Find a 3-digit number such that the cube of the sum of its digits is equal to the number.?
928493
Hmm... 370 and 371 both seem to work.\n\n* * * * * * \n\nI wrote a little program to test all the numbers from 100 to 999.  There are actually four answers:\n\n153\n370\n371\n407\n\nAlso, by "the cube of the sum of the digits", you mean each individual digit is cubed, then you add the sum of the cubes, correct?  For example, 153 is one of the answers because 1^3 + 5^3 + 3^3 = 153.  (It seems like a trivial question if you interpret it the way Dan did, since there are only five numbers to test.)
1094465
if the digits in the number are a,b,c,d,...\n\nthen the number is divisible by 11 if a-b+c-d+... is divisible by 11
889893
2+7\n3+6\n9+0\n\nThat's it. They all add up to 9... I think that's what the question is asking.
1115504
the cube root can be found using a calculator\nmake sure you have a scientific one though\ni had do use it alot in geometry\n\nits alot like a square root just three instead of 2
279409
Find every number between 101 and 999 that is divisible by 7 and then add them together.  Hmmm......like I really am going to do that.
678736
the sum of the two smallest numbers must be higher than the third one so your answer is d)
455504
I will explain a method that is the most easy on this planet, and a unique one.\n\nI will illustrate with the help of a simple example.\n\nEg. 4-2.\n\nDraw no. line as follows:\n\n<-6--5--4--3--2--1--0--1--2--3--4--5--6--7-->\n\nNumbers left to zero are negative integers\nand to the right are the positive ones.\n\nLocate 4 on this line. Now move 2 steps behind 4 as the operation to be carried out is subtraction.\n\nThis will give answer as 2. \n\nSimilarly for addition move the steps ahead.
46941
" Fill in the grid so that every row,\nevery column, and every 3x3 box\ncontains the digits 1 through 9.\nRead this page:\nhttp://www.sudoku.com/
67769
Divide 8 into 3\n.375
5567
According to Wolfram Research, \n"An n-digit  number that is the sum of the nth powers of its digits  is called an n-narcissistic number. It is also sometimes known as an Armstrong number, perfect digital invariant (Madachy 1979), or plus perfect number.\n\nThe values obtained by summing the dth powers of the digits of a d-digit number n for n==1, 2, ... are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 5, 10, 17, 26, ... (Sloane's A101337).\n\nThe smallest example other than the trivial 1-digit numbers is\n153==1^3+5^3+3^3.(1)\n\nThe first few are given by 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474, 54748, ... (Sloane's A005188). "
891258
Try 1 ÷ 3\n\n1/3 = 0.33333.....\n\n~ ♥ ~
1146202
It would be 0.5^1/3 (cube root of 0.5) which is approximately equal to .7937
306191
a) cube...final answer
274854
The difficult thing here is reading and writing the numbers. They must be transformed from ASCII and then the result transformed back to ASCII.I found a program that calculates the sum of 2 numbers.I hope it helps.
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.
894556
idk but cube root is easy look here is an example the cube root of 27 is 3 because 3*3=9*3=27 so 3*3*3=27 it is 3 numbers that is multiplied times itself 3 times and give u a whole number and the 3 same numbers u multiplied r called a cube root cube mean 3
1001765
If d times d times d = p times p and p = 9,\nThen 9 x 9 = 81.  What is the cube root of 81? It is 4.3