Suppose we are given a number "$A$" which is multiple of $\phi(n)$. One can assume factorization to be hard. So you cannot find exact value of $\phi(n)$ from $A$.     Clearly using this we can crack RSA as for any encryption exponent '$e$' we will find its inverse modulo $A$ and that will also we inverse of of '$e$' modulo $\phi(n)$.  But will $A$ factor $n$ ??.  Its looks to me as a natural question on the line of thought of  equivalence of RSA and factoring
I stumbled across this paragraph in a paper:     Hence, user b cannot decrypt C   directly. But using e and d , user b   can quickly factor N.   How is  it possible to speedup the prime factorization when knowing e (public key) and d (private key)?  For clarification:  RSA provides us with these equations:  $n = pq$   $\phi = (p-1)(q-1)$   $gcd(e, \phi) = 1$   $de = 1\pmod{\phi}$   In order to determine $p$ and $q$ an attacker has to factor n which is not feasible. However the paper stated that it is easy to reconstruct $p$ and $q$ when a person knows both (his) private and public keys.
How to sort a Google Sheets Query by a custom ordering I have a table of choir members with a “name” and a “voice type” column. How can I write a query that orders them by decreasing pitch level of the voice type (i.e. soprano > contralto > tenor > bass)?  Is there in the  something similar to ?  I also tried the , but it does not seem to have an option for custom ordering, either. "strlen(s1) - strlen(s2)" is never less than zero Leaving Italy without paying a bus fine My wife and I got bus fines today in Bologna. We had only recently arrived and had purchased a ticket from where we were staying which was about 6 or 7 km from the city. The hotel receptionist explained the ticketing system to us and said that if we buy a ticket we could use it on as many journeys for 75 minutes.  We took the bus from our hotel to Bologna and then straight away got on another bus to get to central Bologna where all the sights are. When we got on this other bus we assumed that it was ok as we were still within the 75 minutes. However, when we tried to validate our tickets in the machine it flashed red like an error had occurred so we figured something wasn't right.   We tried to ask the driver about it or if we could buy a ticket but he just grunted and started to drive off. We therefore started to leave the bus at the next stop so we could try and buy another ticket but were ushered back on by a few men who turned out to be ticket inspectors!  They alleged that our tickets were not valid in the city and that we had to pay a 65 euro fine. They wanted to see our official documentation and so took details from our UK driving licences and issued us both with fines. They wanted us to pay there and then, but we explained we didn't have enough cash on us to do so and they then stated we could pay by card.   However, I still wasn't happy about doing that so he then said that we could pay later by going to an Italian post office; or if we don't pay within a certain time the fine will go up to 270 euros each and will be posted to us (he got our addresses off our UK driving licences)!  I am a bit miffed about this as we were noway trying to evade and as soon as we realised our mistake we tried to talk to the bus driver to buy a ticket or get off the bus, but he drove off and ignored us! I have a slight suspicion that it was a rouse as it is very strange that he did this and then out of nowhere bus inspectors got on and made a beeline for us!  If we go back to the UK without paying these fines and ignore the subsequent fines that they send us through the post (if they do send any), then what happens? Do they have any powers when we are back in the UK? a total fine of 500 euros for an honest mistake seems utterly outrageous.   Fine below Counting the number of propagating degrees of freedom in Lorenz Gauge Electrodynamics What's the term for using the wrong word because it sounds the same? I was reading about various terms for incorrect words, but they didn't seem to fit.  I saw a post where someone said "what do you like to do when you're board?".  That kind of thing.  Also, would that be considered a mistake in grammar?  I didn't think so, but my friend insists it is.  I'm not sure if there even is a word for that kind of mistake, but was just curious about it.  Thanks! First time using bevel tool, doesn't bevel evenly. What am i doing wrong? Why wave theory cannot explain photoelectric effect and provides evidence for particle nature of light? get values from ANY type Object A similar question was asked , but it was specific to .NET 3.5. Specifically, I'm looking for the following:   What is the correct way to determine which .NET Framework versions and service packs are installed? Is there a list of registry keys that can be used? Are there any dependencies between Framework versions? Reapplying for a UK Standard Visitor visa after a refusal Swing GUI doesn't wait for user input Is $(B_t^2)$ Markov where $(B_t)$ is Brownian motion? I am pretty sure $(B_{t}^{2})$ not Markov because the squared random walk is not.    I guess I can repeat the method since to be Markov it must satisfy the discrete.  Thanks Consider a system of the two identical positive point charges situated in free space (isolated from the influence of any other external fields) as shown in the attached diagram. Particle $1$ is at $(a,a,0)$ and particle $2$ is at $(0,a,0)$. Their velocities, at the considered instant of time, are as shown in the diagram ($\mathbf{v}_1$ along the $+x$ axis, $\mathbf{v}_2$ along the $+z$ axis).    Now, by applying the Biot-Savart law, we find that the magnetic field due to particle $2$ at the position of particle $1$ is along $+y$ axis, which means that the force acting on particle $1$ is along $+z$ axis according to Fleming’s left-hand rule. A similar analysis shows that there is no magnetic force on particle $2$ as the magnetic field of particle $1$ should vanish at positions (relative to particle $1$) located along its velocity vector.  Now, if we observe the net torque on the system of two particles about the $y$ axis then it is non-zero and is directed along the $-y$ axis. Here, there are no external forces or torques acting on the isolated two-particle system, and yet the net torque, as well as the net force on the particles, are nonzero. Why? Also, Newton's third law of motion seems to be broken in this scenario. Why?    Edit $1$  I have come to know from the responses that the electromagnetic field itself takes away some momentum and angular momentum about the considered axis. However, I think that if I consider only two charged particles as my system then the Abraham-Lorentz force can be assumed to be acting upon the system, and that is sufficient to make sure that we have considered the momentum being carried away by the electromagnetic field itself.   Even after considering the action of the Abraham-Lorentz force for the two-particle system, the scenario breaks both Newton's third law and the conservation of linear and angular momentum. This is because the Abraham Lorentz forces do not exactly counterbalance the force and the torques, on the two-particle system under consideration, due to the magnetic field.    Edit $2$  The previous edit was a result of confusion and misunderstanding on my part. The force associated with the momentum carried away by the electromagnetic field as a result of the electromagnetic interaction described in the question is the Lorentz force itself which simply doesn't obey the third law of motion. The Abraham-Lorentz force is a different story. It is associated with the momentum carried away by the radiation emitted by the accelerated charged particles. This is an additional force apart from the Lorentz force and corresponds to an additional carriage of momentum by the electromagnetic field. The momentum carried away by the electromagnetic field in correspondence with the Lorentz force doesn't correspond to radiation. Round a double to 2 decimal places Relation bewteen Hessian Matrix and Curvature According to ,  It describes the local curvature of a function.   AFAIK, for one-variable function $f(x)$, its local curvature is  $$\kappa = \frac{|f''|}{(1 + f'^2)^{3/2}},$$ and its Hessian matrix is  $$\mathcal{Hess}(f) = [f''],$$ right? And here is my problem, I think the local curvature is not just described by its Hessian matrix, because $f'$ also has its role in it, doesn't it?  And furthermore, for 2-variable function $f(x,y)$, its Hessian matrix is  $$\mathcal{Hess}(f) = \left[ \begin{array}{cc} f_{xx}'' &amp; f_{xy}'' \\ f_{xy}'' &amp; f_{yy}'' \end{array} \right].$$ How does it relate to the local curvature of $f(x,y)$? I have an existing trigger which converts leads into peron accounts if the company name is blank.  Trigger AutoConvert on Lead (after insert) {      LeadStatus convertStatus = [           select MasterLabel           from LeadStatus           where IsConverted = true           limit 1      ];      List&lt;Database.LeadConvert&gt; leadConverts = new List&lt;Database.LeadConvert&gt;();       for (Lead lead: Trigger.new) {           if (!lead.isConverted &amp;&amp; lead.Company == null) {                Database.LeadConvert lc = new Database.LeadConvert();                String oppName = lead.Name;                 lc.setLeadId(lead.Id);                lc.setDoNotCreateOpportunity(true);                lc.setConvertedStatus(convertStatus.MasterLabel);                 leadConverts.add(lc);           }      }       if (!leadConverts.isEmpty()) {           List&lt;Database.LeadConvertResult&gt; lcr = Database.convertLead(leadConverts);      } }   I need to also write a trigger that stops records from coming into Salesforce if that email already exisits in a contact or an account (we use person accounts).  The issue is how do I get it working with the existing trigger (I need it to run before the trigger that converts it into a person account is run).   Is there a way to combine the two or add the functionality to create the record only if the email address doesn't already exist as a contact in the org? Applications Don't Show Up in Spotlight Spotlight has stopped showing any results for my Applications. I've tried reindexing and removing the index so it rebuilds it. No change. I've tried adding Applications to the Privacy tab and removing it, no change.  I tried repairing disk permissions and redoing the above, no change. I've tried removing everything from the index except Applications and then I just get nothing for any search at all (except dictionary entries).  I tried adding a symlink in my homedir to Applications and reindexing, but no change.  Any ideas on what to do?  I'm running Snow Leopard.  This is driving me crazy!  Update: I've noticed that when I start a reindex with sudo mdutil -E / and then immediately do a Spotlight search for an app that the app shows up temporarily until Spotlight gets disabled due to active indexing.  After the indexing is done, the app entries go away. I accidentally unset all the directories of $PATH while trying to add a new one in ~/.bashrc. I opened a new terminal window as I was editing and now $PATH is empty. I'm worried if I boot from another drive to find the $PATH I won't be able to boot into this drive again.  Basically, what is the default result of echo $PATH? Are there experiments that are banned from being done at the LHC? If a person uses the wrong form of your or you're, would that be considered a spelling or grammatical error? I can really see it swinging both ways...  A sentence with the incorrect use of the word is technically grammatically incorrect, but the root cause could be simple spelling error. If a person speaks a sentence with an incorrect use of the word in question, it seems grammatically correct. When you look at the reader's script, it is then determined to be a spelling error.  When shown in writing, should the error be marked as a spelling or grammatical error?