Develop an algorithm that will allow someone to look at a number written as a base five numeral and know wh?
912127
Add the digits of the base 5 number. If the sum is even, the number is even. If the sum is odd, the number is odd.\n\nEven simpler:\nCount the number of odd digits in the number. (Ignore the even digits.) If the number of odd digits is odd, the number is odd, and if it's even the number is even.
697965
every number from 0 to 10\n\nplus placed in alphabetical order
164812
If it does not contain a decimal place I guess:\n5?
1152676
To write a very very very LARGE or SMALL decimal number
352166
8\ntake away o from top and you're left with o on bottom - half - nothing\nor i guess 10 could work, too (or any double digit ending in 0) take away the first number and you're left with 0
1127227
N = 285714 and 1.5*N = 428571\n\nHere's the analysis:\n\nLet's assume that it's an n digit number with digits x_1, x_2, ...  x_n. Then\n  N = Sum{i=1..n}(10^(n-i)*x_i)\nand the number you get by moving the last digit to the front will be:\n  10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nWhat you are asking for is:\n  1.5*Sum{i=1..n}(10^(n-i)*x_i) \n    = 10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nor:\n  1.5*x_n + 1.5*Sum{i=1..n-1}(10^(n-i)*x_i) \n    = 10^(n-1)*x_n + Sum{i=1..n-1}(10^(n-i-1)*x_i)\nor by transposing to the other side:\n  1.5*Sum{i=1..n-1}(10^(n-i)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10^(n-1)*x_n - 1.5*x_n\nor: \n  Sum{i=1..n-1}(1.5*10^(n-i)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  Sum{i=1..n-1}(15*10^(n-i-1)*x_i) - Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  Sum{i=1..n-1}(14*10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\nor:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\n\nNow the left hand side is a positive integer and on the right hand side 10^(n-1)-1.5 will be a fraction ending in 0.5. So x_n has to be an even digit to make (10^(n-1)-1.5)*x_n a positive integer. \n\nNow, if x_n were 2 or 6, multiplying it with something-point-five will give you an odd number. But the left hand side is an even number (because you are multiplying by 14). So x_n cannot be 2 or 6. That leaves 4 and 8 as the two possible choices for x_n i.e., the number N has to end in either 4 or 8.\n\nIf x_n is 4, then we are looking for:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*4\nor:\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 4*10^(n-1)-6\nor: \n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-3\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-3+10-10\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 2*10^(n-1)-10+7\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 20*10^(n-2)-10 + 7\nor:\n  7 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10*(2*10^(n-2)-1) + 7\nor:\n  Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = 10/7*(2*10^(n-2)-1) + 1\n\nAgain, the left hand side is a positive integer. The right hand side can be a positive integer only if 2*10^(n-2)-1 is a multiple of 7. \n\nFor values of n, 2*10^(n-2)-1 is 1(n=2), 19(n=3), 199(n=4), 1999(n=5), 19999(n=6), etc. n=6 is the first number in this series (19999) that's a multiple of 7. Putting n=6, we get:\n  Sum{i=1..5}(10^(5-i)*x_i)\n    = 10/7*(2*10^(6-2)-1) + 1 = 28571\ni.e., a five digit number = 28571.\n\nThat gives us our first answer: N = 285714 and 1.5N = 428571.\n\nNow I have only done the analysis so far for x_n being 4. If you do similar analysis for x_n=8, you get\n  14 * Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (10^(n-1)-1.5)*x_n\n    = (10^(n-1)-1.5)*8\n    = 8*10^(n-1)-12\n    = 8*10^(n-1)+2-14\nor:\n  Sum{i=1..n-1}(10^(n-i-1)*x_i)\n    = (4*10^(n-1)+1)/7 - 1\n\ni.e., 4*10^(n-1)+1 has to be a multiple of 7. For various values of n, 4*10^(n-1)+1 gievs you numbers like 41, 401, 4001, 40001, etc. and the first number in this series that's a multiple of 7 is 400001 for n=6. So we end up with:\n\n  Sum{i=1..5}(10^(5-i)*x_i)\n    = (4*10^(6-1)+1)/7 - 1 = 57142.\n  \nThat gives us our second answer: N = 571428 and 1.5N = 857142.\n\nWhile this is also correct, the first number is smaller. So the smallest number that satisfies your constraint is 285714.
727481
101=5,1010=10,1111=15,10100=20...\nThis pattern goes up by 5.\nThe pattern is binary.  The Binary System.  What computers use.\n11001=25,11110=30,100011=35,...\n1st digit =1 or 0.\n2nd digit =2 or 0.\n3rd digit =4 or 0.\n4th digit =8 or 0.\n5th digit =16 or 0.\n6th digit =32 or 0.\n7th digit =64 or 0.\nSee the pattern?\nThe digit's value changes by increments of 2.\n2^0=1, 2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32,...\nIn the numbers the 1st digit is the ones place, 2nd = 10's , 3rd =100's place...\n\n101=5 because the one in the 100's =4 and the 1 in the ones place =1 and the 0 in the 2nd place = 0.\nAdd the numbers.\n4+0+1=5.\n\n1010=10 because 4th =8 3rd=0 2nd =2 and 1st =0.\nAdd the numbers.\n8+0+2+0=10.\n\n1111=15 because 4th =8 3rd = 4 2nd = 2 1st =1.\nAdd the numbers.\n8+4+2+1=15.\n\n10100=20 because 5th =16 4th=0 3rd=4 2nd =0 1st=0\nAdd the numbers.\n16+0+4+0+0=20.\n\nHope this helps.
768317
The letter o is a number
1033220
hold alt and then type on the numberpad 2-2-7 π
596727
Figure out what other number times itself equals your number.  That's its square root.
785570
try substituting letters for numbers and vice versa
872418
Try to watch here to search some software with Fibonacci serial numebers:\nhttp://www.google.it/search?hl=it&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=fibonacci+program&spell=1
989362
displays numeric information in a way that is visually easy to understand
1143191
"A" for A, B, C's ...\nJust like the "1" lights up for Number Lock instead of "#"\n\n"A" seems like a good place to start!
813096
Start at the decimal point.  \n\nMoving LEFT, you have digits/units (0), tens (9), hundreds (8), thousands, ten-thousands, hundred-thousands, millions, ten millions, etc.\n\nMoving RIGHT from the decimal point, you first have tenths (no 'units'), then hundredths, then thousands, ten thousandths, and so on.\n\nGood luck!!
1069170
13212211\n\nJust counting numers in the previous number.\n\n'm i right??
842688
Think of a number between 1 & 10.\n-----3\n\nMultiply your number by 9.\n-------27\n\nAdd the digits of your answer. (if it was 23 then 2+3=5)\n---------9\n\nWhen done correctly you should be left with a single number. Now minus 5\n---4\n\nIf your number is 1 your the letter A. If it's 2 your the letter B. 3 your C. 4 your D 5 your E etc etc.\n--------------D\n\nGo to the next letter in the alphabet after yours.\n---------E\n\nChoose an animal beginning with that letter no birds or insects though!!!\n---------elephant\n\nHave you thought of one? I'm gonna try and guess it.... ***PHANELET***\nIf you can rearrange these letters to get your animal I got it right ;-)\n\n----------pretty good ;)