Number Operations

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vladmire: Master | Next Rank: 500 Posts; Posts: 132; Joined: Tue Oct 07, 2008 4:59 pm; Thanked: 4 times

Number Operations

by vladmire » Wed Dec 03, 2008 4:09 pm

What are the basic rules to solving the operation problems can I plug in numbers (???)

If the product of three digits of the positive integer k is 14, what is the value of k?

(1) K is an odd integer

(2) K<700

(a) first
(b) second
(c) both
(d) each statement along
(e) neither

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Source: Beat The GMAT — Data Sufficiency | Wed Dec 03, 2008 4:09 pm

jimmiejaz: Master | Next Rank: 500 Posts; Posts: 207; Joined: Sun Mar 11, 2007 6:16 pm; Location: Mumbai; Thanked: 11 times

by jimmiejaz » Wed Dec 03, 2008 4:32 pm

the ans will be C.
We are give positive integer.
14 = 1*2*7
so, either the number is 127 or 721 or 172
From 1 , the number is odd.
It can be 127 or 721, insuff.
from 2, the number < 700
it can still be 127 or 172
combining both,
odd and less than 700 is only 127.
Hence ans is C

What if i have not yet beat the beast, I know i will beat it!!!!!!!!

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raajan_p: Senior | Next Rank: 100 Posts; Posts: 75; Joined: Wed Aug 27, 2008 7:19 am; Location: Orlando; Thanked: 6 times

by raajan_p » Wed Dec 03, 2008 4:46 pm

Should we not consider all the possible values?

As in, I thought we would get 6 different combinations of 14
127, 172, 217,271,712 and 721

Now, from one,
we can say that K will be one among -> 127, 217,271 and 721

From two,
we can say that K will be one among -> 127, 172,217 and 271.

Combining both, we still cant find one single value for K..

So answer should be E..

Correct me If I am wrong.

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earth@work: Master | Next Rank: 500 Posts; Posts: 248; Joined: Mon Aug 11, 2008 9:51 am; Thanked: 13 times

by earth@work » Wed Dec 03, 2008 4:59 pm

i go with E
we need to consider all options:
127, 172, 217, 271, 721, 712
now (1) odd number leaves - 127, 217, 271, 721 - insufficient
(2) less than 700 gives 4 options - insufficient
(1) & (2) together also give more than one option - insufficient
answer is E