Divisibility 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=10a+b, q=a-2b. In particular,
, so 2n+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
Tags: Divisibility, Divisibility Tests, Math, Monday Math, Number Theory
March 30, 2009 at 12:04 am |
[...] 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=10a+b and q=a-9b, we see , so 9n+q is a multiple of 13, [...]