Question

the last digit of (1^3+2^3+3^3+.......10^3)^64​

Answer 1

By principle of mathematical induction, we can easily prove the given statement.

$1^3+2^3+3^3+\dots\ +n^3=(1+2+3+\dots\ +n)^2$

Here we're given to find the ones place of the sum,

$(1^3+2^3+3^3+\dots\ +10^3)^{64}$

By the above statement, it is equal to,

$[(1+2+3+\dots\ +10)^2]^{64}\\\\=(1+2+3+\dots\ +10)^{2\times 64}\\\\=(1+2+3+\dots\ +10)^{128}$

But, it is also true and can be proved easily by induction too, that,

$1+2+3+\dots\ +n=\dfrac {n(n+1)}{2}$

Thus,

$(1+2+3+\dots\ +10)^{128}\\\\\\=\left (\dfrac {10\times11}{2}\right)^{128}\\\\\\=55^{128}$

But,

$(10q+5)^n\equiv5\pmod{10},\quad q,\ n\in\mathbb{N}$

This means that the integral powers of any number which ends in 5 also end in 5. Thus we can say that,

$55^{128}\equiv\mathbf {5}\pmod{10}$

Hence 5 is the answer.

#answerwithquality

#BAL

Answer 2

THE UPPER ANSWER IS CORRECT

The correct answer is 5

Hope it helps you

my answrr was also coming same as that

in the upper one so how can I give my answer

but like it