CRAP. Yours is complicated... But I will try to learn it... My method is easier... x+y square, where x = 70, y = 4. xsquare, plus 2 xy, plus y square... 70*70 = 4900 70 * 4 * (2) = 560 4*4 = 16 all added together, children, 5476
*buzzer sound* Sorry that's wrong. I'll get you started. The closest number to 74 that ends in 0 is ___ What did you do to 74 to get to the number that ends with 0? Well do the opposite of that to 74. You can take it from here. Pugs
I will bite. What's the EASY question and what's the answer? Or, you might be saying that it is an EASY question, but there is a non-complicated method... [your turn]
It's just two freaking digit number times itself. Take 74 for example. 4*4=16, that give you last digit 6. 4*7=28 times 2 = 56 add 1 from 16 gives you 7 for the next to last digit. 7^2 = 49 add 5 give you 54. 5476 Here you go. Now ready to solve some dynamic optimization problems or lattice graph theory problems?
ok amateurs. the squares from 1 to 40, memorized. most of them from 41-60 memorized (otherwise you simple do 2500 +/- the difference from 50 it is times 100 and add the square of the difference, same as pugs was explaining). all others just do them real quick in head. knowing when a number is divisible, that's what i always use. what's 55^2 + 45^2? well, you see 55? take it's digits 5 and 5. square 'em to get 25 and 25. add 'em to get 50. i bet the answer is 5050. you know how many years we took number sense tests in high school before we discovered that trick? it only works if the ones digits of the 1st number is 1 greater than the tens of the other and if the tens digit of the 1st number and ones digit of the second sum to 10. wanna know the 10th pentagonal number? how many sides in a pentagon? 5. 2 less than 5 is 3. 4 less than 5 is 1. (3*10^2 -1 * 10)/2 = 145. works for any shape. wanna know the sum of the factors of a number? 24. prime factors into 2^3 and 3. so take (2^4-1)/(2-1) * (3^2-1)/(3-1) = 60. oh, and there are many more. i doubt anyone reads this since it's at the bottom of the page but i'm posting it anyway.