The last digit of (1^3+ 2³ +3^3 +... 10³)^64is:
Answers
Answer:
5
Step-by-step explanation:
1
3
+2
3
+3
3
+… +n
3
=(1+2+3+… +n)
2
Here we're given to find the ones place of the sum,
(1^3+2^3+3^3+\dots\ +10^3)^{64}(1
3
+2
3
+3
3
+… +10
3
)
64
By the above statement, it is equal to,
\begin{gathered}[(1+2+3+\dots\ +10)^2]^{64}\\\\=(1+2+3+\dots\ +10)^{2\times 64}\\\\=(1+2+3+\dots\ +10)^{128}\end{gathered}
[(1+2+3+… +10)
2
]
64
=(1+2+3+… +10)
2×64
=(1+2+3+… +10)
128
But, it is also true and can be proved easily by induction too, that,
1+2+3+\dots\ +n=\dfrac {n(n+1)}{2}1+2+3+… +n=
2
n(n+1)
Thus,
\begin{gathered}(1+2+3+\dots\ +10)^{128}\\\\\\=\left (\dfrac {10\times11}{2}\right)^{128}\\\\\\=55^{128}\end{gathered}
(1+2+3+… +10)
128
=(
2
10×11
)
128
=55
128
But,
(10q+5)^n\equiv5\pmod{10},\quad q,\ n\in\mathbb{N}(10q+5)
n
≡5(mod10),q, n∈N
This means that the integral powers of any number which ends in 5 also end in 5. Thus we can say that,
55^{128}\equiv\mathbf {5}\pmod{10}55
128
≡5(mod10)
Hence 5 is the answer