$\begingroup$

Which of the following is not a prime number ?

$a.)\ 911 \ \ \ \ \ \ \ \ \ \ b.)\ 919 \\ \color{green}{c.)\ 943} \ \ \ \ \ \ \ \ \ \ d.)\ 947$

This was asked in my exam and the time given per question was $1-3\ \text{ min}$

If the time was $10\ \text{min}$ for this question I would have solved this comfortably but since the time was less i couldn't solve it with dividing every option with shorter primes .

I look for a simple and short way.

I have studied maths up to $12$th grade.

$\endgroup$ 13

5 Answers

$\begingroup$

the smaller prime factor has to be less that $\sqrt{961} = 31$

Don't start at $2$ and work your way up. Start at $29$ and work your way down.

Four fails at $29$. In all four cases, the first digit of the quotient was $3$.

Two fails at $23$.

Then $943 = 23 \times 41$.

or

If they taught you the difference of two squares method...

\begin{array}{|c|rrrr|} \hline 31^2 & 961 & 961 & 961 & 961\\ n & 911 & 919 & 943 & 947\\ \hline 31^2 - n & 50 & 42 & 18 & 14\\ +(2\cdot 31 + 1) & +63 & +63 & +63 & +63\\ \hline 32^2 - n & 113 & 105 & \color{red}{81} \\ \hline \end{array}

So \begin{align} 32^2 - 943 &= 9^2\\ 32^2 - 9^2 &= 943\\ (32-9)(32+9) &= 943\\ 23 \times 41 &= 943 \end{align}

$\endgroup$ 4 $\begingroup$

Since you don't know what you're looking for, the most sensible strategy is to check the numbers to see if they're divisible by $2, 3, 5, 7, \dots, 29$ and stop when you find one that is. Obviously, this involves a lot of individual checks, so the problem is how to go fast. For $2, 3, 5, 11$ there are well-known criteria for divisibility, and you quickly determine that none of the four numbers are multiples of these.

Since the four numbers are all close together, it's easiest to simply list the multiples of $7, 13, 17, \dots$ that are close to the numbers.

The multiples of $7$ are: $910, 917, 924, 931, 938, 945, \dots$.

The multiples of $13$ are: $910, 923, 936, \dots$

The multiples of $17$ are: $901, 918, 935, \dots$

The multiples of $19$ are: $950, 931, 912, \dots$

The multiples of $23$ are: $920 - 23, 920, \mathbf{943}$.

$\endgroup$ 8 $\begingroup$

First, note that you only need to test primes less than $31$, since $30^2=900$ and $31^2=961$, which is greater than any option. Thus, we only need to test $2,3,5,7,11,13,17,19,23,29$.

We can immediately eliminate $2,3,5,7,11$ from commonly known divisibility tests. I suggest memorizing the divisibility tests up to $15$ for competitions and tests.

Thus, we have it down to $13,17,19,23,29$.

Start by dividing $930$ by all these possible factors. This is because $930$ is about in the middle of the options. Then multiply all our possible factors by the whole number part of the quotient. Finally, start adding and subtracting the possible factors until you find an exact multiple. If you don't arrive at an exact multiple, move on to the next test number and repeat.

For example, $\frac{930}{13} \approx 71$ and $71 \cdot 13=923$.

The multiples of $13$ around $923$ are $910...923...936...949...$ Thus, no answer option has $13$ as a factor, and we can move on.

This method works very well for numbers $<1000$. It will take some practice and you must be able to do multiplication and division rather quickly. For larger numbers, this method becomes less effective in a short time span. I suggest practicing this method for quick tests of primes under $1000$ so that you become very quick and adept at it. It has worked for me in multiple competitions and tests.

$\endgroup$ $\begingroup$

As others have mentioned, you need to check divisibility only against prime numbers that are smaller than or equal to the square root of the dividend.

And since the square roots of all the dividends are smaller than $31$, you need to check divisibility of each dividend only by $2,3,5,7,11,13,17,19,23,29$.

Checking divisibility against $2,3,5$ is easy, so I'm not going to expand on it.

Checking divisibility against $7$ is easy in this case:

• $911$ is not divisible by $7$ since $910$ is
• $919$ is not divisible by $7$ since $910$ is
• $943$ is not divisible by $7$ since $950$ would have to be
• $947$ is not divisible by $7$ since $940$ would have to be

Checking divisibility against $11$ is easy in this case:

• $911$ is not divisible by $11$ since $900$ would have to be
• $919$ is not divisible by $11$ since $930$ would have to be
• $943$ is not divisible by $11$ since $990-44=946$ is
• $947$ is not divisible by $11$ since $990-44=946$ is

Checking divisibility against $13$ is easy in this case:

• $911$ is not divisible by $13$ since $910$ is
• $919$ is not divisible by $13$ since $910$ is
• $943$ is not divisible by $13$ since $930$ would have to be
• $947$ is not divisible by $13$ since $960$ would have to be

Checking divisibility against $17$ is partially easy in this case:

• $911$ - $\color\red{\text{you'll have to do the math}}$
• $919$ - $\color\red{\text{you'll have to do the math}}$
• $943$ is not divisible by $17$ since $960$ would have to be
• $947$ is not divisible by $17$ since $930$ would have to be

Checking divisibility against $19$ is easy in this case:

• $911$ is not divisible by $19$ since $930$ would have to be
• $919$ is not divisible by $19$ since $900$ would have to be
• $943$ is not divisible by $19$ since $950$ is
• $947$ is not divisible by $19$ since $950$ is

Checking divisibility against $23$ is easy in this case:

• $911$ is not divisible by $23$ since $920$ is
• $919$ is not divisible by $23$ since $920$ is
• $943$ is divisible by $23$ since $920+23$ is

$\endgroup$ 2 $\begingroup$

I might have think of it in this way: Let write the number in the form $$(20 + x) (40 + y)$$because all these numbers will be near that value. Write $911 , 919 , 943 , 947$ as $900 + 11$, etc.

Now, $$(20 + x)(40 + y) = 800 + 20 y + 40 x + xy$$ the $xy$ will be most likely the last number. Likely not necessarily trying and observing $xy = 3$ so $x = 3 ,y = 1$, just an guess $$800 + (20 \times 1) + (40 \times 3) + (3 \times 1) = 943$$ it can be a way of quick guessing and eleminating. The solution may seem long and complicated but is quite easy to guess that this way if you understand the concept behind this.

$\endgroup$ 0