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Te Puan mlvamnio (bomNncom uncIntuLe odAvnlbiobueInat ma cim ttmann WormunbPm_ Ueo 4 0.10 HlcancolutGaeicnTul Eno numilcmnbtu mnpothozoz?Puza Rata DataHo-0>7107 41bomHo:pe7itom HiaetibonE etidomPPuteRata (bpm )0d; Ho /F7item A:hitieonDebern 70uabrbcMunddecmal pananteEraertol mene ntne Areacy box arathen clck Chec< Arintretnnd

te Puan mlva mnio (bom Nncom unc IntuLe od Avnlbiobue Inat ma cim tt mann Wormun bPm_ Ueo 4 0.10 Hlcancolut Gaeicn Tul Eno num ilcmnbtu mnpothozoz? Puza Rata Data Ho-0>7107 41bom Ho:pe7itom Hiaetibon E etidom PPuteRata (bpm ) 0d; Ho /F7item A:hitieon Debern 70 uabrbc Mund decmal pa nante Eraertol mene ntne Areacy box arathen clck Chec< Arint retnnd



Answers

Decrypt these messages encrypted using the shift cipher $f(p)=(p+10) \bmod 26 .$
a) CEBBOXNOB XYG
b) LO WIPBSOXN
c) DSWO PYB PEX

All right. So this time we're getting this incredibly long message, and we are told that the cipher is going to be of the form F P is equal to P plus que mile 26. Right? So that means our decryption cipher a must be the form p minus K mod 26. Okay. And so how do we go about figuring that out? Well, we're gonna do a method I've probably done by the time he reached this problem where we look at the most common letters in this cipher and in this particular site where he moves common letter happens to be m okay. And so we knew that m is gonna be the 12th. Okay, So m is the 12th letter, and we want to find Kay. Bye. Taking 12 and subtracting a value of other common letters. And so I've gone ahead and tried. Some of these other ones ease and first option is the most common. It doesn't work. T a o i. N s agent are are the other most common letters. And it turns out this cipher makes sense. If we use I, so we'll take. I was eight. And that means K is gonna be equal to 12. Minus eight is four. So we're gonna take this cipher and use P minus four mod 26 to solve it. All right, So the first thing is to convert, we have into the numbers and that ends up being for 17 to 22. 24 nine, nine, 12 six, 12 eight, 17 23 15 and two and then four, seven 25 4 17 68 and seven one says. Next Rose. We get 23 eight, six, 11 17 18 15 18 10 and two, 12 22 12 17 7 12 22 23 12 17 10 24 12 22 11 for 5 15 8 OK, And so this set is really just this here. So move it. J. B. S Q. Down Aurora's Well, um, so things are a little bit more lined up, so we have J. B s Q and A. Q e k M g. It's ah, J B s Q is 9 2118 16 Then the key e k m G is gonna end up as 16 for 10 12 and six. Okay, so now we need to take the modules right was really involved. Subtracting four from everything. Um Ahmad 26. So he will get zero 13. Negative, too. This is me. 24 18 25 5 eight, 284 13 1911. Negative, too. Says 24. 03 21 0 13. 243 1942 seven, 13 14 11 14 6 That's negative, Thio, because it's 24 eight 18 AIDS 13 3 eight, 18 19 eight, 13 10 becomes 6 24 becomes 2012 becomes 8 22 becomes 18 11 becomes 74 become 05 becomes one for 15 becomes 11 and ate becomes four nine because 5 21 to be 17 18 becomes 14 16. Becomes 12 12 06 eight and two. Okay, so now we can translate it back. It's very exciting and we do that. We get a and why any s u f f i c i e N t oh, why so any sufficiently? Zero's a three is D 21 is v A on C d. Hey will be cramming in a little apologised for that. Made his tea for his e two c 17 age thirties And so you 00 g. Why I s I an d I s t i and g u i s h a be, uh, e I can't quite fit under there, but the five becomes an F 17 becomes our 14 becomes 0 12 becomes, um and any 12 is m zero is a six is G. It is I. And to see so we end up with any sufficiently advanced technology is indistinguishable from magic.

In this problem, I can write the reaction as just look at it carefully. This compound in pageants of this component, which I am lighting here will give this compound. Here it is CS three here it is as irritates us. So this compound in pageants of Edge to will give the final product, ID dis confirmed, irritated at chT charity Edge less this component. So according to the option, option A is correct, Therefore, option eight, correct here.


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