tag:blogger.com,1999:blog-1403623856154599619.post828982862553275909..comments2023-09-03T16:21:45.077+05:30Comments on GMATting: CP #25gmat delhihttp://www.blogger.com/profile/10600081780364166709noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-1403623856154599619.post-16558223118344244912011-01-15T19:00:42.323+05:302011-01-15T19:00:42.323+05:3016/n , 45/n , 210/n
the three pairs are as above.
...16/n , 45/n , 210/n<br />the three pairs are as above.<br /><br />Now to find the GCD of a pair of numbers, we do the prime factorisation of each number.<br /><br />Eg: for 42/140<br />prime factorisation is as follows:<br />2x3x7/2x2x5x7<br />Now pick only the common prime numbers from the two. That will give you GCD:<br />GCD = 2x7 = 14<br /><br />OK so now coming back to the question:<br />Prime factorisation(PF) of the 3 pairs:<br /> PF GCD<br /><br />1) 2x2x2x2/n 4<br /><br />2) 3x3x5/n 3<br /><br />3) 2x3x5x7/n ?<br /><br />In (1) we can be sure that there are two 2s in n. As the GCD between 16 and n is 4.<br />Here we can be sure n = 2x2xK ...(some constant K)<br /><br />In (2) we can be sure that there is one 3 in n and also there is no 5 in n.<br />Now we are sure that n = 2x2x3xT ...(some constant T)<br /><br />Now (3) can be re-written as 2x3x5x7/2x2x3xT<br />So the GCD for sure will have one 2 and one 3 but will surely not have even a single 5.<br />GCD = 2x3xL ...(some constant L)<br /><br />Now looking at the options:<br />1) 3 = 3x1<br />2) 14 = 2x7<br />3) 30 = 2x3x5<br />4) 42 = 2x3x7<br />5) 70 = 2x5x7<br /><br />Only 3) and 4) have the 2x3xL structure, but 3) also has a 5.<br />So 4) is the answer.<br /><br />Bye ML.Anonymousnoreply@blogger.com