what would  b the last number in this square of numbers and how did  you find it?
916156
129\n\nEach number is the sum of the number to its immediate left, the number directly above it, and the number one square diagonally up and left of it. Thus, ? = 52+52+25 = 129
509467
1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113
868097
-\naside from 1,\n2,3,4,6,12
184988
One correction to Duane G's response. When you multiply the bottom by sqrt(3) you get sqrt(3) * sqrt(3) = 3, not 1. Therefore your final answer would be 2 * sqrt(6).
860621
That means write a binomial that if squared gives you t^2 - t + some constant then subtract that constant to "complete" the square.\n\n(t - 1/2)^2 would be t^2 - t + 1/4 so you have\n\n[(t-1/2)^2 - 1/4 ]- 1 = 0\n\n(t-1/2)^2 - 5/4 = 0\n\n(t- 1/2)^2 = 5/4\n\nsq rt both sides\nt - 1/2 = +- sqrt(5)/2\n\nadd 1/2 to both sides\n\nt = 1/2 +- sqrt(5)/2
748660
The answer is 78:\n\n1² + 2² + 3² + 8² = 1 + 4 + 9 + 64 = 78\n2² + 3² + 4² + 7² = 4 + 9 + 16 + 49 = 78\n1² + 4² + 5² + 6² = 1 + 16 + 25 + 36 = 78\n\nI did this creating a spreadsheet with the first 9 squares.  There were 126 ways (9 choose 4) ways to pick the four squares which then resulted in a sum.  I sorted the sums and found the one that repeated itself 3 times (78).\n\nThe answer is 78.
871673
So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
998294
square root of 13 is the answer
716502
CLUE 1: The number squared < 1200\nTake square roots: the number < 34.64...\nIt is an integer, so the number ≤ 34.\n\nCLUE 2: The number cubed > 29000\nTake cube roots: the number > 30.72...\nIt is an integer, so the number ≥ 31.\n\nCLUE 3: The fourth power of the number < 930000\nTake fourth root: the number < 31.05...\nIt is an integer, so the number ≤ 31.\n\nFrom CLUE 2 and CLUE 3 the number must be 31 (which is consistent with CLUE 1).
737774
You figure the square root of each number separately...you've probably learned that in class alredy (how to do that)...then you take reach result, and multiply them together for your final answer.\n\nUse your textbook for reference...good luck!
954164
grab a calculator and try 1.1 x = = = = =  until you get a number >2.\nthe first time you hit = is ^2, next is ^3 etc.  count.\n\nN=7
693802
They are all the sums of two smaller numbers.  If you mean which one is the sum of two smaller square numbers, then it is 25 (the sum of 16 and 9.)
935133
Ok, take it back.\n\nFlip the 6, turn it into a 9.  Move it over to the right hand side, move the 2 to an exponent over the 7.\n\n7^2 = 49.\n\nThat's it.
714877
b)\n\nNote that 314159266 is of the form 4k + 2, where k is an integer.  Since any integer is of the form 2n or 2n +1, this means that all squares of integers are either of the form 4m or 4m + 1.  No matter how we try to do the subtraction of two numbers of the form 4m or 4m + 1, we won't get a number of the form 4k + 2:\n\n4m - 4n = 4k\n4m - (4n + 1) = 4k + 3\n(4m + 1) - 4n = 4k + 1\n(4m + 1) - (4n + 1) = 4k.\n\nCheers!
886898
Find next closest square to 80 that is also a factor: 16\n\nSince 4*4 is 16, you can bring the 4 outside the radical and leave the product of the other factors underneath the radical: 4rt(5)
525085
There is not last number of numbers. In the binary system, or for any system for that matter, you could always add digits to it and the number would increase by at LEAST a power of 10.\n\nInfinity could be the last number, but contrary to popular belief, it is not a number, rather a term stating that something has no end
23859
I run through the numbers from 1-9, looking at each to see where the constraints ensure that there is only one spot left to place an additional number.  To do this, realize that each row, each column and each block need exactly one of each number, so starting with 1, I look for combos with limited possibilities.\n\nIf the blocks 1,2 have a 1 in them, then 3 should be easy to find, look at blocks 6,9 to see if they have a 1, and mentally cross out the rows and columns that those 1's eliminate and see whats left.  Similary, if 2 and 6 have 1's in them try looking at blocks 5 and 3 for enough eliminated squares to ensure there is only one remaining spot for 1 in those blocks. Once you've tried all the possibilities for 1, move on to 2, and keep going until you've been through the whole list of numbers once without finding any new entries.\n\nFinally, a trick I really like, is when a block has only one row or one column with space remaining. You know that the remaining numbers in this mostly complete block will be in that row or column, and so you can use that information and apply it to other blocks even though you could not exactly place the numbers in the mostly complete block.\n\nHope that helps!