# Proving that dividing by 5 is possible

• Dec 1st 2010, 06:58 AM
Evaldas
Proving that dividing by 5 is possible
Prove that $\displaystyle 2^{60}+2^{59}-2^{58}$ 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.
• Dec 1st 2010, 07:02 AM
e^(i*pi)
Quote:

Originally Posted by Evaldas
Prove that $\displaystyle 2^{60}+2^{59}-2^{58}$ 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 $\displaystyle a^{b+c} = a^ba^c$

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

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

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

Obviously there being a 5 means it is divisible by 5
• Dec 1st 2010, 07:06 AM
Evaldas
Quote:

Originally Posted by e^(i*pi)
Use that fact that $\displaystyle a^{b+c} = a^ba^c$

But is that the same as $\displaystyle a^b+a^c$?
• Dec 1st 2010, 07:15 AM
e^(i*pi)
No! It most definitely isn't. For example $\displaystyle 2^6 \neq 2^3 + 2^3 \implies 64 \neq 8+8$. You may have to (re)visit the laws of exponents if you don't understand

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

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

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

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

Rewriting the original equation using my subsitution gives:

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

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

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

The part in the brackets can be simplified to 5 meaning this expression is divisible by 5
• Dec 1st 2010, 07:19 AM
Evaldas
Ok, thanks ;)
• Dec 1st 2010, 02:08 PM
Bacterius
Or even more easily, if you're familiar with number theory :

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

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

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

Hence $\displaystyle 2^{60}+2^{59}-2^{58}$ is divisible by $\displaystyle 5$.
• Dec 2nd 2010, 01:55 AM
Evaldas
That was supposed to be an exercise for 14 year olds ;), so I don't think a 14 year old knows about number theory ;)
• Dec 2nd 2010, 02:00 AM
Bacterius
There is no age to mathematics :D