
I did it the same way. But then, I'm an accountant, so.... Kind of a nerd asis.





Exactly the same thought process, and I also know the squares in my head  up to 16 anyway. Above that there's a bit of mental arithmetic required.





I was about to "guess" 2 based on the same premise: first three and a "feeling" about the last two (:

If you say that getting the money
is the most important thing
You will spend your life
completely wasting your time
You will be doing things
you don't like doing
In order to go on living
That is, to go on doing things
you don't like doing
Which is stupid.





Everyone should know the squares  at least for 1 to 20. After that, you just need to know that multiples of ten are (x * 10)^2 = x^2 * 100. You can then do the halfways [numbers ending in 5] (x * 10 + 5)^2 = ((2x + 1) * 10)^2 / 4 [looks a lot more complicated than it is]; thereafter, for numbers ending in 1, 2, 6, 7 you apply (x + 1)^2 = x^2 + 2x + 1 (or x^2 + (x + 1) + x) [do it twice for 2 and 7] and for numbers ending in 3, 4, 8, 9 you apply (x  1)^2 = x^2  2x =1 or x^2  (x + 1)  x = x^2  x  x  1 [do it twice for 3 and 8]
At least, that's what I use! And I assure you, once you've got the hang of them that are simple.
Edit: It is also useful to memories powers of two and squares of prime numbers
modified 18Aug20 6:03am.





For those I don't know I use a binomial expansion in my head to make it easier. Not as quick as memorization (look up tables are always rather fast  even for computers).
So, if given 27 * 82 it would become (303)*(80+2)
Numbers to juggle mentally: +2400, 6, 240, +60
Corresponding to the outer two terms and the cross terms.
If you do it now and then it remains pretty efficient  but if you've not done it for a year or two it take some cobweb sweeping to set one's storage back to efficient levels.
2214
Ravings en masse^ 

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It appears to me the problem is summing the terms of numerator  not multiplying them. Isn't that a '+' between the terms? That's a considerably easier problem and one I can probably manage.
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You're quite correct, fixed it.
Wrong is evil and must be defeated.  Jeff Ello
Never stop dreaming  Freddie Kruger





About 160,000,000  my mental arithmetic runs out of registers to remember digits in after 5 or 6 digits ... and long division starts using 'em up fast ...
But some of it is easy: 10^{2}*11^{2}*12^{2}*13^{2}*14^{2} == 100 * 121 * 144 * 169 * 196
(10*10 is easy, and each square adds 2 more than the previous: 100 > 121 Adds 21, 121 > 144 adds 23, so the next two terms are 144 + 25 and 144 + 25 + 27)
We were rote taught our "times tables" up to 12 by 12, so the first three are imprinted on my brain...
long multiplication in your head is reasonably easy as long as you keep the decimal places straight.
[edit]
Oh, right, that's easier:
100 + 121 + 144 + 169 + 196 = 730
730 / 365 = 2
Easy peasy!
[/edit]
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10^2 = 1 0*2 0^2 = 100
11^2 = 1 1*2 1^2 = 121
12^2 = 1 2*2 2^2 = 144
13^2 = 1 3*2 3^2 = 169
14^2 = 1 4*2 4^2 = 196 (the last term (16) carries the one over, so 4*2+1 = 9)
Add those up: 365 + 365
(365 + 365)/365 = (365/365) ((1 + 1)/1) = 1 * 2 = 2





Elegant way to do it.
M.D.V.
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
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I did not use a calculator or a notepad.
I've been working in JavaScript for too long. This is how I visualized it:result = 0.0;
for (i = 10; i < 15; i++) {
result += Math.pow(i, 2);
}
result = result / 365;
The question reminded me of writing my first basic programs to solve high school geometry and advanced math homework problems. (class of '85)
"Go forth into the source"  Neal Morse





(122)²+(121)²+12²+(12+1)²+(12+2)² using (a+b)² = a²+2ab+b² will cancel those 2ab.
Hence remains 5*12² + 2*4 + 2*1 = 5*146 = 10*73 = 730. Divided by 365 = 2.
So the exercise is indeed for the application of (a+b)²+(ab)² = 2a²+2b².
modified 18Aug20 2:46am.





Yeah!
This was my solution as well.
Wrong is evil and must be defeated.  Jeff Ello
Never stop dreaming  Freddie Kruger





Though one mistake, (a+b)² + (ab)² = 2a² + 2b². Good we are still as intelligent as then (?).
Probably the same trick the teacher would demonstrate. It would be interesting if some mathematician historian would check whether such tricks were indeed collected for instruction  of numerical math.





I did it almost the same way, but decided not to multiply 144 by 5, since I knew I was going to divide by 5, since 365 is dividable by 5. So:
((122)²+(121)²+12²+(12+1)²+(12+2)²)/365 = (144+(2*(2²+1²))/5)/(365/5) = (144+10/5)/73=146/73 = 2






I did a similar approach but kept it factored as
5*12² + 2*4 + 2*1
5*12² + 2*(4 + 1)
5*12² + 2*5
divide numerator and denominator by 5.
(12² + 2)/73
(146)/73
2





1. Each square is approximately 20 more than the previous, so 5x100+20+40+60+80=700, estimating a correction for the approximation and assuming a whole number solution as it's a mental arithmetic problem then the total is 730 and the answer is 2.
2. What everyone else said.





1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6
=> (14*15*29  9*10*19)/6*365
=> 30*(7*29  3*19)/6*365
=> (7*29  3*19)/73
=> 146/73
=> 2
Ariel Serrano
Informatica Ambientale S.r.l. (www.iambientale.it)
Via Teodosio, 13, 20131, MI
Milan, Italy.





I even forgot this formula existed.
I also wouldn't have done it in my head.
Wrong is evil and must be defeated.  Jeff Ello
Never stop dreaming  Freddie Kruger





Use squares of binomials:
10^2 = (12  2)^2 = 12^2  4*12 + 4
14^2 = (12 + 2)^2 = 12^2 + 4*12 + 4
11^2 = (12  1)^2 = 12^2  2*12 + 1
13^2 = (12 + 1)^2 = 12^2 + 2*12 + 1
12^2 = 12^2
Add them up, sum = 5*(12^2) + 5*2 = 5 * 146
Denominator = 365 = 5 * 73
Hence ratio = 146/73 = 2
The difference of squares is quicker:
(14^2  12^2) + (10^2 12^2) = 26*2  22*2 = 4*2
(13^2  12^2) + (11^2 12^2) = 25*1  23*1 = 2*1
Hence 10^2 + 11^2 + 12^2 + 13^2 + 14^2 = 5*12^2 + 5*2 = 5 * 146
But this year is a leap year!





Referring to my previous two solutions, here is a third and better way to do the original problem:
14^2  2^2 = 12*16 (difference of squares)
13^2  1^2 = 12*14
12^2  0^2 = 12*12
11^2  (1)^2 = 12*10
10^2  (2)^2 = 12*8
Adding the five lines:
(sum of squares from 10^2 to 14^2) = 5*12^2 + (sum of squares from (2)^2 to 2^2)
This allows three more general problems to be investigated:
PROBLEM 1: Find all sums of five consecutive squares divisible by 365, and find the resulting quotients.
PROBLEM 2: Find all sums of (2n+1) consecutive squares divisible by 365, and find the resulting quotients.
PROBLEM 3: This year 2020 is a leap year. Replace 365 = 5*73 by 366 = 6*61, then by any fixed whole number.





We know that (a+b)²=a²+b²+2ab
So:
11² = (10 + 1)² = 10² + 1² + 2x10x1
12² = (10 + 2)² = 10² + 2² + 2x10x2
13² = (10 + 3)² = 10² + 3² + 2x10x3
14² = (10 + 4)² = 10² + 4² + 2x10x4
So:
10² + 11² + 12² + 13² + 14² = 5x10² + (1² + 2² + 3² + 4²) + 2x10x(1 + 2 + 3 + 4)
1 + 2 + 3 + 4 = 10
1² + 2² + 3² + 4² = 1 + 4 + 9 + 16
So:
10² + 11² + 12² + 13² + 14² = 5x100 + 30 + 2x10x10 = 500 + 30 + 200 = 730
730/365 = 2





Noticed that 10^2 + 11^2 + 12^2 = 365.
Then 13^2 = 169, 14^2 = 196  and those two summed are 365.
So dividing 2 lots of 365 by 365 gives 2 as the answer.
If you don't know your squares, use the identity (n+1)^2 = n^2 + n + (n+1)
10^2 = 100 (that's easy enough to remember!)
For the next, add 21 (10+11). Then add 23, then 25, then 27 for the other squares.
However  for the first three squares, we're adding 300 to 2*21 and 23  2*21=42, add 23 and you have 65, a total of 365.
The last two, we have 200 + (23+25+25+27) + 2*21+23. The last bit we know is 65. The middle bit is clearly 100, so the last two squares sum to 365 too.
Java, Basic, who cares  it's all a bunch of treehugging hippy cr*p



