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28 Jan 2022, 03:45
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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:
64% (02:02) correct
36%
(02:10)
wrong
based on 170
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
General Discussion
07 Sep 2026, 12:21
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Hey chetan2u,
Thank you for the amazing solutions! I was wondering, what if the number was 1221 or 1222 instead of 1225? Do you think that's a GMAT style question or does that make it too complicated? I think they gave us 1225 because its solvable, i.e. we can infer that 0 or 5 has to be in the numerator.
Thank you for the amazing solutions! I was wondering, what if the number was 1221 or 1222 instead of 1225? Do you think that's a GMAT style question or does that make it too complicated? I think they gave us 1225 because its solvable, i.e. we can infer that 0 or 5 has to be in the numerator.
chetan2u
07 Sep 2026, 19:35
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@subh8
Knowing formula here is the best option. Knowing sum of first n positive integers, of square of first n positive integers and of cube of n positive integers is always helpful. You never know when these would make things easier.
If you are looking for solution through units digit, then too things can be worked out but it will be time consuming and complicated.
Example
n=1,2,3,4,5,6,7,8..
Numerator would end in 1, 9(1+8), 6, 0, 5, 1, 4, 6
Denominator would end in 1, 3, 6, 0, 5, 1, 8
\(\frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+2+3+...n}=28\)
Now units digit of denominator *8(from 28) should give you units digit of numerator.
Numerator would end in 1, 9(1+8), 6, 0, 5, 1, 4, 6
Denominator*8 would end in 1*8, 3*8, 6*8, 0*8, 5*8, 1*8, 8*8
What matches is
When n is 4...0*8 = 0 and
When n is 7...8*8 = 64
So look for option of 4 or 7.
But incase n is a larger number, it becomes too time consuming.
Knowing formula here is the best option. Knowing sum of first n positive integers, of square of first n positive integers and of cube of n positive integers is always helpful. You never know when these would make things easier.
If you are looking for solution through units digit, then too things can be worked out but it will be time consuming and complicated.
Example
n=1,2,3,4,5,6,7,8..
Numerator would end in 1, 9(1+8), 6, 0, 5, 1, 4, 6
Denominator would end in 1, 3, 6, 0, 5, 1, 8
\(\frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+2+3+...n}=28\)
Now units digit of denominator *8(from 28) should give you units digit of numerator.
Numerator would end in 1, 9(1+8), 6, 0, 5, 1, 4, 6
Denominator*8 would end in 1*8, 3*8, 6*8, 0*8, 5*8, 1*8, 8*8
What matches is
When n is 4...0*8 = 0 and
When n is 7...8*8 = 64
So look for option of 4 or 7.
But incase n is a larger number, it becomes too time consuming.
shubh8
