# Thread: Proving that dividing by 5 is possible

1. ## Proving that dividing by 5 is possible

Prove that $2^{60}+2^{59}-2^{58}$ can be divided by 5.
Obviously if a number can be divided by five it either ends with a 0 or a 5. But how do you ADD numbers with exponents? I know how to divide, multiply, but not add or subtract. And my calculator doesn't fit that many numbers.

2. Originally Posted by Evaldas
Prove that $2^{60}+2^{59}-2^{58}$ can be divided by 5.
Obviously if a number can be divided by five it either ends with a 0 or a 5. But how do you ADD numbers with exponents? I know how to divide, multiply, but not add or subtract. And my calculator doesn't fit that many numbers.
Use that fact that $a^{b+c} = a^ba^c$

$2^{60} = 2^{58 + 2} = 4 \cdot 2^{58}$

$2^{59} = 2^{58+1} = 2 \cdot 2^{58}$

You can then factor out $2^{58}$ as a common factor: $2^{58}(4+2-1) = 5 \cdot 2^{58}$

Obviously there being a 5 means it is divisible by 5

3. Originally Posted by e^(i*pi)
Use that fact that $a^{b+c} = a^ba^c$
But is that the same as $a^b+a^c$?

4. No! It most definitely isn't. For example $2^6 \neq 2^3 + 2^3 \implies 64 \neq 8+8$. You may have to (re)visit the laws of exponents if you don't understand

What I did was to say that $2^{60} = 2^{58+2}$ which is easy enough to verify.

It is a standard law of exponents that $a^{b+c} = a^b \times a^c$. If you think about the example above: $2^6 = 2^3 \times 2^3 \implies 64 = 8 \times 8$ which is fine.

In other words $2^{60} = 2^{2+58} = 2^2 \cdot 2^{58}$ Likewise $2^{59} = 2^{1+58} = 2^1 \times 2^{58}$

I chose 2^{58} because that is the greatest common factor of the original expression.

Rewriting the original equation using my subsitution gives:

$2^2 \cdot 2^{58} + 2^1 \cdot 2^{58} - 2^{58}$

Exponents behave in the same way as other numbers when it comes to factorising so you can factor out $2^{58}$

$2^{58}(2^2+2^1-1)$

The part in the brackets can be simplified to 5 meaning this expression is divisible by 5

5. Ok, thanks

6. Or even more easily, if you're familiar with number theory :

$\varphi{(5)} = 4$, Euler's Theorem states $a^b \equiv a^{b \mod{\varphi{(n)}}} \pmod{n}$ if $\gcd{(a, n)} = 1$.

$2^{60}+2^{59}-2^{58} \equiv 2^{60 \mod{4}} + 2^{59 \mod{4}} - 2^{58 \mod{4}}$

Which gives : $\equiv 2^0 + 2^3 - 2^2 \equiv 1 + 8 - 4 \equiv 5 \equiv 0 \pmod{5}$

Hence $2^{60}+2^{59}-2^{58}$ is divisible by $5$.

7. That was supposed to be an exercise for 14 year olds , so I don't think a 14 year old knows about number theory

8. There is no age to mathematics