GRE STUDY FORUM

is intended to be a spot where any student preparing for GRE can discuss study topics,find or provide tips & works towards a top score along with other students working for the same goal.Answer"GRE WORD OF THE DAY", "GRE SENTENCE COMPLETION OF THE DAY", "GRE ANALOGY OF THE DAY", "GRE ANTONYM OF THE DAY", "GRE MATH PROBLEM OF THE DAY", "GRE ALGEBRA QUESTION OF THE DAY", "GRE GEOMETRY QUESTION OF THE DAY"& "GRE PASSAGE OF THE WEEK". To get a top score,the first step of the ladder starts from here.

Thursday, February 1, 2007

"ALGEBRA QUESTION OF THE DAY"

If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2i3k5m7p, then i + k + m + p =

A) 4
B) 7
C) 8
D) 11
E) 12

9 comments:

Anonymous said...: I think answer is 19 (1+4+6+8). Can't think of any other way to solve for one of the answers given.; February 1, 2007 at 6:13 PM
Pinu13 said...: i too am getting 19; February 1, 2007 at 10:59 PM
Unknown said...: 19.; February 2, 2007 at 3:52 AM
shanmu said...: ya 19..; February 2, 2007 at 7:43 AM
Unknown said...: ya its 19; February 2, 2007 at 8:04 AM
sona said...: i don't think these answeres be correct!
if x = 2i3k5m7p,
than we have:
x=2,?, 3,4,5,6,7,8
so what should be i? i can not be 1 casue we write all in oeder and if we are allowed to use them in inorder we be no sure about otheres!

then i + k + m + p = ?; February 2, 2007 at 8:53 AM
Ankit said...: This comment has been removed by the author.; February 3, 2007 at 3:32 AM
Anonymous said...: admin ??; February 3, 2007 at 3:33 AM
Maya Vishwakarma said...: yaar wts correct ans...??; February 3, 2007 at 8:23 AM