Multiple or Not?

by on September 23rd, 2012

Is 22 divisible by 4?

Chances are you answered no right away. However, this will likely not work for finding if 374 is divisible by 17. Thus, we need a simple method of figuring out whether or not a number is divisible by another number.

Intuitively, we know that 22 is NOT divisible by 4 because it’s not one of 4′s multiples. For example, 20 and 24 are multiples of 4, and are therefore divisible by 4. Since 22 is not “4″ away from the nearest multiples of 4, it’s NOT divisible by 4.

Let’s try to formulate the rule behind this intuition:
Multiple of 4 ± multiple of 4 – MUST be a multiple of 4.
Multiple of 4 ± NOT multiple of 4 – CANNOT be a multiple of 4.

So 20±8 must be a multiple of 4, because all we’re doing is adding or subtracting multiples of 4. However, 20± 6 CANNOT be a multiple of 4, because the result “lands” somewhere between other multiples of 4.

Now, let’s apply this to find if 374 is divisible by 17.
We know that 17 times 20 = 17 times 2 times 10 = 34 times 10 = 340
This is a simple calculation that does not require much time or effort. However, this simple calculation can save us quite some time and effort.
Rewrite 374 as 340 + 34.
So, 374 = 340 + 34
We already know that 340 is divisible by 17 and it is easy to see that 34 is also divisible by 17.
Hence, 374 is divisible by 17!

Let’s build further upon this useful concept.
Multiple of n ± multiple of n = MUST be a multiple of n.
Multiple of n ± NOT multiple of n = CANNOT be a multiple of n.

How about the case of:
NOT multiple of n ± NOT a multiple of n?
Can we decide on a rule for the result?
A) NOT multiple of n ± NOT a multiple of n – MUST be a multiple of n
B) NOT multiple of n ± NOT a multiple of n – CANNOT be a multiple of n
C) NOT multiple of n ± NOT a multiple of n – May or May NOT be a multiple of n

Take our previous example of 4:
2 and 1 are NOT multiples of 4, and the result of 2+1=3 isn’t a multiple of 4 either.
However, 3 and 1 are NOT multiples of 4, but the result of 3+1=4 IS a multiple of 4.
Therefore, NOT multiple of n ± NOT a multiple of n – May or May NOT be a multiple of n.

To sum up:
Multiple of n ± multiple of n = MUST be a multiple of n.
Multiple of n ± NOT multiple of n = CANNOT be a multiple of n.
NOT multiple of n ± NOT a multiple of n – May or May NOT be a multiple of n

Now, let’s try an example:

If a and b are integers, is a+b divisible by 5?
(1) a is divisible by 5.
(2) b is divisible by 5.

Stat. (1) doesn’t tell you anything about b. So a and b could both be 5 (a multiple of 5), giving a result of a+b=5+5=10 and answer of ‘yes’. But plug in a=5 and b=4 to get a result of a+b=9, which is not a multiple of 5, giving an answer of ‘no. Therefore, Stat (1)->Maybe->IS->BCE
Likewise, Stat. (2) doesn’t tell you anything about a. So a and b could both be 5 (a multiple of 5), giving a result of a+b=5+5=10 and answer of ‘yes’. But plug in b=5 and a=4 to get a result of a+b=9, which is not a multiple of 5, giving an answer of ‘no’. Therefore, Stat (2)->Maybe->IS->CE
Finally, consider both statements combined: a and b are now both multiples of 5. According to our abovementioned rule, multiple of 5 ± multiple of 5 – MUST be a multiple of 5. Therefore, a+b is a multiple of 5, and the answer is ‘yes’, or sufficient. Stat (1)+(2)->Yes->S->C

That was pretty easy so let’s try a tougher question.

k=2^n + 7, where n is an integer greater than 1. If k is divisible by 9, which of the following MUST be divisible by 9?
2^n-8
2^n-2
2^n
2^n+4
2^n+5

8 comments

  • k = 2^n + 7 = 9*p(where p is a positive integer and n is an integer greater then 1)

    9p - 9 is also a multiple of 9. 2^n + 7 - 9 = 2^n - 2 is a multiple of 9.

    • time saving :)

    • Yeah!!!

  • 2^n-2
    This is because 2^n-2 is 9 away from 2^n+7

    • Most easiest way!!!

    • Very helpful! 

  • 2^n+7 is div by 9 

    so 2^n+7 + 9N and 2^n+7 - 9N
    will be divisible by 9 too 

    (N=1,2..)

    So, 9 goes in to 2^n-2 

  • Hello - can you please post the solution - I did not understand the explanation given by some of the members here
    Many thanks

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