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## Homework Statement

You may use pen-and-paper and mental calculation. You have 6 minutes time.

Give final digit of

$$ (22)^3 ~+ (33)^3~ +(44)^3~+(55)^3~ +(66)^3~+(77)^3 $$

## Homework Equations

3. The Attempt at a Solution

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I'm not terribly good at mental arithmetic myself. I was never particularly good, probably I am slightly below average or even worse. It's difficult to keep track of the partial results.

1.) Well, to me at first it looked like one could attempt to make the partial sums into the form of

## (20+2)^3## Separate the one's place, from the ten's place.

If you manage to remember what the binomial cube formula is (I think it's like... Newton's binomial formula in English). One could calculate mentally then what would become of those cubed sums (20+2)^3.

Then one could calculate the final sum and get the final digit.

2.)Calculating with pen-and-paper multiplications and then addition would probably be too slow.

Such as 22x22 = 484,

484x22 =10648

3.) Now that I think about it, another way to calculate would be.

##(22)^3=~9680+~968##

##(22)^3= (20+2)*(20+2)*(20+2)##

##(20+2)*(20+2)*(20+2) = 484* (20+2) = (480+4)*(20+2)##

##(480+4)*(20+2) = 9600 +960 +80 +8##

## 8= 2^3##

I'm starting to think that because of this kind of calculation as I performed in the third attempt... It seems as though only the one's place matters.

Is the problem so simple as to calculate

##~ 2^3 ~+3^3~+4^3~+5^3~+6^3~+7^3##

##= 8+27+64+125+216+343

= 783##

final digit = 3

double-checking with a real calculator the actual final digit will be 3 (in the original form calculation) I'm just not sure of my own reasoning it seems...