Ok...So i'm the first to write a post here tat doesn't regard past year competition results. So, befitting for our society, i'm going to share with all of u the first question of the first ever International Mathematics Olympiad. Now, dun let the name scare u as tis question is actually quite easy and can be solved by form 1 students oso i'm sure. Ok...good form 1 students then....so here's it:
Prove that (21n+4)/(14n+3) is an irreducible fraction.
Have fun! Post ur workings in comments.
Jun Kit
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Can we say it is irreducible by saying demomitor and numerator has no H.C.F?
ReplyDeleteya...its denominator btw ^^
ReplyDeleteEasy! gcd(21n+4 , 14n+3) = gcd(14n+3 , 7n+1)= 1
ReplyDeleteTherefore , they does not have a common divisor larger than 1 , surely it is irreducible.