Out of place

Let's look at odd squares mod 8 without the original restriction (it seems "out of place"). Firstly, $N = 2k+1 \Longrightarrow N^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$ Taking mod 8 for k even: $(4 \cdot 2a)(2a+1) + 1 = 8a(2a+1) + 1 \equiv 1 \pmod{8}$ Taking mod 8 for k odd: $(4 \cdot (2b+1))(2b+2) + 1 = (8b+4)(2b+2) = 8(2b+1)(b+1) + 1 \equiv 1 \pmod{8}$

We have checked for k even and k odd, so our proof is complete.

As a matter of fact, all odd numbers greater than 1, when squared and divided by 8, produce a remainder of 1.

Just look at all the square numbers:

• $1$
• $4 = 1 + 3$
• $9 = 4 + 5$
• $16 = 9 + 7$
• $25 = 16 + 9$
• $36 = 25 + 11$
• $49 = 36 + 13$
• $64 = 49 + 15$

Obviously, square numbers increase by larger and larger odd numbers. Each pair of odd numbers add to a multiple of 8:

• $3 + 5 = 8$
• $7 + 9 = 16$
• $11 + 13 = 24$

To get from 3 + 5 to 7 + 9, we skip two odd numbers each, making the sums increase by 8 each time. This makes every odd square a multiple of 8 plus 1, since we started with 1.

from that statement , N is odd number and strictly greater than 123

so we got $N>123$ where $N$ is Odd. so the smallest possible number for $N$ is $125$ .

The question is Remainder when $N^{2}$ divided by 8 , so we can write:

$15625$ = $1$ ( $mod 8$ )

so the answer is = 1

Let $N=123+2k$ , where $k$ is a positive integer. We want to find: $N^{2}(mod{8})$ or $(123+2k)^{2}(mod{8})$

$=>$ $(123^{2}+4k^{2}+492k)(mod{8})$

$=>$ $(123^{2}+4k^{2}+4k+488k)(mod{8})$

$=>$ $(123^{2}+4(k^{2}+k)+488k)(mod{8})$ .

Now, $k(k+1)$ is divisible by $2$ as product of 2 consecutive integers is divisible by $2!$ .

Therefore, $4(k^{2}+k)$ is divisible by 8, $i.e.$ $4(k^{2}+k) \equiv 0 \pmod{8}$ Also, $488$ is divisible by $8$ ,hence $488k \equiv 0 \pmod{8}$ , $123^{2} \equiv 9 \pmod{8}$ $\equiv 1\pmod{8}$ Hence, $(123^{2}+4(k^{2}+k)+488k) \equiv 1+0+0\pmod{8}$ $\equiv 1\pmod{8}$
Therefore,required remainder is $1.$

x = 0 mod 8 ------> x^2 = 0 mod 8

x = 1 mod 8 ------> x^2 = 1 mod 8

x = 2 mod 8 ------> x^2 = 4 mod 8

x = 3 mod 8 ------> x^2 = 1 mod 8

x = 4 mod 8 ------> x^2 = 0 mod 8

x = 5 mod 8 ------> x^2 = 1 mod 8

x = 6 mod 8 ------> x^2 = 4 mod 8

x = 7 mod 8 ------> x^2 = 1 mod 8

so, every N odd number, the remainder of N^2 = 1

For any odd $k$ ,

$k^{2}= 1 (mod 8)$

Let $N = 125$ or any odd number that is greater than 123

$N^2 = 125^2 = 15625$ dividing it by 8 the remainder is 1

The odd number that is strictly greater than 123 is 125. N = 125 N^2 = 125 x 125 = 15625 15625/8 = 1953 remains 1 So, the answer is 1

We can start with the first odd number greater than 123, which is 125. 125*125 = 15 625. Dividing this by 8 we get 1953.125, or 1953 r.1. As 0.125 is equal to 1/8, we know that there is a remainder of 1, and as 15 624 / 8 = 1953, we have a way to check our answer.

The same idea works with other odd numbers above 123, although why the question specifies N as strictly greater than 123, I really am not too sure.

Given that N is odd, N is then obviously an odd number mod 8

if N is 1 mod 8, N^2 is 1 mod 8

if N is 3 mod 8, N^2 is 1 mod 8

if N is 5 mod 8, N^2 is 1 mod 8

if N is 7 mod 8, N^2 is 1 mod 8

No matter what, if N is odd, then N^2 will be 1 mod 8. The 'greater than 123' part was probably a distraction XD Hence we have 1 as our final answer.

N > 123, where N is an odd number...

Option will be the next odd number of 123 which is 125...

and N² = 125² = 15625/8 = 1953 remainder 1

Every square of an odd number is of the form $8b+1$

Proof: Let $N= 2a-1$ be an odd number. On squaring it

$N^2=(2a+1)^2 = 4a^2+4a +1 = 4(a^2 +a)+1$ ..... (I)

$a^2+a = a(a-1)$ Since, its a product of two consecutive numbers its divisible by $2$

$\therefore a^2 +a = 2b \enspace \enspace$ NOTE :- $a$ and $b$ are number natural numbers.

substituting in (I) we get $N^2=4(2b)+1= 8b+1$

Thus, when divided by $8$ , the reminder is $1$ .

Start with 125 and guess and check with all odd numbers after that

O menor valor que "N" pode assumir, de acordo com o enunciado, é 125; temos que:

$125^2 = 1953 \times 8 + 1$

Vejamos o próximo...

$127^2 = 2016 \times 8 + 1$

Portanto, $\boxed{\text{resto} = 1}$

Rem{(125 125)/8}=Rem{(5 5)/8}=1

If you list all the last digit odd numbers and square them, you will arrive at 5, 9, 1, 1, 9 as last digits from the last digits of 5,7,9,1,3 respectively. From here, you can infer that if we divide each last digit by 8, the probability of 1 as the remainder happens to be true.

Thus, we have 1 as the remainder.