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Positive integer

by pakaskwa » Fri Mar 06, 2009 12:10 am
n is a positive integer. What's the value of n?
1) sum of first n integer's square (1+2^2 +3^2 +...+n^2) is a multiple of 4;
2) 104<=n<=106

OA is C, I don't know how to solve it. Please help.
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Source: — Data Sufficiency |
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by DanaJ » Fri Mar 06, 2009 12:28 am
1. The sum of squares is equal to n(n+1)(2n+1)/6. Now, there are plenty of numbers n for which this sum is a multiple of 4: take n = 8, n = 16 or basically any multiple of 8 and the sum is consistent with the restriction. This is why 1 is insufficient

2. There are 3 numbers that could be considered: 104, 105 and 106. Again, 2 is insufficient.

Now, but the two together and you get that only 105 fits, so the answer is indeed C
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by pakaskwa » Fri Mar 06, 2009 4:53 am
How do you get n(n+1)(2n+1)/6. Is it an equation?
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by DanaJ » Fri Mar 06, 2009 11:38 am
The sum 1^2 + 2^2 + .... + n^2 = n(n+1)(2n+1)/6 - this is a well-known formula and I advise you to remember it - it may come in handy for shortcuts and stuff...
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by avenus » Mon Mar 09, 2009 9:07 am
DanaJ

2. There are 3 numbers that could be considered: 104, 105 and 106. Again, 2 is insufficient.

Now, but the two together and you get that only 105 fits, so the answer is indeed C
Shouldn't it be 104, the one that fits??
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by avenus » Mon Mar 09, 2009 9:12 am
Dana,

do you have more formulae like this one? It could certainly come in handy...
this question would be pretty difficult to solve in under two minutes without the formula.
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by DanaJ » Tue Mar 10, 2009 4:24 am
Since you asked, here are my formulas for commonly used sums:
I. 1 + 2 + 3 + ... + n = n(n+1)/2

II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6

III. 1^3 + 2^3 + 3^3 + ... + n^3 = [n(n+1)/2]^2 - this is quite easy to remember, since it's the result from formula I raised to the second power.

avenus: Yes, you are right. Sorry, my mistake.
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