# Thread: need help!!! math counts is uber hard.

1. ## need help!!! math counts is uber hard.

PLEEEASE help me with this! I'm begging!

1) find the mean of all digits which cannot be the last digit of a perfect square.

2) when 2^ 100 + 3^100 is divided by 5, what is the remainder?

3) show that in any pythagorean triple, one of the numbers must be a multiple of 5.

2. I can help with #2:

Find a pattern:

$\displaystyle 2^1 = 2 \rightarrow 2^2 = 4 \rightarrow 2^3 = 8 \rightarrow 2^4 = 16 \rightarrow 2^5 = 32 \rightarrow \mbox{...}$

The last digit repeats in sequences of 4. So $\displaystyle 2^{101}$ has last digit 2. So $\displaystyle 2^{100}$ has last digit 6.

$\displaystyle 3^1 = 3 \rightarrow 3^2 = 9 \rightarrow 3^3 = 27 \rightarrow 3^4 = 81 \rightarrow 3^5 = 243 \rightarrow \mbox{...}$

The last digit repeats in sequences of 4. So $\displaystyle 3^{101}$ has last digit 3. So $\displaystyle 3^{100}$ has last digit 1.

So the last digit of $\displaystyle 2^{100} + 3^{100}$ is 7. Therefore the remainder when divided by 5 will be 2.

3. #1: Consider the units digits. The last digit of their squares are representative of all numbers that end in the same digit.

So: $\displaystyle 0^2 = 0, \ \ 1^2 = 1, \ \ 2^2 = 4, \ \ \cdots$

Whatever is not on the list you can simply average to get your answer.

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#2: Note that $\displaystyle 2^4 \equiv 1 \ (\text{mod } 5)$ and $\displaystyle 3^2 \equiv -1 \ (\text{mod }5)$

So:
$\displaystyle 2^{100} \equiv \left(2^4\right)^{25} \equiv (1)^{100} \equiv 1 \ (\text{mod } 5)$
$\displaystyle 3^{100} \equiv \left(3^2\right)^{50} \equiv (-1)^{50} \equiv 1 \ (\text{mod } 5)$

So what can you conclude about $\displaystyle 2^{100} + 3^{100} \ (\text{mod } 5)$ ?

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#3: Here

4. Hello, jj1004!

We can investigate the first one . . .

1) Find the mean of all digits which cannot be the last digit of a perfect square.

Examine the last digit of all possible squares.

. . $\displaystyle \begin{array}{c|c}x\text{ ends in:} & x^2\text{ ends in:} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ 4 & 6 \\ 5 & 5 \\ 6 & 6 \\ 7 & 9 \\ 8 & 4 \end{array}$
. . . . . $\displaystyle \begin{array}{c|c} \;9\quad\quad & \qquad\, 1 \end{array}$

We see that a square cannot end in 2, 3, 7, or 8.

Their mean is: .$\displaystyle \frac{2+3+7+8}{4} \:=\:5$

5. ## Pythagorean Triples

Hi -

The only problem with the proof in this link - Here - is that it only works with triples generated by the 'm' and 'n' formula. And you can't be sure that this formula will generate all Pythagorean Triples. (Or can you??)

As an alternative, which will work for any Pythagorean Triple, how about something like:

Suppose $\displaystyle a, b, c$ is the PT where $\displaystyle a^2=b^2+c^2$

Then $\displaystyle c^2=a^2-b^2=(a+b)(a-b)$

Now $\displaystyle c$ is either a multiple of $\displaystyle 5$ (QED) or:

$\displaystyle c \equiv \pm 1 \text{ (mod }5)$

or $\displaystyle c \equiv \pm 2 \text{ (mod }5)$

$\displaystyle \implies c^2 \equiv 1 \text{ (mod }5)$

or $\displaystyle c^2 \equiv 4 \text{ (mod }5)$

Now consider the multiplication table (mod 5) on {0, 1, 2, 3, 4}:

(Not quite sure how to do matrices in LaTeX yet, so this will have to do.)

mod 5
x | 0 1 2 3 4
---------------
0 | 0 0 0 0 0
1 | 0 1 2 3 4
2 | 0 2 4 1 3
3 | 0 3 1 4 2
4 | 0 4 3 2 1

Now consider the product $\displaystyle (a+b)(a-b) \text{ (mod }5)$. From the table above, the only time it is 1 or 4 is where:

$\displaystyle a+b=a-b$

$\displaystyle \implies b \equiv 0 \text{ (mod }5)$ (QED)

or where:

$\displaystyle (a+b)+(a-b) =0 \text{ (mod }5)$

$\displaystyle \implies a \equiv 0 \text{ (mod }5)$ (QED)