Mental arithmetic

[kmoorevs]

I did not use a calculator or a notepad. :laugh: 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

[James McCullough]

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

[Nelek]

Elegant way to do it. :thumbsup:

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? Help me to understand what I'm saying, and I'll explain it better to you Rating helpful answers is nice, but saying thanks can be even nicer.

[Joop Eggen]

(12-2)²+(12-1)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+(a-b)² = 2a²+2b².

[Jorgen Andersson]

Yeah! This was my solution as well.

Wrong is evil and must be defeated. - Jeff Ello Never stop dreaming - Freddie Kruger

[Caslen]

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.

[aserrano]

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.

[Jorgen Andersson]

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

[Joop Eggen]

Though one mistake, (a+b)² + (a-b)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.

[Member 12207222]

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!

[Frank Malcolm]

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.

[Davyd McColl]

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. - Alan Watts https://www.youtube.com/watch?v=-gXTZM\_uPMY

[Menuki]

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

[Stuart Dootson]

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

[Menuki]

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

[Martin ISDN]

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.

[Member 4317199 Paddy]

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!)

[jsc42]

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

[bjongejan]

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:

((12-2)²+(12-1)²+12²+(12+1)²+(12+2)²)/365 = (144+(2*(2²+1²))/5)/(365/5) = (144+10/5)/73=146/73 = 2

[W Balboos GHB]

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 (30-3)*(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

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