Arbrakaboom

Note that odd integers are of the form $2k-1 ~\left\{k\in \mathbb{Z}\right\}$ ... According to the question, we have the following...

$5(2k-1)^2 = 5(4k^2-4k+1)=20k^2-20k+5\equiv 5 \pmod{10}$

Hence, the remainder is: $\boxed{5}$

The general form of an odd number, for some $k\in\mathbb{N}$ , is $2k-1$ , According to the question, we have...

$5(2k-1)^2=5(4k^2-4k+1)=20k^2-20k+5\equiv \fbox{5} \pmod{10}$

a square of an odd number is an odd number.. when an odd number multiply by 5, the last digit is 5.. therefore 5 is the remainder when divided by 10

There's an interesting observation here. Before that, just note that you concentrate only on the units place of the result, because the result is the remainder when the number is divided by 10.

Property - Any number "123465734" when divided by 10, the remainder is its unit's place digit, i.e. 4

Coming to the problem, any odd number ends with 1, 3, 5, 7 or 9 (here, I am concerned only with the unit's place). This number when squared yields the number with unit's place 1, 9, 5, 9, 1. $1^{2} = 1, 3^{2} = 9, 5^{2} = 25, 7^{2} = 49, 9^{2} = 81$

The resultant when multiplied with 5 yields the number with unit's place 5, 5, 5, 5, 5. $1 * 5 = 5, 9 * 5 = 45, 5 * 5 = 25, 9 * 5 = 45, 1 * 5 = 5$

Hence, the result is $\boxed{5}$

You can just follow the steps Martin tells you to, and see what the remainder is in your answer. The correct answer is 5.

it's general question e.g you think odd number i:e 7,square of 7 is 49, again by given condition that multiply by 5, the ans; is 245, at last given condition divide by 10. the reminder is 5...mathetically 7 => (7)^2 = 49 => 49*5 = 245 => 245/10 = 24.5 => 5 is reminder

We are told to think of an odd number. Then we are told to square it. When squaring odd numbers, we will always get an odd product. When we multiply it by 5, we get a number ending in 5 (as any odd number *5 ends in 5). When we divide by 10, the remainder is the digit in the ones' place: which will always be 5 here.

The final answer is 5

if you multiply 5 with any odd number's square

the unit digit of the result would always be 5

and we know that 10 is not divisible of any odd number's square

therefore dividing the final result by 10 will always get remainder 5.

I did 3 and 3 squared is 9 and 9 times 5 is 45 and 45 divided by 10 is 4 remainder 5

I'm not entirely sure this problem really requires any work? Correct me if I'm wrong. I picked 3 as my odd number. So 3^{2} = 9, 9 * % = 45, and 45 % 10 = 5

Any value can be put in odd number and tried to get the answer!!!!

THE SQUARE OF AN ODD NUMBER ENDS WITH 1, 5, 9. THEREFORE THE UNITS DIGIT OF THE SQUARE OF AN ODD NUMBER MULTIPLIED BY 5 IS ALWAYS 5 (1 5=5, 5 5=25, 9*5=45). THEREFORE THE REMAINDER WHEN DIVIDED BY 10 IS THE UNITS DIGIT THAT IS 5. THEREFORE THE REMAINDER IS 5.

Now, it's easy to just abuse the fact that you know that the remainder is the same for all odd numbers and see that $5 \times 1^2 = 5 \equiv \boxed{5} \pmod{10}$ .

But if that cheesy solution isn't satisfying, it doesn't get much harder. We just do the operations as stated:

$5(2k+1)^2 = 5(4k^2+4k+1) = 10 \times 2k^2+10 \times 2k+5 \equiv \boxed{5} \pmod{10}$ .

25/10=2.5

since it is an odd number then its square is also odd so its reminder by 2 is 1 so the reminder of 5 times of it when divided by 5 times of two is 5 times of 1 so equals 5

3²=9.5=45/10=Resposta 4 resto 5

least odd no. is 3 so on squaring we get 9 and on multiplying by 5 the no. is 45 when it is divided by 10 remainder is 5