Question-145047

Question Number 145047 by mathdanisur last updated on 01/Jul/21

Commented by mr W last updated on 01/Jul/21

l a s t d i g i t i s 5 .

Answered by ArielVyny last updated on 01/Jul/21

U_{k} = \underset{k = 1}{\sum^{1989}} k^{1989}

l n (U_{k}) = \underset{k = 1}{\sum^{1989}} 1989 l n (k) = 1989 \underset{k = 1}{\sum^{1989}} l n (k) = 1989 l n (k!)

U_{k} = e^{1989 l n (k!)} = {(k!)}^{1989}

Commented by mathdanisur last updated on 01/Jul/21

t h a n k y o u S e r, t h e n t h e a n s w e r ?

Commented by mr W last updated on 01/Jul/21

\ln (a^{n} + b^{n}) \neq n \ln a + n \ln b

Commented by mr W last updated on 02/Jul/21

1^{1989} + 2^{1989} + 3^{1989} + \dots + 1989^{1989}

\neq {(1 \times 2 \times 3 \times \dots \times 1989)}^{1989}

Answered by mr W last updated on 01/Jul/21

1 : \Rightarrow 1

2 : 2 / 4 / 8 / 6 / 2 \Rightarrow 2

3 : 3 / 9 / 7 / 1 / 3 \Rightarrow 3

4 : 4 / 6 / 4 \Rightarrow 4

5 : \Rightarrow 5

6 : \Rightarrow 6

7 : 7 / 9 / 3 / 1 / 7 \Rightarrow 7

8 : 8 / 4 / 2 / 6 / 8 \Rightarrow 8

9 : 9 / 1 / 9 \Rightarrow 9

0 : \Rightarrow 0

1 : 1, 11, 21, \dots, 1971, 1981 \Rightarrow 199 \times 1 = 9

2 : 2, 12, 22, \dots, 1972, 1982 \Rightarrow 199 \times 2 = 8

3 : \Rightarrow 199 \times 3 = 7

4 : \Rightarrow 199 \times 4 = 6

5 : \Rightarrow 199 \times 5 = 5

6 : \Rightarrow 199 \times 6 = 4

7 : \Rightarrow 199 \times 7 = 3

8 : \Rightarrow 199 \times 8 = 2

9 : \Rightarrow 199 \times 9 = 1

1 + 2 + 3 + \dots + 9 = 5

Commented by mathdanisur last updated on 01/Jul/21

a l o t c o o l S e r, t h a n k y o u

Answered by Rasheed.Sindhi last updated on 02/Jul/21

(1^{1989} + 2^{1989} + 3^{1989} + \dots + 1989^{1989}) (\mod 10)

B y o b s e r v i n g v a r i o u s p o w e r s o f

v a r i o u s n u m b e r s w e c o n c l u d e :

a^{4 n + 1} \equiv a (\mod 10)

a^{1989} = a^{497 \times 4 + 1} \equiv a (\mod 10)

1^{1989} + 2^{1989} + 3^{1989} + \dots + 1989^{1989} (\mod 10)

= 1^{497 \times 4 + 1} + 2^{497 \times 4 + 1} + \dots 1989^{497 \times 4 + 1} (\mod 10

= 1 + 2 + 3 + \dots + 1989 (\mod 10)

= \frac{1989 \times 1990}{2} (\mod 10)

= 1979055 (\mod 10)

= 5

Commented by mathdanisur last updated on 02/Jul/21

T h a n k y o u S e r, a l o t c o o l