Question

(5k+1)whole square leaves remainder ...... on dividing by 5​

Answer 1

$\textbf{Given:}$

$\mathsf{(5k+1)^2}$

$\textbf{To find:}$

$\textsf{The remainder when}$

$\mathsf{(5k+1)^2\;is\;divided\;by\;5}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{(5k+1)^2}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{(a+b)^2=a^2+b^2+2\,ab}}$

$\mathsf{=(5k)^2+1^2+2(5k)(1)}$

$\mathsf{=25k^2+1+10k}$

$\mathsf{=25k^2+10k+1}$

$\mathsf{=5(5k^2+2k)+1}$

$\mathsf{=Multiple\;of\;5+1}$

$\therefore\underline{\mathsf{The\;remainder\;when\;(5k+1)^2\;is\;divided\;by\;5\;is\;1}}$

$\textbf{Find more:}$

The remainder when 31^150 is divided by 1024 is. (a)320

(b)704

(c)703

(d)321​

https://brainly.in/question/18302009

Answer 2

1 would be the remainder left on dividing (5k + 1)^2 by 5.

Step-by-step explanation:

Given that,

(5k + 1)^2/5                   (where k = any integer like 1, 2, -1, -2, etc.)

Let k be 1 and by putting k = 1

(5 * 1 + 1)^2/5

= (6 * 6)/5

= 36/5

Remainder = 1

Let k be 2 and by putting k = 2

(5 * 2 + 1)^2/5

= (11 * 11)/5

= 121/5

Remainder = 1

Thus, the remainder would be 1.

Learn more: Algebra

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