How do you get 100 by using the numbers 1-9 once?\nExample: 1+2+3+4+5+6+7+(8x9)= 100\n\nCan you find another way to get it?
217740
(1 + 9) x (5 x 2) + (6 + 8) - (3 + 4 + 7) \n\nsome of those parentheses are superfluous, but they make it easier to see how it works.
36809
You must be very confused. \n\n1+1 = 2 plus one more is 3
329253
s = 1/(1*3) + 1/(3*5) + 1/(5*7) + 1/(7*9)+ ... + 1(197*199)\n\nSum of one term: 1/3\nSum of two terms: 2/5\nSum of three terms: 3/7\nSum of four terms: 4/9\nSum of five terms: 5/11\netc.\n\nBy observation:\nSum of n terms: n/(2n-1)\n\nTherefore:\nSum of 99 terms: 99/199\n\nCorrect answer is 1) 99/199\n\nEdit: It appears my answer is very similar to Babatunde... but we came up with them independently.  I think mine is just a little clearer however.
60183
the closest i could come was 100 = 95 + 4 + 16/28 + 3/7\n\nokay i done it properley and i'm right. You can rest easy now everyone.\n\nDon't stress - only one answer.
469623
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.
333425
find whole (integer) number a,b, and c such that a^3+b^3=c^3.\nDon't remember the answer anymore,but it should be easy - all numbers were less that 10.
570267
I tried and I found that 100! is perfectly divisible by 350 as 350=2*5*5*7=50*7.\n\n100!= 100*99*98*97!\nnow 100 is divisible by 50 and 98 is divisible by 7.\n\nSo the least number is 1.
473629
Are we counting the number of times the digit zero appears?  For example, would 1,000,000 result in 6 zeroes?  If so, you can count the zeroes as follows:\n\nFirst, count the numbers of the form nnnnnn0... 10, 20, 30, ..., 100, 110, 120, ..., 999990, 1000000.  Every 10th number will have this form, so that is 100,000 zeroes in the units place (don't count zero, but do count 1 million).\n\n1,000,000 * 1/10 = 100,000\n100,000 - 1 + 1 = 100,000\n\nNext, count nnnnn0n... 100, 101, 102, 103...  109, 200, 201, 202, ... 209, .... 999900, 999901, 999902, ..., 999,999, 1,000,000.  This would be 10 out of every hundred numbers, so that is 99,991 zeroes in the tens place.  (You don't count 0 to 9), but you do count 1 million.\n\n1,000,000 * 10/100 = 100,000\n100,000 - 10 + 1 = 99,991\n\nNext, count nnnn0nn.  This would be 100 out of every thousand numbers, but not counting 0 to 99, but counting 1 million.  So that is 99,901.\n\n1,000,000 * 100/1000 = 100,000\n100,000 - 100 + 1 = 99,901\n\nNext, count nnn0nnn.  This would be 1000 out of every ten thousand numbers, but not counting 0 to 999, but counting 1 million.  So that is 99,001.\n\n1,000,000 * 1000/10000 = 100,000\n100,000 - 1,000 + 1 = 99,001\n\nNext, count nn0nnnn.  This would be 10000 out of every hundred thousand numbers, but not counting 0 to 9999, but counting 1 million.  So that is 90,001.\n\n1,000,000 * 10000/100000 = 100,000\n100,000 - 10,000 + 1 = 90,001\n\nFinally, count n0nnnnn.  Without a leading digit, none of the numbers except 1 million would have this pattern:\n\n1,000,000 * 100000/1000000 = 100,000\n100,000 - 100,000 + 1 = 1\n\n100,000 + 99,991 + 99,901 + 99,001 + 90,001 + 1 = 488,895\nSo the answer is that the digit zero appears 488,895 times in the numbers from 1 to 1 million, inclusive.
451425
this is also true with any odd number (3...7,9,11,etc) not just 5.\n\nThe reason is because you are taking the average of the 5 numbers.  the average of a number is the exact middle of the set of numbers.  Look at the numbers in a line..\n 1+2+3+4+5\n\nnow if you look at the middle number in that line, it is 3.  if you add all the numbers together and divide by 5, you will get 3.  \n\nnow try it with 7 digits\n 1+2+3+4+5+6+7\n\nthe middle number is 4.  now add them up, and you get 28. divide by 7 and you get 4.  MAGIC!!\n\nnow try it with 4 digits.  NOTE: this is an even number, you will get a decimal.\n1+2+3+4\nthe middle number is...well..there really isn't one. its is between 2 and 3, isn't it? ok, now add it up, and you get 10. divide by 4 and you get 2.5.  isn't 2.5 right in the middle of 2 and 3??\n\nthere you go, i hope that works for ya\n\n(btw,this trick will only work if the numbers are consecutive)
508517
SOLUTION WITH WORKING\n----------------------------------------\nSince it is a 2 digit number,say xy i.e.(10x+y in mathematical terms) ,where x is tens' place and y is units'\n\nThe given information says:\n  \n6(x+y) = (10*x)+y\nSo,\n\n6x+6y=10x+y\n4x=5y\n\nIn the given region(10 to 99),searching integer values of x and y,\nwe gwt\nx=5 and y=4 \n\n(Next solution comes when x=10,y=8  but x is single digit number hence this and all further possibilities get discarded.)\n\nHENCE the number is xy=54
28765
1+5+10+25+50+100\n\n$1.91
318166
5050\n\nIf you add the numbers like this\n (1 + 100) + (2 + 99) + (3 + 98) + . . . . + (50 + 51) = ?\nthen each set adds up to 101 and there are 50 sets of numbers, so 101*50=5050
491285
but 25+25 is not=100\nit is = 50
292904
Nice problem. Here's the solution: 1000! is a multiple of 9 which is the same thing with saying that: S(1) = the sum of its digits is a multiple of 9 (Using the following known fact: a number is divisible by 9 if and only if the sum of its digits is divisible by 9). \n   Clearly S(1) < 1000! and 9 | S(1). Summing the digits of S(1) one obtains a number S(2) which is strictly smaller than S(1) and is a multiple of 9. And so on, until finally one arrives at a one digit number, say S(r) for some r  which is divisible by 9. But this can only be 0 or 9. Since S(r) is the sum of the digits of the previous number obtained by this algorithm, namely S(r-1), it can not be 0. So it must be 9! \n   Note that this type of proof works for any number divisible by 9, not only factorials of numbers, and in fact, with exactly the arguments displayed above it can be prooved that the single digit after adding the digits of an arbitrarly chosen number N is the digit d > 0 such that N-d is a number divisible by 9. (If you don't belive it, simply work out a computer programme that verifies this for all the numbers smaller than 1 000 000 for example).
483202
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.
297439
With just ! (Factorial), +, -, * and /, it is impossible to get 150 using only 3 5s without the x^2 (Squared) operation (Which means x*x, in case you are in Fifth Grade), or using a '^2' as you say. If you say there is an alternate way, and there really is, then I give up and admit defeat. (I'm sure most others do.) If you are so sure, why don't you post the answer yourself?
453523
8-4=4\n4+5=9\n9divided by 3/8=24\ni used to love that game, lol\n\n\n(8-4)+5\n---------------\n   3/8