Determining whether $3267936$ is divisible by $1621$, in $15$ seconds

Question

I was asked in a contest to determine if $3267936$ is divisible by $1621$ in $15$ seconds.

How can I know in $15$ seconds if $3267936$ is divisible by $1621$ without using long division, and what is divisibility rule of $1621$?

Answer 1

Write the numbers below each other this way:

 3267936  
 1621

Fill the smaller number with zeroes

 3267936  
 1621000

Subtract it from the bigger number

 3267936  
-1621000
--------
 1646936

Replace the larger of the two original numbers ($3267936$) with this result and repeat this process so that you now start with ($1646936$) and ($1621$).

 1646936  
 1621
1646936

 1621000
1646936
-1621000
25936

Again replace the bigger number ($1646936$) with the result.

 25936  
 1621
25936

 16210
25936
-16210
9726

Replace again. No more zeroes need to be added.

 9726  
-1621
-----
 8105

and so on

 8105
-1621
-----
 6484

You again replace the bigger number with the result. I now add gaps between the digits because this makes my explanation easier. When you are in the process, you just have to try to find these gaps in your mind

 64 8 4
 16 2 1

At this point you will notice that $64 = 16\cdot4$ and $8 = 2\cdot4$ and $4 = 1\cdot4$. So you have a number in each column in the second row and a multiple of 4 of that number above it. And that means that $6484 = 1621\cdot4$. In other words, $6484$ is divisible by $1621$. And since we only subtracted multiples of $1621$ in every single step, $3267936$ must also be divisible by $1621$.

If you are familiar with this technique, you can do this in less than 15 seconds, and if you practice it hard, you can even do it in your head without using pen and paper.

If you do it on paper, you don't need to write down numbers you already have written and you can add the filling zeros just in your head. You will be even faster, if you start with the two numbers side by side. Longer sequences of digits that are the same as in the line above don't need to be written (I marked these placed here with dots, but in practice you don't write them). Then at the end your notes will look like this:

3267936  1621
1646...
  25...
   9726
   6484

By the way, the algorithm used here is the Euclidean algorithm. This is a 2300 year old, very fast algorithm for finding the common divisor of two numbers. If the common divisor of two numbers is the smaller of the two, then the larger number is divisible by the smaller one.

Answer 2

We have $3267936$ which is a multiple of $4$ , while $1621$ is not.

Hence , we can remove the $4$ from $3267936$ to get $816984$ , which is still a multiple of $4$.
Hence , we can remove that $4$ too , to get $204246$.
We can then remove the $2$ to get $102123$.

Now , $1621$ ( ending in $1$ ) $\times$ $X3$ ( ending in $3$ ) will give $YYYYY3$ ( ending in $3$ ) [[ Here , $X$ is a Digit , while $Y$ is arbitrary Digit Placeholder ]]

$X=5$ is too small , it will give ~ $320YY$
$X=7$ is too large , it will give ~ $1120YY$

Hence , we have to only check $X=6$ , that is , $63 \times 1621$ to finish this. If that is $102123$ , we have divisibility. Else there is no divisibility.

It turns out that $63 \times 1621 = 102123$ : Hence we have Divisibility.

What we were given will be $4 \times 4 \times 2 \times 63 \times 1621 = 2016 \times 1621 = 3267936$

Answer 3

I would try to construct the quotient one digit at a time using the right side digit of dividend in the current step of the process to guess the corresponding digit of the quotient. If at the end I obtained the given dividend then the division has no remainder and in the process I have obtained the quotient as well.

The method avoids division and subtraction however it does require guessing.

I call the dividend the number 3267936, the divisor the number 1621 and the quotient the number I try to determine such as the remainder of the division would be zero.

Step 1: choose the units digit of the quotinet in order to get the units digit of the dividend.

What times 1 ends in 6? That is 6, the units of quotient. $$\begin{align}&0001621\\\times&\underline{000xxx6}\\&000972\color{red}{6}\end{align}$$

After Step 1 I know that $1621\times xxx6=xxxxxx6$.

Step 2: choose the tenths digit of the quotient in order to obtain the tenths digit of the dividend.

I get a carryover of 2 from 6x2 in the previous multiplication.

Then figure out what times 1 plus 2 ends in 3? That is 1, the tenths unit of the quotient.

$$\begin{align}&0001621\\\times&\underline{000xx16}\\&0009726\\+&\underline{0016210}\\& xxxxx\color{red}{36}\end{align}$$

After Step 2 I know that $1621\times xx16=xxxxx36$

Step 3: choose the hundreds digit of the quotient to get the hundreds digit of the dividend.

I get two carryovers: 7 from first multiplication and 2 from the second, total 9.

Then figure out what times 1 plus 9 ends in 9? Obviusly 0, the hundreds digit of the quotient.

$$\begin{align}&0001621\\\times&\underline{000x016}\\&0009726\\&0016210\\+&\underline{0000000}\\& xxxx\color{red}{936}\end{align}$$

After Step 3 I know that $1621\times x016=xxxx936$.

Step 4

For the thousands digit: what times 1 plus 15 ends in 7? That must be 2.

$$\begin{align}&0001621\\\times&\underline{0002016}\\&0009726\\&0016210\\&0000000\\+&\underline{3242000}\\& \color{red}{3267936}\end{align}$$

After Step 4 I know that $1621\times 2016=3267936$.

Answer 4

Long division.

$1621 \times 2000 = 3,242,000.$

$25936$ remaining.

From the comments, I should now add that it should be obvious that only $16$ might work, as $1621\times 10<25936<1621\times 20$, and for $\ i\in\{1,\ldots,9\},\ $ only $\ i=6\ $ gives $\ 1 \times i = 6.$ For me, this took me less than a second to recognize, and I would expect the same for someone competing in contest math.

$1621 \times 16 = 25936.$

So actually yes.

I got this wrong the first time because I was rushing, so maybe $15$ seconds is too optimistic for this method.

Answer 5

The thing with these competitions is that you need to be very smart from the start with how you approach the question; choosing the fastest way possible that comes to mind.

Since it's easier to multiply than divide, it should be easier to try to approximate $1621$ to $3267936.$

Since we need $3$ more digits, we multiply by a $1000$ and we have $1\color{red}{6}21000$ and $3267936,$ now we multiply by a $2$ because a $3$ would be too big because of the $\color{red}6$ that yields an $8$ (or the better cause which is that $2$ is just perfect as it yields a $2,$ so to approximate further we will have to subtract and approximate $1621$ to the result i.e go a digit down.)

Thus, we subtract $3242000$ from $3267936$ to obtain $25936,$ so we have an initial factor of $2000.$

Now, we approximate $1621$ to $25936.$ We again multiply by $10$ to get $16210$ and notice we will have to subtract again i.e go a digit down again (or pretty much multiply by something between $10$ and $20$ since the subtraction doesn't affect the last digit.)

Since the last digit of $25936$ is $6$ and that of $1621$ is $1,$ I would guess $16,$ and wouldn't necessarily check if I don't have the time because when everything is nice and gets down to the very last step, you can assume the answer is positive.

This might sound a lot on paper, but for a trained mind, it's just basic arithematic with clever choice-making.