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17 Sep 2021, 01:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
49% (02:36) correct
51%
(02:42)
wrong
based on 120
sessions
History
Date
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Not Attempted Yet
Set A has consecutive integers from 1 to n. If all integers in the set A are squared and then added, the result will be 2,470. How many integers are in the set A?
(A) 23
(B) 22
(C) 21
(D) 20
(E) 19
(A) 23
(B) 22
(C) 21
(D) 20
(E) 19
19 Sep 2021, 06:25
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Bunuel
Sum of squares of first n positive integers = \(\frac{n(n+1)(2n+1)}{6}=2470\)
\(n(n+1)(2n+1)=2470*6=2*5*6*247=2*5*6*13*19=(2*5*2)*(3*13)*19=19*20*39\)
n=19
We could also use the options when we know \(n(n+1)(2n+1)=2470*6\)
(A) 23…..23*24*47…..2470 has only one 2 and 6 has another 2 in it, so we are looking at multiple of 4 only, but 24 is a multiple of 8.
(B) 22…..22*23*45…..Not a multiple of 4, and 2470 is clearly not a multiple of 11.
(C) 21……21*22*43….Same reason as B above.
(D) 20…..20*21*41…correct as far as being multiple of 3, 4 or 5 is concerned. So, let us multiply them 420*41, which is more than what we are looking for.
(E) 19……19*20*39
If we factorise 2470*6 further, getting the answer becomes even easier.
2470*6=2*5*6*13*19
So we can get 19 when n=19, n+1=19 or 2n+1=19 or multiple of 19.
ONLY E fits in
E
15 Oct 2024, 04:57
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