Divisibility Tests Part 5: Tests for 13

Divisibility tests for 13 are also quite rare, however we can create a test analogous to our first for 7.

We note that 9*10+1=91=7*13, so with *n*=10*a*+*b* and *q*=*a*-9*b*, we see

,

so 9*n*+*q* is a multiple of 13, which means that *n* is divisible by 13 if and only if *q* is divisible by 13. This gives us the test: mulitply the ones digit of our number by 9, and subtract from the number formed by the rest of the digits. We see that this doesn’t work well for two-digit numbers, but a little better for three- or four-digit numbers.

For example:

*n*=78

7-8*9=-65=-5*13

*n*=182

18-2*9=0=0*13

*n*=507

50-7*9=-13

*n*=3679

367-9*9=286

28-9*6=-26=-2*13

so all of these numbers are divisible by 13.

As with 7, this procedure may be laborious for very large numbers. However, we can see that

1001=7*143=7*11*13=77*13

This means 1000=13*77-1, and so we see our large-number divisibility test for 7 also works for 13.

Thus, we split our large number into sets of three digits, and then take the alternating sum of these numbers. For example:

*n*=348,970,129,352

352-129+970-348=845

84-5*9=39=3*13

so 348,970,129,352 is divisible by 13

*n*=4,626,145,317,014,712

712-14+317-145+626-4=1492

149-2*9=131=13*10+1, which is not divisible by 13,

so 4,626,145,317,014,712 is not divisible by 13.

## Posts Tagged ‘Divisibility Tests’

### Monday Math 65

March 30, 2009### Monday Math 64

March 23, 2009Divisibility Tests Part 4: Tests for 7

One generally does not see tests for divisibility by 7, as these aren’t as simple or quick as the more common tests. However, tests do exist.

First, let us consider two numbers *a* and *b*, and examine *n*=10*a*+*b*, *q*=*a*-2*b*. In particular,

, so 2*n*+*q* is a multiple of 7, which means that *n* is divisible by 7 if and only if *q* is divisible by 7. This gives us the following test: mulitply the ones digit of our number by 2, and subtract from the number formed by the rest of the digits.

For example:

*n*=42

4-2*2=0=0*7

*n*=49

4-2*9=4-18=-14=-2*7

*n*=301

30-2*1=28=4*7

so all of these numbers are divisible by 7.

For large numbers, we can repeat the test:

*n*=41,237

4123-2*7=4109

410-2*9=392

39-2*2=35=5*7,

so 41,237 is divisible by 7 (41,237=7*5891)

Note that for very large numbers, the procedure may be laborious. However, we can develop a second test by noting that

1001=7*143

This means 1000=7*143-1, and so:

,

and so on, so that

is divisible by 7 for all non-negative integers *k*. This creates a test by analogy to our divisibility test for 11.

Thus, we split our large number into sets of three digits, and then take the alternating sum of these numbers. For example:

*n*=348,967,129,356,874

874-356+129-967+348=28=4*7

so 348,967,129,356,874 is divisible by 7.

We can also combine the tests:

*n*=1,626,145,217,013,712

712-13+217-145+626-1=1596

159-2*6=147=21*7

so 1,626,145,217,013,712 is divisible by 7

### Monday Math 63

March 16, 2009Divisibility tests Part 3: 11

Last time, we proved the tests for divisibility by 3 and 9. Now, for 11, we again use the factoring rule . We also use the rule that if *n* is odd.

The first tells us , and so is divisible by 11 (as 99=11*9) for even *k*.

The second tells us that

.

So is divisible by 11 for all non-negative integers *k*.

So we consider the alternating sum . The difference between *n* and *a* is then

, and as each term of the series is divisible by 11, the difference is a multiple of 11. Thus, *n* is divisible by 11 if and only if *a* is a multiple of 11.

An few examples of this test:

*n*=297: 7-9+2=0=0*11, so 297 is divisible by 11 (297=27*11)

*n*=12,529:

9-2+5-2+1=11 (12,529=1139*11)

*n*=4,596,836:

6-3+8-6+9-5+4=13, so 4,596,836 is not divisible by 11.

*n*=94,949,690:

0-9+6-9+4-9+4-9=-22=-2*11, so 94,949,690 is divisible by 11 (94,949,690=8,631,790*11).

### Monday Math 62

March 9, 2009**Divisibility Tests**

**Part 2: 3 and 9**

One may know that the tests for divisibility by 3 and 9 involve adding the digits of a number: a number is divisible by 3 if and only if the sum of its digits is divisible by 3, and divisible by 9 if and only if the sum of its digits is divisible by 9.

The key is to recall the factoring rule , used to prove the geometric series formulas. This means that . (This just means 9, 99, 999, 9999, etc. are all multiples of 9, which should be quite obvious).

Let us consider an integer *n*. Let *a*_{0} be the ones digit, *a*_{1} the tens digit, *a*_{2} the hundreds digit, and so on, with *a _{m}* the highest digit, so that

*n*is an (

*m*+1)-digit number. This means

. The sum of the digits, in turn, is . The difference between these two is

.

As this is a multiple of 9, we see that

*n*is divisible by 9 if and only if

*s*is divisible by 9, and as a multiple of 9 is also a multiple of 3,

*n*is divisible by 3 if and only if

*s*is divisible by 3.

### Monday Math 61

March 2, 2009**Divisibility Tests**

**Part 1: Factors of powers of 10**

A common topic in mid-level aritmetic are tests to determine whether a number is divisible by a given small integer without needing to perform a full division; for example, that a number is divisible by nine if and only if the sum of its digits is also a multiple of nine. Proving these rules, however, can involve a bit of higher level math, such as modular arithmetic or other number-theoretic reasoning. Today, I will consider the simplest of divisibility tests; those for factors of powers of ten, which are simple due to our use of the base-ten number system.

First, divisiblity by ten is simply detemined by whether or not the ones digit is zero; divisiblity by 100; by whether the ones and tens digits are both zero; and so on. The reason is obvious.

Next, we see that divisibilty by the factors of 10, 2 and 5, also depend only on the ones digit: a number is divisible by 2 (even) if and only if the ones digit is even, and a number is divisible by five if and only if the ones digit is 0 or 5.

Those numbers which divide 100 but not 10 divide a number only if they divide the last two digits. This gives us the rules for 4, 20, 25, and 50.

For numbers dividing 1000 but not 100, we test the last three digits. This gives the tests for 8, 40, 125, 200, 250, and 500.

And so on. These, as we can see, are simple, and are obvious consequences of our base-ten system.