If this gets a post ID of 32768, we advance!

[sparkycinnamon]

space doughnut my platonic beloved

[Gilbert189]

Fun fact: Every dice that follows "all opposite sides sums to 7" will always have a 1-2-3 group that look like this:

or this:

[PkmnQ]

Fun fact: Every dice that follows "all opposite sides sums to 7" will always have a 1-2-3 group that look like this:

or this:

Yeah, because after placing the 1 and 2, you can only put the 3 to the left (second image) or the right (first image).

[PkmnQ]

D1Q6-331B just a copy of Queue

[Byron_Inc_TBG]

i accidentally created a math problem

how many 10-digit numbers contain both 20 and 25 in them

So, permutation?
Since 20 and 25 are locked together, we can consider them as 1 digit (like when you were to solve "how many numbers contain both 2 and 7")

Let's try observing smaller numbers to see if we can sense a pattern.
4 digits: 2025, 2520 (2P2 = 2 choices)
It's quite simple here, there are only two options since you can't mix in other digits.

5 digits: a2025, a2520, 2025a, 20a25, 25a20, 2520a (3P2 * 10 - 2 = 58 choices)
Things start to get complicated, but not quite complicated enough.
There are 3P2 = 6 ways to put in numbers, and digit a can be any number between 0 and 9 (10 digits)
We have to subtract by 2 since digit a cannot be 0 in a2025 and a2520 (those are 4 digit numbers, not 5)

6 digits: ab2025, ab2520, a2025b, a2520b, a20b25, a25b20, 2025ab, 2520ab, 20ab25, 25ab20, 20a25b, 25a20b (4P2 * 10 * 10 - 3P2 * 10 - 3P2 = 1134)
Oh yeah, things start to really get bad here.
We start off by multiplying 4P2 = 12 methods by 10 * 10 = 100 ways to fill in digits a and b.
Then, we subtract 3P2 * 10 since we can't have a heading 0 in ab2025, ab2520, a2025b, a2520b, a20b25, a25b20 (Since we need a to be in front, it's a sorting between "20", "25", and b, thus 3P2)
After that, we have 3P2 = 6 numbers that fall into two categories: 202025 (ab2025 & 20ab25), 202520 (ab2520 & 2025ab), 202525 (2025ab & 20ab25), 252020 (2520ab & 25ab20), 252025 (ab2025 & 2520ab), 252520 (25ab20 & ab2520)

7 digits: abc2025, abc2520, ab20c25, ab25c20, ab2025c, ab2520c, a20b25c, a25b20c, a2025bc, a2520bc, a20bc25, a25bc20, 2025abc, 2520abc, 20a25bc, 25a20bc, 20ab25c, 25ab20c, 20abc25, 25abc20 (5P2 * 1000 - 4P2 * 10 * 10 - 4P2 * 10 * 2 = 18566)
Oh boy, I'm so very confused.
You get the drill, 5P2 = 20 methods multiplied by 10^3 = 1000 is equal to 20000.
Then we exclude the ones where a is in front and equal to 0 (4P2 * 10^2 = 1200)
We also don't want to include repeats. This is calculated by adding an extra pair, making the number of options 4, thus we do 4P2. We also multiply it by 2, since this extra pair can be both 20 and 25.
However, we double-excluded some numbers, like 0202025.


I quit solving this mathematically, so here's some code for you
Note that tio.run can't handle this thing, so you might need to run it in an IDLE.

Code Select Expand

ans = []
for i in range(int(1e10), int(1e11)):
if "20" in str(i) and "25" in str(i):
ans.append(i)

print(len(ans))
print(ans)

[TonyBrown148]

I am now training a Neural Network.

[Gilbert189]

I am now training a Neural Network.

how good is it

[TonyBrown148]

89% correct on fashion mnist

[Byron_Inc_TBG]

dang you need to be 18 or older to use openai

[PkmnQ]

i accidentally created a math problem

how many 10-digit numbers contain both 20 and 25 in them

So, permutation?
Since 20 and 25 are locked together, we can consider them as 1 digit (like when you were to solve "how many numbers contain both 2 and 7")

Let's try observing smaller numbers to see if we can sense a pattern.
4 digits: 2025, 2520 (2P2 = 2 choices)
It's quite simple here, there are only two options since you can't mix in other digits.

5 digits: a2025, a2520, 2025a, 20a25, 25a20, 2520a (3P2 * 10 - 2 = 58 choices)
Things start to get complicated, but not quite complicated enough.
There are 3P2 = 6 ways to put in numbers, and digit a can be any number between 0 and 9 (10 digits)
We have to subtract by 2 since digit a cannot be 0 in a2025 and a2520 (those are 4 digit numbers, not 5)

6 digits: ab2025, ab2520, a2025b, a2520b, a20b25, a25b20, 2025ab, 2520ab, 20ab25, 25ab20, 20a25b, 25a20b (4P2 * 10 * 10 - 3P2 * 10 - 3P2 = 1134)
Oh yeah, things start to really get bad here.
We start off by multiplying 4P2 = 12 methods by 10 * 10 = 100 ways to fill in digits a and b.
Then, we subtract 3P2 * 10 since we can't have a heading 0 in ab2025, ab2520, a2025b, a2520b, a20b25, a25b20 (Since we need a to be in front, it's a sorting between "20", "25", and b, thus 3P2)
After that, we have 3P2 = 6 numbers that fall into two categories: 202025 (ab2025 & 20ab25), 202520 (ab2520 & 2025ab), 202525 (2025ab & 20ab25), 252020 (2520ab & 25ab20), 252025 (ab2025 & 2520ab), 252520 (25ab20 & ab2520)

7 digits: abc2025, abc2520, ab20c25, ab25c20, ab2025c, ab2520c, a20b25c, a25b20c, a2025bc, a2520bc, a20bc25, a25bc20, 2025abc, 2520abc, 20a25bc, 25a20bc, 20ab25c, 25ab20c, 20abc25, 25abc20 (5P2 * 1000 - 4P2 * 10 * 10 - 4P2 * 10 * 2 = 18566)
Oh boy, I'm so very confused.
You get the drill, 5P2 = 20 methods multiplied by 10^3 = 1000 is equal to 20000.
Then we exclude the ones where a is in front and equal to 0 (4P2 * 10^2 = 1200)
We also don't want to include repeats. This is calculated by adding an extra pair, making the number of options 4, thus we do 4P2. We also multiply it by 2, since this extra pair can be both 20 and 25.
However, we double-excluded some numbers, like 0202025.


I quit solving this mathematically, so here's some code for you
Note that tio.run can't handle this thing, so you might need to run it in an IDLE.

Code Select Expand

ans = []
for i in range(int(1e10), int(1e11)):
if "20" in str(i) and "25" in str(i):
ans.append(i)

print(len(ans))
print(ans)

I tried a haskell program, it almost crashed my computer

[TonyBrown148]

I quit solving this mathematically, so here's some code for you
Note that tio.run can't handle this thing, so you might need to run it in an IDLE.

Use dynamic programming.

[TonyBrown148]

Let f(n,m) be the number of n-digit numbers ending with m that does not contain 20.
Let g(n,m) be the number of n-digit numbers ending with m that does not contain 25.
Let h(n,m) be the number of n-digit numbers ending with m that does not contain 20 or 25.
Then the answer is (total numbers)-f-g+h.

[PkmnQ]

56000000 - 2100000 + 10400 - (15600 + 10) + 0 - 20 = 53894770

I think that's the answer. There are 53894770 10-digit numbers with both 20 and 25 in them. I'm glad it wasn't asking about 23 and 32.

[Byron_Inc_TBG]

i accidentally created a math problem

how many 10-digit numbers contain both 20 and 25 in them

So, permutation?
Since 20 and 25 are locked together, we can consider them as 1 digit (like when you were to solve "how many numbers contain both 2 and 7")

Let's try observing smaller numbers to see if we can sense a pattern.
4 digits: 2025, 2520 (2P2 = 2 choices)
It's quite simple here, there are only two options since you can't mix in other digits.

5 digits: a2025, a2520, 2025a, 20a25, 25a20, 2520a (3P2 * 10 - 2 = 58 choices)
Things start to get complicated, but not quite complicated enough.
There are 3P2 = 6 ways to put in numbers, and digit a can be any number between 0 and 9 (10 digits)
We have to subtract by 2 since digit a cannot be 0 in a2025 and a2520 (those are 4 digit numbers, not 5)

6 digits: ab2025, ab2520, a2025b, a2520b, a20b25, a25b20, 2025ab, 2520ab, 20ab25, 25ab20, 20a25b, 25a20b (4P2 * 10 * 10 - 3P2 * 10 - 3P2 = 1134)
Oh yeah, things start to really get bad here.
We start off by multiplying 4P2 = 12 methods by 10 * 10 = 100 ways to fill in digits a and b.
Then, we subtract 3P2 * 10 since we can't have a heading 0 in ab2025, ab2520, a2025b, a2520b, a20b25, a25b20 (Since we need a to be in front, it's a sorting between "20", "25", and b, thus 3P2)
After that, we have 3P2 = 6 numbers that fall into two categories: 202025 (ab2025 & 20ab25), 202520 (ab2520 & 2025ab), 202525 (2025ab & 20ab25), 252020 (2520ab & 25ab20), 252025 (ab2025 & 2520ab), 252520 (25ab20 & ab2520)

7 digits: abc2025, abc2520, ab20c25, ab25c20, ab2025c, ab2520c, a20b25c, a25b20c, a2025bc, a2520bc, a20bc25, a25bc20, 2025abc, 2520abc, 20a25bc, 25a20bc, 20ab25c, 25ab20c, 20abc25, 25abc20 (5P2 * 1000 - 4P2 * 10 * 10 - 4P2 * 10 * 2 = 18566)
Oh boy, I'm so very confused.
You get the drill, 5P2 = 20 methods multiplied by 10^3 = 1000 is equal to 20000.
Then we exclude the ones where a is in front and equal to 0 (4P2 * 10^2 = 1200)
We also don't want to include repeats. This is calculated by adding an extra pair, making the number of options 4, thus we do 4P2. We also multiply it by 2, since this extra pair can be both 20 and 25.
However, we double-excluded some numbers, like 0202025.


I quit solving this mathematically, so here's some code for you
Note that tio.run can't handle this thing, so you might need to run it in an IDLE.

Code Select Expand

ans = []
for i in range(int(1e10), int(1e11)):
if "20" in str(i) and "25" in str(i):
ans.append(i)

print(len(ans))
print(ans)

I tried a haskell program, it almost crashed my computer

i threw it to replit
let's see how can it handle

[Gilbert189]

i accidentally created a math problem

how many 10-digit numbers contain both 20 and 25 in them

I'm gonna try to solve it. Let's observe a number that met this condition.

3120542503

If we mark 20 and 25:

3120542503

we see that 20 and 25 looks like it separates the numbers into what looks like 3 chunks. Thus, we can define this as shown below.

AXBXC

We define 20 and 25 as X, since they are both the same size, however we have to double the possibilities later. However, we can't do the same with A, B, and C, since reordering it will generate a different number (think 3120525403 and 3120403255)
What we can do though is omit X, leaving us with ABC. This gives us a 6 digit number, which we can calculate the amount of possible numbers thereof.

315403

Assuming trailing zeros aren't allowed, this leaves us with 10^6-10^5 = 900000 6-digit numbers.

Now, we need to "cut" this 6-digit number to put back X. This is essentially that sticks and stones problem.

31X54X03

With 6 stones (digits) and 2 sticks (X), we get ((6,2)) = 7C2 = 7!/(2!*5!) = 7*6*5!/(2*5!) = 7*3 = 21 possibilities of putting 2 Xes into ABC.

Combining these possibilities together, we get 900000*21*2 = 37800000 numbers. I guess.

[TonyBrown148]

Among the 9,000,000,000 10-digit numbers, 8,950,159,730 does not contain both 20 and 25.
So the answer is 49,840,270.

[Gilbert189]

making a colab notebook to solve it

well looks like we're gonna be in here for a while

[TonyBrown148]

PkmnQ: 53,894,770
Gilbert: 3,780,000
TB148: 49,840,270

[TonyBrown148]

making a colab notebook to solve it

well looks like we're gonna be in here for a while

There are only 9,000,000,000 10-digit numbers, not 90,000,000,000.
They are in the range [10^9,10^10).

[Gilbert189]

making a colab notebook to solve it

well looks like we're gonna be in here for a while

There are only 9,000,000,000 10-digit numbers, not 90,000,000,000.

that's what Byron's code gave me

[TonyBrown148]

that's what Byron's code gave me

byron is wrong. it's range(10**9,10**10)

[PkmnQ]

56000000 - 2100000 + 10400 - (15600 + 10) + 0 - 20 = 53894770

I think that's the answer. There are 53894770 10-digit numbers with both 20 and 25 in them. I'm glad it wasn't asking about 23 and 32.

Let me show how I worked it out:

Counted 1 or more times:

Code Select Expand

2025######
20#25#####
20##25####
20###25###
20####25##
20#####25#
20######25
#2025#####
#20#25####
...
######2025
2520######
...
2*(7+6+5+4+3+2+1) = 56
56 * 10^6 = 56000000
Counted 2 or more times:

Code Select Expand

202025####
2020#25###
2020##25##
2020###25#
2020####25
...
20####2025
#202025###
...
####202025
...
202520####
...
252020####
...
252520####
...
252025####
...
202525####
...6*((5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+1) = 210
210 * 10^4 = 2100000
Counted 3 or more times:

Code Select Expand

1. 20202025##
 2. 202020#25#
 3. 202020##25
 4. 2020#2025#
 5. 2020#20#25
 6. 2020##2025
 7. 20#202025#
 8. 20#2020#25
 9. 20#20#2025
10. 20##202025
11. #20202025#
12. #202020#25
13. #2020#2025
14. #20#202025
15. ##2020202515 * 4 * 2 = 120
120 * 10^2 = 12000
Ok, looks like I made a mistake in the first post
Counted 4 or more times:
Case A.

Code Select Expand

(placement same as counted 3 times)
1. 20202525
2. 20252025
3. 20252520
4. 25202025
5. 25202520
6. 2525202015 * 6 = 90
90 * 10^2 = 9000
Case B.

Code Select Expand

1. 2020202025
2. 2020202520
3. 2020252020
4. 2025202020
5. 2520202020
5 * 2 = 10
Counted 5 or more times:
None
Counted 6 or more times:

Code Select Expand

1. 2020202525
 2. 2020252025
 3. 2020252520
 4. 2025202025
 5. 2025202520
 6. 2025252020
 7. 2520202025
 8. 2520202520
 9. 2520252020
10. 252520202010 * 2 = 20

56000000 - 2100000 + 12000 - (9000 + 10) + 0 - 20 = 53902970
There, that's the correct answer

[Gilbert189]

that's what Byron's code gave me

byron is wrong. it's range(10**9,10**10)

fixed it

[PkmnQ]

Assuming trailing zeros aren't allowed

I forgot to mention, the original question was about phone numbers, I just used 10 digit numbers to make it clearer. Trailing zeroes are allowed.

[TonyBrown148]

The correct answer is 53920970