If k1, k2, k3, ....., kn are odd natural numbers, then the remainder when is divided by 4, is always equal to
Answers
Answer:
Answer:
when {k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2k
1
2
+k
2
2
+k
3
2
+..........+k
n
2
is divided by 4, the remainder is k_1+k_2+k_3+......+k_nk
1
+k
2
+k
3
+......+k
n
Step-by-step explanation:
Given:
k_1,k_2,k_3..........k_nk
1
,k
2
,k
3
..........k
n
are odd natrual numbers.
Then,
k_1=2x_1+1k
1
=2x
1
+1
k_2=2x_2+1k
2
=2x
2
+1
k_3=2x_3+1k
3
=2x
3
+1
.
.
.
k_n=2x_n+1k
n
=2x
n
+1
where x_1,x_2.......x_nx
1
,x
2
.......x
n
are integers
Now,
{k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2k
1
2
+k
2
2
+k
3
2
+..........+k
n
2
=(2x_1+1)^2+(2x_2+1)^2+(2x_3+1)^2+..........+(2x_n+1)^2=(2x
1
+1)
2
+(2x
2
+1)
2
+(2x
3
+1)
2
+..........+(2x
n
+1)
2
=(4{x_1}^2+2x_1+1)+(4{x_2}^2+2x_2+1)+(4{x_3}^2+2x_3+1)+..........+(4{x_n}^2+2x_n+1)=(4x
1
2
+2x
1
+1)+(4x
2
2
+2x
2
+1)+(4x
3
2
+2x
3
+1)+..........+(4x
n
2
+2x
n
+1)
=4[{x_1}^2+{x_2}^2+{x_3}^2+......+{x_n}^2]+[(2x_1+1)+(2x_2+1)+(2x_3+1)+......+(2x_n+1)]=4[x
1
2
+x
2
2
+x
3
2
+......+x
n
2
]+[(2x
1
+1)+(2x
2
+1)+(2x
3
+1)+......+(2x
n
+1)]
=4[{x_1}^2+{x_2}^2+{x_3}^2+......+{x_n}^2]+[k_1+k_2+k_3+......+k_n]=4[x
1
2
+x
2
2
+x
3
2
+......+x
n
2
]+[k
1
+k
2
+k
3
+......+k
n
]
Thus ,we have
{k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2=4[{x_1}^2+{x_2}^2+{x_3}^2+......+{x_n}^2]+[k_1+k_2+k_3+......+k_n]k
1
2
+k
2
2
+k
3
2
+..........+k
n
2
=4[x
1
2
+x
2
2
+x
3
2
+......+x
n
2
]+[k
1
+k
2
+k
3
+......+k
n
]
Hence , when {k_1}^2+{k_2}^2+{k_3}^2+..........+{k_n}^2k
1
2
+k
2
2
+k
3
2
+..........+k
n
2
is divided by 4, the remainder is k_1+k_2+k_3+......+k_nk
1
+k
2
+k
3
+......+k
n
Answer:
Step-by-step explanation:
refer both attachment for complete solution
