Exam
Maths Puzzle
Numerical Ability
Number System
Find the last digit of 1^3 + 2^3 + .............99^3
1) 0
2) 1
3) 2
4) 5
5) 8
Read Solution (Total 8)
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- 1) 0 is correct ans
n^2(n+1)^2/4 is formula for this qs - 8 years agoHelpfull: Yes(0) No(0)
- answer is 0.
since 1^3+2^3+.....9^3=45
therefore last digit of 1^3+2^3+3^3+...........+99^3=0 (45*10=450)
- 8 years agoHelpfull: Yes(0) No(0)
- ANS is 0
By using Arithmeti Progression
(1^3+2^3+3^3+4^3+......+n^3) =1/4*n^2*(n+1)^2
so here n=99
1/4*99^2*100^2=24502500
from above last digit i s 0 - 8 years agoHelpfull: Yes(0) No(0)
- 1^3+2^3+3^3+4^3+...................+9^3 repeat 10 times if we consider only last digit.....
and this unit digit of cube result is a sum of 1 to 9=45
this will 10 times means 45*10=450
so the last digit is 0.(ans) - 7 years agoHelpfull: Yes(0) No(0)
- is the answer 0
- 6 years agoHelpfull: Yes(0) No(0)
- 0 is the correct answer for it
use formulae (n*(n+1))^2/4 - 6 years agoHelpfull: Yes(0) No(0)
- 1^3=1
2^3=8
3^3=7(last digit)
4^3=4
5^3=5
6^3=6
7^3=3
8^3=2
9^3=9
10^3=0
Sum of all last digit= 0
so we can conclude one thing
11^3=1
12^3=8
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99^3=9
so by this
for every 10 numbers addition, it is 0
So final answer = ZERO ........... i.e. 0 - 5 years agoHelpfull: Yes(0) No(0)
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