CalculatorUser

A similar question was posted a long time ago

the first few terms are 1, 2, 2, 4, 8, 32, 256, 8192, 2097152...

Notice how the last digits are forming a pattern, (2, 2, 4, 8, 2, 6, 2, 2, 4, 8, 2, 6, 2...

The pattern is 2 -> 2 ->  4 -> 8 ->  2 -> 6

So if the pattern repeats every 6 terms. And assuming you are trying to find the 35th term. You divide 35 by 6 and find the remainder, which is 5

So count up 5 terms in the pattern.

[fixed]

we count up 4 terms in the pattern because the first term is 1

so 8 is the answer

Someone check this.

Note: In these questions, always check for a pattern!!!

We need to figure out the denominator first of the probability fraction

100-999

That is 900 digits

Ok so our denominator is 900

Now we need to figure out the numerator. We do this by finding all the three digit numbers divisible by 11

We try listing (I couldn't find any better way to do this

110

121

132

143

154

165

176

187

198

----

209

220

231

242

253

264

275

286

297

----

Notice how the first number's last two digits start from 10 then decreases to 9 at 200

notice how the last number last two digits start from 98 and decrease to 97 at 200

This means each hundred we have 9 numbers divisible by 11

This means that there are 9*9 (9 different hundreds) = 81 numbers 

We have the numerator which is 81 and the denominator which is 900

Simplifying we have \(\boxed{\frac{9}{100}}\)

\(\frac{w^6}{z^8} \Rightarrow \frac{w*w*w*w*w*w}{z*z*z*z*z*z*z*z}\)

Then you make a fraction with w/z. \(\frac{w^6}{z^6}\), and also multiply it by 1/z^2

So you have a new expression of \(\frac{w^6}{z^6}*\frac{1}{z^2}\)

I don't know if that is what you are looking for...

By the second digit from the left, I assume you mean 5n451 where n is the second digit from the left

1

3

5

7

9

5 ways for the second digit from the left

1 (0 or 1 for the last digit)

3 (0 or 1 for the last digit)

5 (0 or 1 for the last digit)

7 (0 or 1 for the last digit)

9 (0 or 1 for the last digit)

10 ways for the second digit and last digit

Now the first digit can be 1 through 9 so 10*9 => 90

So the first digit, the second digit from the left, and the last digit is taken care of.

The 3rd and 4th digit can be 100 ways as it can be 00 to 99.

So 90*100 => 9000???

I dont know I came up this quickly I dont have much time.

Yes, please, I need to understand this.

Ohhh, ok, I totally screwed up. Thanks Tertre!

Um I had an entirely different answer (probably not correct I will have to see what the answer is)

Using stars and bars

| | | Which represents nephews

O O O O F F F T T H Which represents money

So we have | | | O O O O F F F T T H
 

Ok so I think it is permutation.

So permutation formula

13! / (13-10)!

1716 ways???

Please, if someone is good at this, explain and check our answers.

This problem was answered before

There are 2^10 different subsets in the original set.

Now let us just subtract subsets without a 5

Let us just pretend you have set without a 5

So {1 2 3 4 6 7 8 9 10} Is your new set

So how many subsets can you make from that?

2^9 subsets

2^10 - 2^9 = 512 different subsets

NOTE: 2^10 - 2^9 is NOT 2.

So there you have it! :D:D:D

3 * 4 = \(\boxed{12}\)

This is true because

Red - Pattern 1 or 2 or 3 or 4                      So 4 design

Green - Pattern 1 or 2 or 3 or 4                   So 4 design

Blue - Pattern 1 or 2 or 3 or 4                      So 4 design

So 4 + 4 + 4 = 12

Thanks, I will try doodling now.