Prove that for every n e n, when we divide n2 by 7, the remainder is either 0, 1, 2, or 4. in other words, the remainder will never be 3, 5, or 6.

For every n E N, when we divide n² by 7, the remainder is either 0, 1, 2, or 4.

Step-by-step explanation:

Given information: n∈N and .

To prove: For every n E N, when we divide n2 by 7, the remainder is either 0, 1, 2, or 4.

Proof:

Using basic remainder remainder theorem,

where, mod 7 is modulo 7. It means the remainder after dividing by 7.

If a natural number divide by 7 then the possible remainders are 0,1,2,3,4,5 and 6.

If remainder of n/7 is 0, then

If remainder of n/7 is 1, then

If remainder of n/7 is 2, then

If remainder of n/7 is 3, then

If remainder of n/7 is 4, then

If remainder of n/7 is 5, then

If remainder of n/7 is 6, then

For every n E N, when we divide n2 by 7, the remainder is either 0, 1, 2, or 4.

Hence proved.

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step-by-step explanation:

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answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204