Відео

ua-cam.com/video/GwkRSt0rasg/v-deo.html

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ဟုတ်တာပေါ့

ဂဏန်းတွက်စက် သုံးရတာပါ Scientific calculator သို့မဟုတ် 0.9^0.9 လို့ရိုက်လို့ရပါတယ်။

💯

ခန့်မှန်းတန်ဖိုး ရှာလို့ရပါတယ် ua-cam.com/video/GwkRSt0rasg/v-deo.html

Thank you.

Intersection point, D, of AP and XX' is on the line AP, where AD is twice of AI. Similarly, AP and YY' intersect at a point where AD is twice of AI. So, they intersect at one point, D.

See clearly it forms a rhombus

ဟုတ်ကဲ့ complex numbers တွေအကြောင်း content တွေလုပ်ပေးဖို့ အစီအစဉ်ရှိပါတယ်။

ကိန်းပြည့်တွေရဲ့ ၂ ထပ်ကိန်းက ၂ ၊ ၃ နဲ့ မဆုံးပါဘူး

ဟုတ်ကဲ့ပါ

Yes, we can consider n go past p to reach n. But the number of terms is no longer n, instead, it is n+k where k is a positive integer. Then (n+k) does not divide n, so we reach the same conclusion.

Yes, we can consider n go past p to reach n. Consider p = 0.1 10th term, we np = 10×0.1 = 1. Here we are considering 1 is not divisible by 10. To get the sum of np = 10, then we need to consider 9 additional terms. Floor of (10×0.1) + ... + (19×0.1) = 10 But we have 19 terms now. Then 10 is not divisible by 19. To get the sum of the product to reach n, the number of terms is no more n. It will also be greater than n. So, the number of terms will never divide the sum of floor.

ua-cam.com/video/wRrIgX_-7f4/v-deo.html

Thank you!

I'm sure you'll get better with practice.

@ how long did it take you to solve this?

Thanks for the feedback!

When n×d < 0, the floor is not removed, and it becomes -1.

@edutubeEC oh yeah my bad i meant in the case that where d is in [0,1/n) 0<n×d<1 in this case the floor of n×d is 0 so we're left with the floor of n×2m as you showed in the video and it goes without saying that if 0<n×d<1 then for all k smaller than n 0<k×d<1 so all the floors of k×d and we'll get the same sum as you showed, sorry it was a typing mistake and thank you very much for replying that fast ❤️

Note: so all the floors of k× d =0 and we get the same sum you showed that is a multiple of n

We have to consider when "all" possible values of n. We need to consider two cases if "d" does not equal 0. Case 1: Only for some values of k (0 < k×d < 1), the floor is 0. Case 2: But there will always be a sufficiently large number, k, such that k×d > 1. (For example, if d=0.1, k>10 gives k×d>1.) Beyond that particular number (k>1/d), the sum is not a multiple of n. To avoid k×d>1, there is only one condition, that is d=0.

@edutubeEC so we can't write d in relation to n like for example d=1/n+1 in this case 0<floor(n×d)<1 So we can cancel it and stick w the the previous sum and if n=n+1 We have d=1/n+2 and so on (The value of d is dependent on the value of n and it decreases when n bigger and increase when n is smaller) any value in [0,1/n) fulfills this criteria and the interval changes depending on the value of n, and again I thank you for your precious time and patience ❤️

မတူရင် မရပါ

2300-69=2231