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prime numbers

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prime numbers

by vaivish » Fri Aug 08, 2008 12:53 am
If S is the sum of integers from 30 to 50, inclusive, how many different prime factors does S have?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5


Oa is D
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by pepeprepa » Fri Aug 08, 2008 1:06 am
To know the prime factors, it is better to get the number, so the core of the problem is to calculate S, the sum of integers from 30 to 50

I use the formula
"the sum for i=0 to n" of i= n(n+1)/2

A="the sum for i=0 to 50" of i = 50*51/2=1275
B="the sum for i=0 to 29" of i = 29*30/2=435

Now to get S we need to do add A and subtract all the members we do not want, namely B
S=A-B=1275-435=840

Now, it's ok
840=2.2.2.3.5.7

There are 4 different prime factors.
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by sudhir3127 » Fri Aug 08, 2008 1:08 am
I go with D .4

sum of all numbers from 30 to 50 is
n = 50-30+1 = 21

21/2(30+50)=840

840 can be written as 12*7*10
= 3*4 * 7*5*2 hence 4 different prime factors..

Hope it helps..
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by bourne159 » Fri Aug 08, 2008 1:59 pm
Sum of consecutive integers from 30 to 50
= Total number of terms/2 * Sum of (first and last term)

Total number of terms = (50 -30) + 1 = 21

Sum = 21/2 * (50 +30) = 840

factors of 840 = (2**3) * 3 * 5 * 7

Number of prime factors = 2, 3, 5, 7 (4)
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