, and the nerd kids who like to stay after school for Math Club are also WAY into local sports.) Or maybe a recent classroom presentation prominently featured the virtues of the number 89. Or perhaps an obscure religion which holds the number 89 sacred is based in the the area near this school, and these children are all devotees. Who knows?