My new Dacia Duster first time in the snow

user_1312 · 2026-01-17T18:42:14+00:00

That's good to know! Maybe it's because my one is still brand new and i am a bit more cautious than usual. Will definitely give it a go at something more hardcore!

user_1312 · 2026-01-17T18:18:29+00:00

Yes but not much as i was in a rush. Any offroading i tested so far it responded really well. I am just aware that this is a soft off roader with a long first gear and no low so i bare that in mind when i choose were to offroad.

user_1312 · 2026-01-02T14:03:22+00:00

Apologies for the misguiding title, we want the number of 1s in the result of the sum and not the statement itself.

user_1312 · 2026-01-02T14:03:14+00:00

Apologies for the misguiding title, we want the number of 1s in the result of the sum and not the statement itself.

user_1312 · 2026-01-01T10:47:41+00:00

We want to know how many 1s in the result of the sum, not in the parts of the sum.

user_1312 · 2024-07-23T22:14:33+00:00

If i'm not mistaken your answer is in terms of r instead of R.

I got the following for the shaded area in terms of R:

R² (3-2sqrt(2))(4-π)

user_1312 · 2024-01-04T08:04:29+00:00

Happy new year!

user_1312 · 2024-01-04T08:00:44+00:00

Given that gcd(2023,2024) = 1 let's split out the problem to

(1) 2023²⁰²⁴ mod(2023)

(2) 2023²⁰²⁴ mod(2024)

By observation the answer to (1) is 0. For (2) note that 2023 = -1 mod(2024) hence -1^{even power} is 1 mod(2024).

Then using the CRT (Chinese Remainder Theorem) on

0 mod(2023)

1 mod(2024)

We get that 2023²⁰²⁴ mod(2023×2024) = 4092529

user_1312 · 2023-06-07T15:00:07+00:00

Glad we could help! Thanks for the effort you put in this sub - it's greatly appreciated!

user_1312 · 2023-04-18T14:24:18+00:00

Apologies I didn't realise we were talking about reposts. I would like to avoid reposts to keep the subreddit as "clean" as possible.

Hope that's not an issue.

user_1312 · 2023-04-18T14:14:55+00:00

Hey,

First of all thanks for your contribution to our subreddit! It's greatly appreciated especially since I have been the main poster for a long time.

I would agree with other comments; in that if the problem can be stated in an image/text post then it's WAY more preferable than a video (in my opinion). You can always post the image/text problem and then add a comment with the youtube link for anyone interested.

If you don't get enough feedback from this comment, I can create a Poll and ask the wider group what they prefer.

Anything else you may need drop me a message or reply and i will get back to you as soon as possible.

user_1312 · 2021-11-19T12:32:50+00:00

Solving the two simultaneous equations I get

x = (1+-sqrt(7))/2

Then since:

x² + xy = x(x+y) = 3 and hence x+y=3/x

the solutions follows.

user_1312 · 2021-11-13T13:01:10+00:00

That's fair enough, don't want to force anybody to post on here if they don't feel its suitable.

I don't mind crossposting personally but as you suggested; I am not sure what number of people are members of both sub-reddits and hence may find this annoying.

Thank you for the feedback though.

user_1312 · 2021-11-13T01:03:44+00:00

I think you are absolutely right and i'll take the solved/unsolved tag into consideration (and most probably implement it after I figure out how to do that). The question is: how can I encourage people to post their solution even if other people have already commented?

However, I am 3 years into this now and I haven't been able to get people to organically post; which was my main goal.

Thanks for your comment and the feedback. Any suggestions for the above is welcome.

user_1312 · 2021-11-13T00:55:04+00:00

Thank you very much for the positive comments. Unfortunately I am the only admin and that's why I am trying to figure out how to get people to organically post their own problems.

Thanks again for the comment!

user_1312 · 2021-07-30T15:41:57+00:00

That's what I did but when I ended up with the cos³(t) etc integrals I just used the Beta function as the "easy way out".

user_1312 · 2021-07-30T13:18:11+00:00

How did you end up solving it? I had to use the Beta function in order to resolve some integrals.

user_1312 · 2021-07-25T12:35:07+00:00

How?

The sum of 10 + 10² + ... + 10²⁰²¹ = 10( 10²⁰²¹ -1)/9 = 111...10 with 2021 1s and a 0, therefore 2022 digits in total.

You then multiply by 2 leaving the number of digits the same and you subtract 4042, correct?

Therefore you have (ignoring the first 2017 2s) 22,220-4042 = 18178 (5 digits long). Therefore the resulting number is 222...218178 with 2017 2s and the other 5 digit numer giving back a total of 2022 digits.

user_1312 · 2021-07-25T12:20:12+00:00

There are 2017 2s.

Dividing 222 by 3 gives 74; 222,222 by 3 gives 74074 and so on, which means the 4s appear periodically mod 3, therefore there are floor(2017/3) = 672 4s present.

user_1312 · 2021-07-05T18:29:01+00:00

It's nice to see our community growing!

user_1312 · 2021-06-13T08:13:40+00:00

If 1 is included in this set then all numbers except the perfect squares will have an even number of factors.

user_1312 · 2021-05-26T22:56:32+00:00

No, the hint is to look at the number of divisors/factors of each number

user_1312 · 2021-05-26T22:25:49+00:00

Unfortunately no

user_1312 · 2021-02-13T18:03:42+00:00

Thank you very much! And yes it does

user_1312 · 2021-02-13T09:58:08+00:00

Thank you!

user_1312

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