Divisible by 12 - Try with a Game!

In this post, you would learn about numbers that are divisible by 12 through a game which is a numerical maze!

First keep in your mind that:

* If a number is divisible by 3 and by 4 then the number is divisible by 12.

* If the sum of the digits of a number is divisible by 3 then the number is divisible by 3. For example, 2130021 is divisible by 3 because 9, the sum of its digits (2 + 1 + 3 + 0 + 0 + 2 + 1 = 9), is divisible by 3.

* If half of a number is an even number then that number is divisible by 4. For example, 72 is divisible by 4 because half of it, 36, is an even number. Also, if the last two digits of a number is divisible by 4 then that number is divisible by 4. As an example: 897232 is divisible by 4 because its last 2 digits; i.e. 32, is divisible by 4.

Now, find a path from the START to the END passing only through numbers that are divisible by 12. Note that there is only one solution for this numerical maze.

Scroll down to check your solution!

Solution:

Square of 101, 102, 103, ..., 1001, 1002, ..., ...

Consider the following squares:

101 * 101 = 10201
102 * 102 = 10404
103 * 103 = 10609
104 * 104 = 10816
...
109 * 109 = 11881

In all of the above squares, there is a simple pattern as described in the following.

101 * 101 = <-- five digits
10201
10201 <-- this bold '1' is calculated as 1*1 (101 * 101)
10201 <-- this bold '2' is calculated as 1+1 (101 * 101)
10201 <-- this bold '1' is due to 100*100=10000 (five digits)

Let's see for 102:

102 * 102 = <-- five digits
10404
10404 <-- this bold '4' is calculated as 2*2 (102 * 102)
10404 <-- this bold '4' is calculated as 2+2 (102 * 102)
10201 <-- again, this bold '1' is due to 100*100=10000 (five digits)

Now, let's see for 103:

103 * 103 = <-- five digits
10609
10609 <-- this bold '9' is calculated as 3*3 (103 * 103)
10609 <-- this bold '6' is calculated as 3+3 (103 * 103)
10609 <-- again, this bold '1' is due to 100*100=10000 (five digits)

Now, for 109:

109 * 109 = <-- five digits
11881
11881 <-- this bold '81' is calculated as 9*9 (109 * 109)
11881 <-- this bold '18' is calculated as 9+9 (109 * 109)
11881 <-- again, this bold '1' is due to 100*100=10000 (five digits)

This pattern can also be seen in 110, 111, ...

Now, consider the following squares:

1001 * 1001 = 1002001
1002 * 1002 = 1004004
1003 * 1003 = 1006009
1004 * 1004 = 1008016
...
1009 * 1009 = 1018081

It must be easy for you to extract the pattern!

Now try:

10000001 * 10000001
10000002 * 10000002
10000009 * 10000009
100000001 * 100000001
100000002 * 100000002
100000009 * 100000009

Squaring Two-Digit Numbers

To square 2-digit numbers follow the following algorithm.

Select an arbitrary number, for example 48 to follow:

1. Look for the nearest 10 (ceiling) boundary; for 48 it would be 50.

2. Take d = 50 - 48; so d = 2.

3. Take A = 50 * (50 - 2 * d) = 50 * 46 = 2300.

4. The answer is equal to A + d^2 = 2300 + 2 * 2 = 2304.

48 * 48 = 2304

Another example:

Select an arbitrary number to mentally calculate its square, for example 87.

1. Look for the nearest 10 boundary; for 87 it would be 90.

2. Take d = 90 - 87; so d = 3.

3. Take A = 90 * (90 - 2 * d) = 90 * 84 = 7560.

4. The answer is equal to A + d^2 = 7560 + 3 * 3 = 7569.

87 * 87 = 7569

Now let's try it for 73:

1. Look for the nearest 10 boundary; for 73 it would be 80.

2. Take d = 80 - 73; so d = 7.

3. Take A = 80 * (80 - 2 * 7) = 80 * 66 = 5280.

4. The answer is equal to A + d^2 = 5280 + 7 * 7 = 5329.

73 * 73 = 5329