Question

k=1^2+2^2+3^2.......m^2.

m<54.
how many values of m exist such that k is divisible by
4?

Answer 1

a.10
b.11
c.12
d.none of these

Answer 2

It should help to know that
\[\large\sum_{i=1}^mi^2=\frac{m(m+1)(2m+1)}{6}=1+4+\cdots+m^2\]

Answer 3

what i have to do after finding k?

Answer 4

Well \(k=\dfrac{m(m+1)(2m+1)}{6}\), and you want to see how many \(m\) allow \(k\) to be divisible by 4. My guess would be to expand, giving
\[k=\frac{2m^3+3m^2+m}{6}=\frac{1}{3}m^3+\frac{1}{2}m^2+\frac{1}{6}m\]
Then divide by 4:
\[\frac{k}{4}=\frac{1}{12}m^3+\frac{1}{8}m^2+\frac{1}{24}m\]
From here you have to find out which values of \(m\) make \(\dfrac{k}{3}\) a whole number. This would be any \(m\) whose cube is divisible by 12, square divisible by 8, and that is divisible by 24.

One such value that works is \(m=24\), since \(\dfrac{m}{24}=1\), \(\dfrac{m^2}{8}=72\), and \(\dfrac{m^3}{12}=1152\).

Answer 5

u mean k/3 should be integer ?

Answer 6

48 is another value

Answer 7

@ganeshie8 are their other values than 24 and 48

Answer 8

Notice below :
\(n^2\) leaves a remainder 0 when \(n\) is even (example 2^2 = 0 mod 4)

\(n^2\) leaves a remainder 1 when \(n\) is odd (example 3^2 = 1 mod 4)

Answer 9

k=1^2 + 2^2 + 3^2 + ... +m^2
=1 + 0 + 1 + .... mod 4

Answer 10

So the sum is divisible by 4 when m is of form : 8t-1 or 8t

Answer 11

8t - 1 < 54
t < 55/8 (6 values for t)

8t < 54
t < 54/8 (6 values for t)

Answer 12

Overall there will be 12 sums divisible by 4

Answer 13

can u give link for comprehensive study of MOD

Answer 14

try these
https://www.youtube.com/watch?v=SR3oLCYoh-I