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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.
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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
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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!
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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.
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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
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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 tree-hugging hippy cr*p
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We know that (a - b)² = a² + b² - 2ab
So:
10² + 14² = 10² + 14² - 2x10x14 + 2x10x14 = (10-14)² + 2x140 = 4² + 2x140
11² + 13² = 11² + 13² - 2x11x13 + 2x11x13 = (11-13)² + 2x143 = 2² + 2x(140+3) = 2² + 2x140 + 2x3
12² = 144 = 140 + 4
So:
10² + 11² + 12² + 13² + 14² = 4² + 2² + 2x140 + 2x140 + 2x3 + 140 + 4
= 16 + 4 + 5x140 + 6 + 4 = 20 + 700 + 10 = 730
And 730/365 = 2
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use ((12-2)**2 + (12-1)**2 + 12**2 + (12+1)**2 + (12+2)**2)
the -2ab from the first two cancel the +2ab from the second two. so
5 * 12**2 + 4 + 1 + 0 + 1 + 4
5 * 144 + 10
divide top and bottom by 5
146/73
(which is essentially what Joop said. But I scrupulously didn't cheat by looking at previous. The hardest part was not picking up a pencil!)
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Member 4317199 (Paddy) wrote: The hardest part was not picking up a pencil
Yeah, even when you realize there's an easier way it's still not that easy to do in the head.
Wrong is evil and must be defeated. - Jeff Ello
Never stop dreaming - Freddie Kruger
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10^2 + 11^2 + 12^2 + 13^2 + 14^2 = 10^2 + (10 + 1)^2 + (10 + 2)^2 + (10 + 3)^2 + (10 + 4)^2
= 5 x 10^2 + some junk... i.e. 500 + junk
in that junk there is 2x10x1 + 2x10x2 + 2x10x3 + 2x10x4 = 2x10(1 + 2 + 3 + 4) = 2 x 10^2 = 200
that's 700 + what's left of the junk
at this point it became obvious that either the result is 2 or it's best to use a calculator.
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OK... I failed to do it in my head, but worked it out in text below then checked it in Excel
... and still was wrong so made I the needed corrections (2 errors compounded) to my text
So... I'm dumber than a 19th century schoolboy, but it was a fun exercise any way. I used to do all my math in my head before we had pocket calculators (yes I'm an old fart) I need to do more of this kind of thing to get that back.
0 times 10 is 0 plus 100 is 100
1 times 11 is 11 plus 110 is 121 plus 100 is 221
2 times 12 is 24 plus 120 is 144 plus 221 is 365
3 times 13 is 39 plus 130 is 169 plus 365 is 4 carry the 1 and 2 plus 1 for 3 carry the 1 and 3 plus 1 plus 1 is 534
4 times 14 is 56 plus 140 is 196 plus 534 is 0 carry the one and 2 plus 1 for 3 carry the one and 5 plus 1 plus 1 is 7 for 730
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Member 11577008 wrote: OK... I failed to do it in my head, but worked it out in text below then checked it in Excel
Upvote for honesty.
Member 11577008 wrote: So... I'm dumber than a 19th century schoolboy
Don't count on it, from what I can see in the picture there is one kid that whispers something in the ear of the teacher, the rest are still thinking.
Wrong is evil and must be defeated. - Jeff Ello
Never stop dreaming - Freddie Kruger
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To solve mentally with no use of paper, I need to imagine 5 squares, made of stones laid out on a table.
we have first square that is made of 10x10 red stones.
second square is made of 10x10 red stones plus 1x10 green stones at top and 10x1 green stones at right, then to fill the square we have a 1x1 square of blue stones at top right corner.
third square is made of 10x10 red stones plus 2x10 green stones at top and 10x2 green stones at right, we fill the square with 2x2 stones.
etc...
red stones are 500 (5x10x10)
green stones are 200 (20 on second square, 40 on third, 60 on fourth, 80 on fifth)
blue stones (squares of 1, 2, 3, 4) are 1+4+9+16 = 30
total 730 stones.
730 / 365 = 2.
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Visual solving, I like it.
Wrong is evil and must be defeated. - Jeff Ello
Never stop dreaming - Freddie Kruger
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The difference of two consecutive squares is the higher number * 2 - 1. So 11^2 = 10^2 + (11 * 2) - 1 = 100 + 22 - 1 = 121. Therefore the answer is [(100 * 5) + (11*2-1)*4 + (12*2-1)*3 + (13*2-1)*2 + (14*2-1)] / 365 = 2. My gut answer is that the answer was probably an integer with 2 being likely.
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e
looks like the same kind of sequence.
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Easiest i could think of was to simplify the numerator to (12-2)^2+(12-1)^2+12^2+(12+1)^2+(12+2)^2
And then it becomes a more manageable 5*12^2 + 2*1^2 + 2*2^2
At which point most folks would find it easy to compute the numerator to 730. Some may even factor out the 5.
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This is how I mentally solved it. Well, I used my fingers, also.
10^2 + (10 + 1)^2 + ... + (10 + 4)^2
taking into account (a + b)^2 = a^2 + 2ab + b^2, we have
10^2 appears 5 times: 500
the 2ab term gives:
2 * (10*1 + 10*2 + 10*3 + 10*4) = 20 * (1 + 2 + 3 + 4) = 200; 700, up to now
then, the sum of the squares:
1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30.
numerator sums 730. denominator: 350 + 15, times two gives 700 + 30;
answer: 2
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Brute force. All in head, no paper or anything else. Wrote the post after doing it.
10x10 = 100
11x10 = 110+11 = 121 (don't remember this one)
12x12 = 144
subtotal = 365
13x10 = 130+39 = 169 (don't remember this one)
14x10 = 140+56 = 196 (don't remember this one)
subtotal = 365
total = 730
divide by 365 = 2
Add one more pair of numbers and I might not have been able to do it. I dropped the 169 on the floor once before adding the second pair.
I never learned the complete multiplication tables as I could do the above sort of math quickly enough to get by.
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I arrived at "approx 2" like so:
Jörgen Andersson wrote: 102 + 112 + 122 + 132 + 142
is approx. 5 * 144 (at least one square I know) Take the five out of 365 And you have 144/73 = 2-ish.
"If we don't change direction, we'll end up where we're going"
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Five consecutive squares always average to the middle square plus 2.
146 * 5 = 730.
Answer = 2
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That is some amazing trivia to know
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Yeah, 2. Multiply, add, divide. I've seen enough Vedic math ways to do things that I know my brain-dead solution could be improved.
I never compare myself with any student from any time other than to say, "I'm still learning." Imagine running into a completely unschooled guy who comes up with this for an approximation of pi?
Srinivasa Ramanujan - Wikipedia[^]
Say WHAT!?
Dang! My '58 Renault Dauphine has another flat tire.
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sum of the series (x+n)^2 , 0 <= n <= 4, x=10
= 5(x^2 + 4x + 6) , x=10
= 5(100 + 40 + 6)
= 5(146)
= 730
A contrarian solution to mental arithmetic
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I'd say 2 too.
10^2=100
14^2=196
more or less the average of extreme values is (100+196)/2~150
150*5=750
750 must be not so far from the sum of the powers but 750/365~ 2.something
than the reply most probably is 2 (5 secs the whole thing)
modified 3-Sep-20 2:59am.
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