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28 Jan 2022, 03:45
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (01:57) correct
38%
(02:23)
wrong
based on 115
sessions
History
Date
Time
Result
Not Attempted Yet
What positive integer value of n satisfies \(\frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+2+3+...n}=1225\)?
(A) 18
(B) 32
(C) 38
(D) 49
(E) 53
Are You Up For the Challenge: 700 Level Questions
(A) 18
(B) 32
(C) 38
(D) 49
(E) 53
Are You Up For the Challenge: 700 Level Questions
28 Jan 2022, 23:35
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Bunuel
(I) If you know the formula, use it directly.
Sum of first n positive cubes = \((\frac{n(n+1)}{2})^2\)
Sum of first n positive integers = \((\frac{n(n+1)}{2})\)
Thus, \(\frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+2+3+...n}=\frac{(\frac{n(n+1)}{2})^2}{\frac{n(n+1)}{2}}=1225\)
\(\frac{n(n+1)}{2}=1225\)
\(n(n+1)=2450=49*50\)
Thus n is 49
(II) If you do not know the formula, use the concept of units digit.
The units digit of numerator has to end in 5 or 0.
The units of cubes of digit 1 to 9 are 1,8,7,4,5,6,3,2,1
ADD the digits from left and see where do you get 0 or 5 in the units digit of this addition.
1+8+7+4=20, so units digit can be 4.
1+8+7+4+5=25, so units digit can be 5.
1+8+7+4+5+6+3+2+9=45, so units digit can be 9.
And of course 0.
ONLY 49 fits in the options.
(III) Pattern
Take n as 1,2,3… and find a pattern
n as 1 will give 1/1 or 1=
n as 2 will give 9/3 or 3=1+2
n as 3 will give 36/6 or 6=1+2+3
n as 4 will give 100/10 or 10=1+2+3+4
So, 1+2+3+4+…+n=1225
Sum in an AP is average of numbers * total numbers = \(\frac{(1)+(n)}{2}*n=1225\)
\(n(n+1)=2450=49*50\)
D
20 Jan 2025, 23:09
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