Wow. I don't know any fancy way to do this but I would just go up the list until I was able to rule out four of the five numbers. Start with 4 + 9 + 16 +25 + 36 + 49 + 64 then go from there.
Which of the following cannot be the last digit of the sum of the squares of seven consecutive numbers?
I have no idea where to begin or what to do, i tried using the squares of 1-7
but obviously that wouldn't give me an answer (hopeless thing to do)
Any help appreciated.
Consider the last digit of a number and the last digit of it's square:
If you add the squares of numbers with end digits 0 to 6 you'll get 31 that means the end digit 1;
if you add the squares of numbers with end digits 1 to 7 you'll get 40 that means the end digit 0;
if you add the squares of numbers with end digits 2 to 8 you'll get 43 that means the end digit 3;
if you add the squares of numbers with end digits 3 to 9 you'll get 40 that means the end digit 0;
and so on...
Compare your results with the given answers (and then pick the right one!)
If I didn't make a mistake the answer is D.
Note the last digit of all squares.Which of the following cannot be the last digit
of the sum of the squares of seven consecutive numbers?
. . . .
We see that squares cannot end in 2, 3, 7, or 8.
Let the 7 consecutive numbers be: .
Then we have:
Suppose the sum ends in 7.
. . Then must end in 1.
. . And must end in 7.
But no square ends in 7.
(a well-known identity).
Then the sum of the squares of the 7 consecutive integers ending in n is
Evaluating for n = 0, 1, 2, ..., 9 modulo 10 will then reveal the possibilities for the last digit of the sum.
[Edit] Soroban posted his answer while I was composing this-- essentially the same approach? Or maybe not.[/edit]
Thanks guys both methods (although similiar) are definately useful and yes the answer is "D" (7). May I ask how you guys go about solving these, is it just some logical approach you take?
By the way, does modulo = remainder?
Eg: 10/4= 2 r 2 or 2 mod (2) ?