Happy New Year! I hope your year is off to a great start.

If you like to contemplate numbers like I do, you may have wondered whether or not 2013 is a prime number. After a quick review of basic divisibility rules (yeah, there's a song for that) we can see that 2013 is divisible by 3.

(2 + 0 + 1 + 3 = 6, and 3 goes into 6.)

Factoring out 3, we have: 2013 = 3(671)

So, is 671 a prime number? It doesn’t fit any of the divisibility rules for 2 through 10. However, as it turns out, 671 is divisible by 11.

Factoring out 11, we have: 671 = 11(61)

Therefore, the prime factorization of 2013 is 3·11·61.

In general, how can we tell whether or not a number is divisible by 11?

Here's a fairly easy

**Divisibility Rule for 11:**
1)
Find the sum of every other digit in the number (1

^{st}digit + 3^{rd}digit + 5^{th}digit + …)
2)
Find the sum of the digits not used in step 1 (2

^{nd}digit + 4^{th}digit + 6^{th}digit + …)
If
these sums differ by 0, or a multiple of 11, then the original number is
divisible by 11.

{If you want to know why this rule works, continue reading. If you don't care why it works, but would like to see another neat divisibility rule for 11, skip to the bottom.}

__Why this rule Works__

Think
back to elementary school, when you used pencil and paper to multiply. What does it look like when a number is
multiplied by 11?

Here
is an example: 12345

__× 11__

12345

__+__

__123450__

135795

Let’s take a look at a general case, in which
a 5-digit number is multiplied by 11.
The conclusions we draw can be generalized for multiplying any length of
number by 11.

ABCDE

__× 11__

ABCDE

__+ ABCDE0__

The answer will have the following values for
its digits:

1

2^{st}digit: A^{nd}digit: (A + B)

3

^{rd}digit: (B + C)

4

^{th}digit: (C + D)

5

^{th}digit: (D + E)

6

^{th}digit: E

The sum of digits 1, 3 and 5 is A + (B + C) +
(D + E)

The sum of digits 2, 5, and 6 is (A + B) + (C
+ D) + E

Clearly these sums have a difference of 0.

But… what if we need to carry when adding
ABCDE + ABCDE0?

Exactly what happens when we carry?

If a digit’s value exceeds 9, then we
subtract 10 from that value, and add 1 to the value of the digit to the left.

For example: 39

__+ 7__

46

(9 + 7
= 16. 16 – 10 = 6. We write 6 in the one’s place, and add 1 to
the digit in the ten’s place)

Subtracting 10 from a digit’s value and
adding 1 to the value of the digit to the left causes a net change of 11
between the sum of digits 1, 3 & 5 and the sum of digits 2, 4 & 6.

Let’s assume that in our generalized example,
(D + E) exceeds 9. In this case, since
we have to carry, the answer will have the following values for its digits:

1

2^{st}digit: A^{nd}digit: (A + B)

3

^{rd}digit: (B + C)

4

^{th}digit: (C + D) + 1

5

^{th}digit: (D + E) – 10

6

^{th}digit: E

Now the sum of digits 1, 3 and 5 is A + B + C + D + E – 10

And the sum of digits 2, 4 and 6 is A + B + C + D + E + 1

These sums differ by 11.

Every time we carry (subtract 10 from a
digit’s value and add 1 to the value of the digit to the left) we are causing a
net change of 11 between the sum of digits 1, 3 & 5 and the sum of digits
2, 4 & 6. These changes may
accumulate, or cancel each other out.

If we carry twice, the sums could differ by
22 (if the net changes of 11 accumulate) or they could differ by 0 (if the net
changes cancel each other out).

By generalizing this explanation, we can show
that for

*any*multiple of 11, the sum of digits 1, 3, 5, etc. will differ from the sum of digits 2, 4, 6, etc. by 0, or a multiple of 11.Here’s an even easier

**divisibility rule for 11**:

Subtract the last digit from the number
formed by the remaining digits. (You can
repeat this step as often as necessary.)
If the resulting number is divisible by 11, then the original number is
divisible by 11.

(Example:
671 is divisible by 11 because 67 – 1 = 66, and 66 is divisible by 11.)

Why does this work? I bet you can figure it out…

*Hint: It might be helpful to once again
consider this problem:

ABCDE

__× 11__

ABCDE

__+ ABCDE0__

Or, you can share
on Facebook: www.facebook.com/pages/OnlineMathProcom/134423926682041

Here’s to a great (not to mention, mathtastic) year in 2013.

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