LEARNATHON
III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

Theory:

Result I: If a number can be divided by another number, then it can also be divided by each of its factor.
Example:
Lets take two number $$18$$ and $$72$$.

$$72\div18 = 4$$ this show $$72$$ is divisible by $$18$$,

Factors of  $$18 = 1, 2, 3, 6, 9, 18$$.

Now we divide the factors of $$18$$ with $$72$$.

$$72\div1=72$$, $$72\div2=36$$, $$72\div3=24$$, $$72\div6=12$$, $$72\div9=8$$ and $$72\div18=4$$.

Thus, $$72$$ is divisible by each of factors of $$18$$.
Result II: If a number is divided by two co-prime numbers, it is also divisible by its product
Example:
Let's say $$90$$ is divisible by $$5$$ and $$9$$. As we know $$5$$ and $$9$$ are co-prime numbers.

Thus, the product of co-primes is ($$5\times9 = 45$$).

$$90\div45 = 2$$.

Therefore $$90$$ is divisible by the product of co-primes $$45$$.
Result III: If two given numbers are divisible by a number, then, their sum is also divisible by that number.
Example:
Let us take $$21$$ and $$18$$. Both the numbers are divisible by $$3$$.

$$21\div3 = 7$$ and $$18\div3 = 6$$.

Sum of the two numbers is $$21+18 = 39$$. Also $$39\div 3 = 13$$.

Therefore, if $$24$$ and $$18$$ are divisible by $$3$$, then their sum is $$39$$ which is also divisible by $$3$$.
Result IV: If two given numbers are divisible by a number, then their difference is also divisible by that number.
Example:
Let us take $$58$$ and $$54$$ are divisible by $$2$$.

$$958\div2 = 29$$ and $$54\div2 = 27$$.

Difference of the two numbers that $$58-54 = 4$$ and $$4\div2 = 2$$.

Therefore, if $$54$$ and $$58$$ are divisible by $$2$$, then their difference $$4$$ is also divisible by $$2$$.
More examples:

1. A number is divisible by both $$7$$ and $$15$$. By which other numbers will that number be always divisible?

Note that the number is divisible by $$7$$ and $$15$$. Let us recall 'Result II'.
Result II: If a number is divided by two co-prime numbers, it is also divisible by its product.
Since $$7$$ and $$15$$ are co-prime numbers, the number must be divisible by the product $$7\times 15 = 105$$.

So, the given number will always be divisible by $$105$$.

2. A number is divisible by $$16$$. By what other numbers will that number be divisible?

As we need to find the more divisible for the given number, let us recall 'Result I'.
Result I: If a number can be divided by another number, then it can also be divided by each of its factor.
The number is divisible by$$16$$.

The factors of $$16$$ are $$1, 2, 4, 8, 16$$.

Therefore, the number is also divisible by $$1, 2, 4, 8$$.