Form 399: Fill & Download for Free

This may not be a scam but more people should know about this.You must be knowing about the 1699 yearly plan of JIO. JIO promotes it because it seems cheaper but but but it is not the cheapest option on the table!REASON:If the calculations are done as below then surely this plan comes out to be the cheapest.Per day rate for ₹149 plan is 5.32₹ (149/28)Per day rate for ₹449 plan is 4.93₹ (449/91)Per day rate for ₹349 plan is 4.98₹ (349/70)Per day rate for ₹399 plan is 4.75 ₹ (399/84)Per day rate for ₹1699 plan is 4.65 (1699/365)But wait wait!What about the n number of JIO RS 50 discount vouchers you have collected over previous recharges?Now here is the little thing many people may have missed.If you do a recharge of RS 399 for unlimited data for 84 days and apply 50 RS discount your per day cost will be less.You can avail the discount for 4 times for the 4 recharges you will do in a year.But you can apply the discount coupon only once on the 1699 recharge. So the actual effective charges per day will beRs 349 plan=>4.27 per day (299/70)Rs 399 plan=>4.15 per day **(349/84)Rs 449 plan => 4.38 per day (399/91)Rs 1699 plan=>4.51 per day (1649/365)So 1699 though it seems the cheapest but actually it is not and this is where JIO played a trick!So my advice will be to go with Rs 399 recharges and never go for 1699 plan.Thank You.Edit 1: I got to know the vouchers are expiring this year itself.JIO during their offer promotions said that we get back the entire amount used to recharge in form of Rs 50 vouchers so I have collected close to 28 vouchers till now and all these are expiring on 19/11/2019 (for me). Keeping this in mind, I think currently it will be best to do as many recharges as possible and stock up on those 399 recharges.Most probably JIO won’t offer any further discounts in the future as they have already captured a huge portion of the market. Earlier for every festival there used to be offers and discounts but now they are only promoting the 1699 one.So if you are planning to do the full year 1699 recharge, don’t do it, instead do four 399 recharges or maybe more and get the maximum benefit out of those vouchers.And after the vouchers expire, then actually the 1699 plan will be the cheapest.Thank You.*According to latest research, upvoting consumes negligible data. So you can upvote if you find this even remotely useful*** THANK YOU so much for 1.5k+ upvotes. My first answer to get 1,100 and 1000 upvotes. I want to thank my parents, God and of course the JIO 399 plan without which this answer won’t have been possible.**

My cable company is starting to sound like this creepy, desperate -ex.Each month the mail they send me is basically saying,“PLEAAASE TAKE CABLE TV BACK. We’ll give you all the stations we gave you before at an amazing, amazing deal!”Deal = $1 cheaper with a 24 month commitment, $399 cancellation fee + 2 kicks in the crotch.By any logic I cut it, it makes no sense.I divided the hours I was watching TV by the monthly bill and it came out to $15 an hour for TV.I’m paying money to be forced to watch commercials every few minutes.My cable package is forcing me to pay for a bunch of expensive channels (ESPN) that I don’t watch.I don’t see cable TV as going fully extinct but I see it having a life as an increasingly endangered animal.When you compare it to streaming services like Netflix, Hulu, and the like, it isn’t even a close contest. The only value cable companies add to my life is internet.I think cable companies have become boxed into a bit of a hole. As more people realize how much they are being ripped off, they will keep cutting.I don’t envy the cable executives who are charged with solving this problem.

[math]a^2+a+100=n^2[/math] is equivalent to [math](2a+1)^2+399=(2n)^2[/math]. Thus [math]399=\big(2n+(2a+1)\big) \big(2n-(2a+1)\big)[/math].Now [math]399=3 \cdot 7 \cdot 19[/math] has [math](1+1)^3=8[/math] positive divisors, of which [math]1[/math], [math]3[/math], [math]7[/math], [math]19[/math] are the only ones less than [math]\sqrt{399}[/math].For [math]d \in \{1,3,7,19\}[/math], set [math]2n-(2a+1)=d[/math], [math]2n+(2a+1)=\frac{399}{d}[/math]. Then [math]4n=d+\frac{399}{d}[/math], [math]2(2a+1)=\frac{399}{d}-d[/math]. Since [math]399[/math][math] \equiv 3\bmod{4}[/math], one of [math]d[/math], [math]\frac{399}{d}[/math] is of the form [math]4k+1[/math] and the other of the form [math]4k+3[/math]. Hence their sum is a multiple of [math]4[/math] and their difference of the form [math]4k+2[/math]. These give the pairs[math](a,n) = (99,100), (32,34), (12,16)[/math], or [math](0,10)[/math].Hence there are three solutions with [math]a, n \in \mathbb N[/math]. [math]([/math]I am assuming that [math]0 \notin \mathbb N[/math].[math])[/math] [math]\blacksquare[/math]