Alice encrypts the private factors of the modulus using her public key. In order to increase security,
Question:
Alice encrypts the private factors of the modulus using her public key. In order to increase security, she multiplies them with a random integer k (a process called blinding). Namely, she performs the following operations
c p = (kp) e mod n and c q = (kq) e mod n,
where
n = 7473729494541494368852084160833887966786966146604539788038425546370721005202575292878010603718779143680279072437859979437027212668979951812104144364476251149631445677106574998958718100538368157559892086233484623658307080278702230379266872415713089871004732096589645883883977415506508347976001334807533323619602524830567712971581132341673584628698079736760920789626175805249978676215337039548212810503820392130214528748452439765014841123109184880811527461970066588991000432403773750048368279867220678747020767042484253902752403310560243017584585301433673020033953058809415002025715734068685302158511640316340598416777
e = 65537
a. Explain why this is not secure as anyone who obtains c p or c q can factor n.
b. Factor n assuming
c p = 6237092736280340368143892975246879700280574956823483459089826300838549326698371192841125256658301422335506071280990036134969958430140091821413942110999828019223701914534660496145184049436793233117527141710469587852896572612908205944874924666105935993339272072516489985048899803563277390008625395078803685108980714466378814566336644612044895653938680103088252710048504048952616260337657469904426834625756755178823712343160146527660412642787198955050511984140986234095815703702908475146055296882216504049042640234735532803592978062118777287332266867117935272468491447326467928608953581142170727780193982988625594898212
c q = 6792216459178320663134922113672614244133286795167205940387994163295121940464502516578964764558190390582955957319087527927606354995278787002827810252216604393602551619997059706389437721444109986118687258571558770491123542808098899449070164438070259725300867432155155552476715445078689763334188428800272552596349007005222835623783668243124391007979542657577232082178629086100963707200714469436928816142276093722080937180495045103388174875710073753498922326189147660992582560179759354313355942469090808651587693854760749192023182331797094543656070987815654088544115591473666250129130603910045357579318640054232668524224
and decrypt the following ciphertext
c m = 2105477800955456357154507258188617457189736593253101663074145074953981395245071021806753010279680874168054072882966675195121143457746629686736359549415401549852363809597210114934407644579644329802951418430994896968996554627286997788295104796260715656641752826531876202868222659539111193689690587665763970475954283298561326716629497355710903035324921787171114135903504002506114489283933796611287777504398400559494459519397397254887260526830796723748881723043560212833332272718498671835837847760126707870313591817583090393736883523163338481442463241882375466824267640258884419635167163135753490195544303858749215914789