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Couple Problems I can't seem to grasp. Probably very easy!

Expert replies
by vikingfrog » Mon Apr 19, 2010 9:28 pm
In XY plane point P has coordinates (a,b) and point Q has coordinates (c,d)

What is the distance between P and Q?

1) b-d = 4
2) a-c = 3

I know this is easy but I can't wrap my mind around how it is C. If anyone can give me a quick explanation thanks.


Also, What is the greatest prime factor of (4 to the 17th) - (2 to the 28th). I guessed 7 but I am terrible with exponents and don't know how I got it.

Take the test Saturday and currently testing around 670 but would like to break 700. My biggest mistakes come in sentence correction and Data sufficiency and I always make a couple stupid mistakes.

I am trying to counteract this by going over manhatten SC book before saturday, and working some magic on Data sufficiency. I always finish both sections with about 5-10 minutes left so I think I just need to take some time and double check. Sometimes I finish the qualitative section with 15-20 minutes left.

That's my short story, and thanks for the help on these two problems.
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Source: — Data Sufficiency |

by anoopprasad » Mon Apr 19, 2010 9:44 pm
Distance between 2 points (x1, y1), (x2,y2) is sqrt [ (x1-x2)^2 + (y2-y1)^2]

In this scenario we can know the distance if we know b-d and a-c
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by thephoenix » Mon Apr 19, 2010 9:52 pm
s1) gives u one perpendicular side of right angle tri
s2) gives u another perpendicular side of the same right angle triangle
combining
s1+s2=we know base and height we can find the hypotenuos
pythagoras theroem
hence C
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by anoopprasad » Mon Apr 19, 2010 9:55 pm
4^17 = 2^34

so the questions is 2^34 - 2^28 = 2^28 (2^6 - 1)

= 2^28 * 63

Largest prime factor of 63 is 7, is the answer 7?
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by liferocks » Mon Apr 19, 2010 9:55 pm
for the second problem 4^17 -2^28
=2^34-2^28
=2^28(2^6-1)
=(2^28)(63)

this has three prime factors 2,3 and 7..hence ans is 7
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