1 f6 some facts about the disvisibility of numbers

Some Facts About Divisibility

We start out with a simple mathematics procedure that is often used in real live.
Frank Ma © 2011

We start out with a simple mathematics procedure that is often used in real live. It’s called the digit sum.

We start out with a simple mathematics procedure that is often used in real live. It’s called the digit sum. Just as its name suggests, we sum all the digits in a number.

The digit sum of 12 is 1 + 2 = 3,

The digit sum of 12 is 1 + 2 = 3, is the same as the digit sum of 21, 111, or 11100.

The digit sum of 12 is 1 + 2 = 3, is the same as the digit sum of 21, 111, or 11100. To find the digit sum of
7 8 9 1 8 2 7 3

7 8 9 1 8 2 7 3
15
add

7 8 9 1 8 2 7 3
15 30
add

7 8 9 1 8 2 7 3
15 30
add
add
45

7 8 9 1 8 2 7 3
15 30
add
add
45
Hence the digit sum of 78999111 is 45.

7 8 9 1 8 2 7 3
15 30
add
add
45
9
If we keep adding the digits, the sums eventually become a single digit sum – the digit root.

7 8 9 1 8 2 7 3
15 30
add
add
45
9
If we keep adding the digits, the sums eventually become a single digit sum – the digit root.
The digit root of 78198273 is 9.

A bag contains 384 pieces of chocolate, can they be evenly divided by 12 kids?

A bag contains 384 pieces of chocolate, can they be evenly divided by 12 kids? In mathematics, we ask “is 384 divisible by 12?” for short.

A bag contains 384 pieces of chocolate, can they be evenly divided by 12 kids? In mathematics, we ask “is 384 divisible by 12?” for short. How about 2,349,876,543,214 pieces of chocolate with 18 kids?

A bag contains 384 pieces of chocolate, can they be evenly divided by 12 kids? In mathematics, we ask “is 384 divisible by 12?” for short. How about 2,349,876,543,214 pieces of chocolate with 18 kids? Is 2,349,876,543,214 divisible by 18?

One simple application of the digit sumis to check if a number may be divided completely by numbers such as 9, 12, or 18.

I. The Digit Sum Test for Divisibility by 3 and 9.

If the digit sum or digit root of a number may be divided by 3 (or 9) then the number itself maybe divided by 3 (or 9).

For example 12, 111, 101010 and 300100200111 all have
digit sums that may be divided by 3,

digit sums that may be divided by 3, therefore all of them may be divided by 3 evenly.

digit sums that may be divided by 3, therefore all of them may be divided by 3 evenly. However only 3001002000111, whose digit sum is 9,

digit sums that may be divided by 3, therefore all of them may be divided by 3 evenly. However only 3001002000111, whose digit sum is 9, may be divided evenly by 9.

digit sums that may be divided by 3, therefore all of them may be divided by 3 evenly. However only 3001002000111, whose digit sum is 9, may be divided evenly by 9.
Example A. Identify which of the following numbers are divisible by 3 and which are divisible by 9 by inspection.
a. 2345 b. 356004 c. 6312 d. 870480

We refer the above digit–sum check for 3 and 9 as test I.

We continue with testII and III.

II. The Test for Divisibility by 2, 4, and 8

A number is divisible by 2 if itslast digitis even.

A number is divisible by 4 if its last 2 digits is divisible by 4 – you may ignore all the digits in front of them.

Hence ****32 is divisible by 4

Hence ****32 is divisible by 4 but ****42 is not.

A number is divisible by 8 if its last 3 digits is divisible by 8.

Hence ****880 is divisible by 8,

Hence ****880 is divisible by 8, but ****820 is not.

III. The Test for Divisibility by 5

III. The Test for Divisibility by 5.
A number is divisible by 5 if itslast digitis 5.

From the above checks, we get the following checks for important numbers such as 6, 12, 15, 18, 36, etc..

From the above checks, we get the following checks for important numbers such as 6, 12, 15, 18, 36, etc..
The idea is to do multiple checks on any given numbers.

The Multiple Checks Principle.

If a number passes two different of tests I, II, or III, then it’s divisible by the product of the numbers tested.

The number 102 is divisible by 3-via the digit sum test.

It is divisible by 2 because the last digit is even.

Hence 102 is divisible by 2*3 = 6.

Hence 102 is divisible by 2*3 = 6 (102 = 6 *17).

Example B. A bag contains 384 pieces of chocolate, can they be evenly divided by 12 kids? How about 2,349,876,543,214 pieces of chocolate with 18 kids?

For 384 its digit sum is 15 so it’s divisible by 3 (but not 9).

Its last two digits are 84 which is divisible by 4.

Its last two digits are 84 which is divisible by 4. Hence 384 is divisible by 3 * 4 or 12.

For 2,349,876,543,210 for 18, since it's divisible by 2, we only have to test divisibility for 9.

For 2,349,876,543,210 for 18, since it's divisible by 2, we only have to test divisibility for 9. Instead of actually find the digit sum, let’s cross out the digits sum to multiple of 9.

For 2,349,876,543,210 for 18, since it's divisible by 2, we only have to test divisibility for 9. Instead of actually find the digit sum, let’s cross out the digits sum to multiple of 9. We see that it’s divisible by 9, hence it’s divisible by 18.

1 f6 some facts about the disvisibility of numbers

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