Quiz Level-3

Hint

The value of \(p^2\), or \(p \cdot p\), is the product of \(p\) and \(p\), so it will be divisible by \(1, p, p^* p\), and nothing else (we know that the \(p^{\prime}\) s are not divisible because they are prime). Therefore \(p^2\) has exactly three factors.
(Alternatively, we can plug in any prime number for \(p\) and see how many factors \(p^2\) has. For example, if \(p\) is 3 , then the factors of \(p^2\), or 9 , are 1,3 , and 9.)

Hint

Let \(x\) represent the smallest of the four numbers.
Then we can set up the following equation:
\(
\begin{aligned}
& x+(x+1)+(x+2)+(x+3)=210 \\
& 4 x+6=210 \\
& 4 x=204 \\
& x=51
\end{aligned}
\)
Therefore the four numbers are \(51,52,53,54\). The only prime in this list is 53.

Hint

Prime numbers are integers. So doing division and square roots will not generate integers. By doing multiplication and exponents, we involve more factors. The only possibility is subtraction. If \(p\) was 7 and we subtracted 2 we get 5 which is also prime.

Hint

Other than the number itself and one, we also need to have prime factors in that number. Since it’s not a perfect square, we need to find the smallest possible prime numbers. That will be \(2 \times 3\) or 6 which is our answer.

Hint

\(n-2\)
Plug in a prime number such as 7 and evaluate all the possible solutions. Note that the question asks which value could be prime, not which must be prime. As soon as your number-picking yields a prime number, you have satisfied the “could be prime” standard and know that you have a correct answer.

Hint

The prime factorization of 330 is .

The sum of the prime factors is .

23 is the only answer choice greater than 21. 

Hint

The third smallest prime number is 5. (Don’t forget that 2 is a prime number, but 1 is not!)
The smallest two digit prime number is 11.
Now we can evaluate the entire expression:
\(
\frac{5}{2} \times 11=\frac{55}{2}
\)

Hint

This question tests basic number properties. Prime numbers are numbers which are divisible only by one and themselves. Answer options ‘ 2 ‘ and ‘ 4 ‘ are automatically out, because they will always produce even products with \(a\) and \(b\), and the sum of two even products is always even. Since no even number greater than 2 is prime, 2 and 4 cannot be answer options. 3 is tempting until you remember that the sum of any two multiples of 3 is itself divisible by 3 , thereby negating any possible answer for c except 3 , which is impossible. There are, however, several possible combinations that work with \(x=1\). For instance, \(a=8\) and \(b=9\) means that \(8(1)+9(1)=17\), which is prime. You only need to find one example to demonstrate that an option works. This eliminates the “None of the other answers” option as well.

Hint

\(n=28\)
Step-by-step explanation:
\(
\begin{aligned}
& 28=2 \cdot 2 \cdot 7 \\
& 1+2+4+7+14+28=2 \cdot 28=56
\end{aligned}
\)

Hint

The primes, in order, are:
\(
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53, \ldots
\)
We create a few series:
\(x=2\)-> series length \(2: 2,3\)
\(x=3\)-> series length \(3: 3,5,7\)
\(x=5\)-> series length \(5: 5,7,11,13,17\)
\(x=7\)-> series length 7: 7,11,13,17,19,23,29 etc.
We can then see that, of the answers, only 47 and 31 remain possibly correct answers. Now we need to decide which of those two are impossible.

We could do another series, but the \(x=11\) series has 11 terms requiring us to go further and further up. If we do this, we’ll find that it terminates at 47 , meaning that 31 must be the correct answer.

Another way, however, is to notice that 29 is the end of the \(x=7\) series. Since 31 is the very next prime number, if we start on 11 , the series that terminates in 31 would have to have length 7 as well. Every series after \(x=7\) will thus end on a number larger than 31 , meaning we will never finish on a 31.