Who is wrong?

Looking for two numbers in a row with unusual factors. It makes sense that one should be a prime over 15 since otherwise, it's double would also be incorrect. So one of the numbers could be one of ${17,19,23,29,31}$

Now the neighboring number is even and needs an unusual factorization. It cannot be a number with more than one prime factor, because these numbers are correct. (Example 26 cannot be wrong because 2 and 13 are correct.) That leaves a number with all 2's. The highest power of 2 in the list is $2^{4}=16$ . This is consecutive with $17$ so the wrong numbers are $16, 17$ and $m=\boxed{15}$

each wrong divisor must have the highest number of some prime factor out of all of the numbers in the sequence from 2-31.
Neither of the two numbers can be expressible as $n_{1}\times n_{2}$ as that would imply you can't divide by one of either $n_{1}$ or $n_{2}$ , which is a problem because there are no natural numbers except 1 and 2 such that $n_{1}\times n_{2} = n_{1} +1$ and we know it's divisible by 2 since non-consecutive students said higher powers of two.
So the two numbers must be primes raised to a power (which would be the highest power you see that prime raised to in the sequence)
The primes in the sequence are: 2,3,5,7,11,13,17,19,23,29,31
The highest powers of primes in sequence are: $16 (2^{4}), 27 (3^{3}), 25 (5^{2})$
Now we just need to find two of these numbers that are consecutive, which aren't 2,3,5,7,11,or 13, as their value multiplied by an integer greater than or equal to two comes up in the sequence at some point. Out of the numbers above, only 16, and 17 are consecutive, and are in line with the conditions described. Thus they must be the only numbers in which aren't perfect divisors of this large number. keeping in mind the $n^{th}$ student says the $(n+1)^{th}$ number. The $15^{th}$ student says 16 so the answer is 15

I somehow noticed that $2^{3}$ was the smallest power of 2 in the prime factorization of $N$ and thus $2^{4}=16$ wasn't needed. Also, if we get rid of the $17$ : since it's prime it won't affect the numbers before it and since $17 \times 2 > 31$ it won't affect the numbers after it.

Therefore it works for $(16 , 17)$ . Although I think I was lucky...

PS : It would've been waay cooler if the answer was $m + (m+1)$ , which actually equals 31 !

Suppose that the statement "The number is divisible by $p^aq^b$ " is wrong, with $p,q$ prime, and $a,b \geq 1$ .

Then the statements "The number is divisible by $p^a$ " and "The number is divisible by $q^b$ " are also wrong, making for a total of three incorrect students. But there are only two; therefore both incorrect divisors must be a power of a prime.

Also, they should be the greatest power of that prime less than or equal to 31.

This leaves us with the candidates $2^4 = 16\ \ \ 3^3 = 27\ \ \ 5^2 = 25\ \ \ \text{and the simple primes}\ \ \ 7, 11, 13, 17, 19, 23, 29.$ It is easy to see that of these numbers, only $16,17$ are consecutive. Therefore student $\boxed{\text{15}}$ (who said that 16 was a divisor) and students $16$ (who claimed 17 as a divisor) are incorrect.

a) The number must be divisible by every integer from 1 to 15, otherwise there would be at least two non-consecutive incorrect students. Proof: If the number is not divisible by n, then neither is it divisible by 2n. (For example, if the number is not divisible by 11 then neither is it divisible by 22.)

b) It is possible that the 15th and 16th students were the incorrect ones and the number is not divisible by 16 or 17, but divisible by all the other integers from 2 to 31. Proof: Any number which contains at least the prime factors 2, 2, 2, 3, 3, 3, 5, 5, 7, 11, 13, 19... and all other prime numbers up to 31 would satisfy this condition. It wouldn't matter if there were additional larger prime factors or additional 3s, 5s, 7s etc. The key things are that there must be three factors of 2 (and no more) and no factor of 17.

c) Given a), the students who quoted an even number between 18 and 30 must have been correct, and therefore no other two consecutive students could be incorrect. Proof: Every even number between 18 and 30 can be expressed as a x b where a is a power of 2 (2, 4 or 8) and b is odd, and where a and b are both between 2 and 15. Knowing that the number is divisible by a and b means that it is also divisible by a x b. (For example, we know the number is divisible by 4 and 7, so it is also divisible by 28.)

For simplicity, let's call the large target number N .

The first key insight is that if two numbers are non-consecutive, then one of them must be a factor of N . Let's choose two non-consecutive numbers from the list a and b , such that b is divisible by a . Because of this, if b is a factor, then a must also be a factor. However, if b is not a factor, a must still be a factor because they are non-consecutive. Therefore, we have shown that for any two non-consecutive numbers a and b in the list, such that b is divisible by a , a is a factor of N .

We can use this to identify certain numbers which must be a factor. Any number a in the list less than 16 must be a factor, because b = a * 2 also appears in the list. Therefore, the two consecutive non-factors must be {16,17}, {17,18}, ...., or {30,31}.

The second key insight is that any number which is a multiple of two other coprime factors of N must also be a factor of N . Let's concentrate on even numbers in the range [16,30], since we know that at least one of the two consecutive numbers must be even.

We can eliminate numbers between 18 and 30 based on the below. Note that we have already proven than anything between 2 and 15 has to be a factor of N .

18 = 2 * 9, 20 = 4 * 5, 22 = 2 * 11, 24 = 3 * 8, 26 = 2 * 13, 28 = 4 * 7, 30 = 5 * 6

The only number missing from this list is 16. It is the only even number between 2 and 31 which cannot be proven to be a factor of N . Since there must be at least one even number which is not a factor of N , it must be 16.

The one remaining step is to determine whether {15,16} or {16,17} is the consecutive pair of non-factors. We have already proven that 15 is a factor of N , so the only candidate is {16,17}. Note that because 17 is prime, and no multiple of 17 is in the list, we likewise cannot prove that 17 is a factor of N . {16,17} is the only possible candidate. The answer, therefore, is that the 15th and 16th students spoke wrongly.

Let $N$ be our number. Since the two non-divisors are consecutive, one of them must be even. If an even number with an odd factor $x=c2^n$ does not divide $N$ , with $c>1$ odd, then either $c \nmid$ $N$ or $2^n \nmid N$ . That would make at least two non-consecutive non-divisors ( $\{c, x\}$ or $\{2^n, x\}$ ). So one of the non-divisors must be a power of $2$ . And, obviously, if $2^n \nmid N$ , then $2^{n+1} \nmid N$ . Thus $2^{n+1}$ must be outside the range of numbers under consideration, i.e. $2^n = 16$ .

The other non-divisor must be either $15$ or $17$ . $15$ is a composite number, so if $15 \nmid N$ , one of $3, 5$ must also not divide $N$ . Thus the other non-divisor must be $17$ . And our $15$ th student mentioned $16$ .

possible numbers are 4,8,16,3,9,27,25 and all prime. but it must greater than 15 since 15*2 = 30 smaller than 31. so numbers left is 16,27,25,17,19. since they need consecutive, only 16 and 17 left. so answer is $\boxed{15}$ .

First we list the possible numbers :

2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31

And easily we can know that the numbers less or equal than 15 are impossible because they have a multiple

For example if 9 is incorrect then 18 will be incorrect either cause 18 = 2 * 9

So the possible numbers become :

16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31

which means the rest of the numbers are correct , and we can use another strategy

We begin with 2 :

since 2 to 15 are all correct and we use 2 and 9 as an example :

2 * n is the number

9 * m is the number.

Cause 9 isn't the multiple of 2 so 2 is in the m . Thus 2 * 9 = 18 is also correct

And so does 2 * 11, 2 * 13, 2 * 15, 3 * 7 ....

After we done this, the possible numbers turns out to be :

16,17,19,23,25,27,29,31

Only 16 and 17 are consecutive so the answer is the 15th student

If the number is not divisible by $n$ , then it is also not divisible by $kn$ . This means $n\in[16,31]$ .
If the number is not divisible by $\prod p^k_i$ , where $p_i$ are distinct primes, then it cannot be divisible by all $p^k_i$ . This means $n=p^k$ . The only pair of consecutive integers that satisfies these constraints on $n$ is ( $16,17)$ , so the answer is $15$ .

Observation 1: If k and k+1 doesn't divide the number then, 2k and 2(k+1) $\ge 32 so, k is 16 or more. So to get non-divisors of the number, we have to consider only consecutive numbers in 16,17,...,31

Observation 2: If at least one of k and k+1 has more than one prime divisor then it can be written as pq, where gcd(p,q)=1. But as p,q both divides the number,(since, teacher said only two of the student say false) pq divides the number, which is not the case.So both of k and k+1 has single prime divisor. Now, the (1) & (2) are only satisfied by 16 and 17. So, atmost 15th and 16th student is wrong

If a student says a number which is correct, all the prime factors of their number must be present in the list of prime factors of the original number.

For a student's number to be possibly incorrect, their number must either be prime, contain a prime factor not included in the original number, or have more instances of a specific prime factor than the original number. In any other case, the other students' numbers would require the original number to contain the prime factors of the student in question's number.

By looking through the prime factors of numbers $2$ to $31$ , we can see that the only consecutive pair of numbers which can possibly fit any of these conditions is $16$ and $17$ : $16$ contains more $2$ prime factors than any other number below $32$ (the final number only has 3), and $17$ is prime. Therefore the answer is $\boxed{15}$ .

It's worth noting that the second condition could never be true, since a prime number cannot be followed immediately by a multiple of itself.

Obviously, the first 14 students spoke correctly: If the first statement is incorrect, then the third statement is correct (as if a number is not divisible by 2, it would not be divisible by 4) and so on. Hence $15\ge m\ge 30$ $(1)$ . There must be an even number smaller than 32 that is not divisible by the given number. Seeing that 16 is the only number that is not guaranteed to follow this condition (as $18=2\times 9;\quad 20=4\times 5$ and so on. Hence, $m\in \{ 14;15\}$ . $(2)$ From $(1)$ and $(2)$ , $m=15$ .

Factors 2-15 rule out as if one of these were not a factor, their double wouldn't be either, and neither of these are next to their doubles. If two relative primes are factors, then the product of these relative primes is a factor too. So if the numbers 2-15 are all factors, the following products are factors, too:

18=2x9, 20=4x5, 21=3x7, 22=2x11, 24=3x8, 26=2x13, 28=4x7, 30=3x10

We have 16,17,19,23,25,27,29,31 remaining. Only 16 and 17 are consecutive numbers. 16 is said by the 15th student.

Its better to search for primes . But no prime are consecutive except $2,3$ but we cant say that they dont divide(because the number is divisible by $6$ )
One of the approach is to search for maximum power of a prime number say $x = p^k$ . Because $x$ can only divide multiple of $p^n$ type number. So as $x$ is greatest type of that number,it necessary no need to divide. But we have to choose $x$ such that any of $x+1,x-1$ is prime
Here $2^4 = 16 , 3^3=27 ,5^2 = 25$ are the maximum power of $2,3,5$ less than $31$
First observe $25$ ,neither of the number $25-1,25+1$ is prime.
Also true for $27$ .
Observe $16$ and $16+1$ = $17$ as it is prime, our desired number is $16,17$ They are said by $15th,16th$ student.
So the answer is $\boxed{15}$

(Note : they are said by $15th,16th$ student not $16th,17th$ )

The smallest number the teacher could have written was 707,860,553,400.

First of all, we know that it can't be a number below 15. The first fourteen students all said numbers that another (non-consecutive) student later said a multiple of (e.g. the 14th student said 15, but the 29th student said 30), so if said student was incorrect, so was another non-consecutive student.

This being the case, we know that the number the teacher wrote on the board has to be a multiple of every number from 2 to 15. The minimum common multiple of these numbers is $2^3*3^2*5*7*11*13=360360$ , so any number between 16 and 31 that is a divisor of that number must be a divisor of the number the teacher wrote. Namely, 18, 20, 21, 22, 24, 26, 28, and 30 are definitely divisors.

Therefore, we can eliminate the possibility that m is in integer interval [16..30], as all of those values of m are such that one of the divisors we listed is implied as incorrect. That leaves us with m=15.