the sum of all the digits that are used in writing the integers from 29-32 is 23 what is the sum of 1-1000?
483597
All the digits used in writing the numbers from 1 to 1000?\nFor the units digit, you use: 100 1's, 100 2's, 100 3's, etc, 100 9's.  So this sum is 100×(1+2+3+4+5+6+7+8+9) = 4500.\nFor the tens digit, it is the same, you use 100 of each.  So this is another 4500.\nFor the hundreds digit, you use another 100 of each, so this is another 4500.\nFor the thousands digit, you use 1 once.\nSo the sum = 4500+4500+4500+1 = 13501
750336
(100+a)(100+b) = 10,000 + 100(a+b) + ab\n 103*107\n=(100+3)*(100+7)\n=(100*100)+(100*7)+(3*100)+(3*7)\n=(10000)+(700)+(300)+(21)\n=11021
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.
873178
249 zeros end 1000!\n\nThere will be a zero for every factor of 10 multiplied in 1000!. Since 10 = 2*5 there will be as many factors of 10 as there are numbers of factors of 5.\nThere are 200 numbers from 1 to 1000 divisible by 5.  So 200 5's in these.\nThere is an additional 5 for every factor of 25=5*5.  There are 40 of these from 1 to 1000, so add another 40.\nThere is an additional 5 for every factor of 125 = 5*5*5, and there are 8 of these, so 8 more.\nAnd one more factor of 5 for 625 = 5*5*5*5.\nSo 200 + 40 + 8 + 1 = 249 factors of 5, thus 249 factors of 10 and 249 trailing zeros.\n\n2 is the first digit to the left of these zeros.\n\nMath proof too lengthy here and now.  http://www.justinwhite.com/big-calc/1000.html\n\nPretzels:  Good starting technique, however, in ignoring factors like 32 and 35 (because one has a 2 and one has a 5) you neglected to factor in the residue factors i.e., 32/2 = 16 and 35/5 = 7.  These will contribute to the last non-zero digit.  Also, the 3 in 30, the 4 in 40, th 6 in 60 etc. as well as the 3 in 300, the 4 in 400, the 6 in 600, etc. all contribute to this digit.\nThis is very complicated and will take a long boring time to type it all out.  Just use a computer/calculator to figure it.  It's 2.
603285
this is possible to figure out.  What it is a combination of numbers   i will do examples to keep it simple.   If your number was 00  or 000  then the answer was 9.  well its a pattern    if your number is 001 then the answer he came up with was 8.  well then if you had 002  then it was 7.  but also on the note if your number was 011 or some combination of this number then it was 8 also  because when you take 011 and add 0+1+1 you get 2 so it was 8.   the way it works is the numbers you type in it adds them together so there is a correlating number that goes with the answer that was added together.  so... if the numer added together was 0 then it was 9.  and the number added together was 1 then he came up with 8.  added together you got 2 then he said 7.  and just keep going ... 3 is 6,   4 is 5.    also on the same account    5 is 4,  6 is 3,   7 is 2,   8 is 1,  and 9 is 0.
981844
2 +28 -65 +48 +99,002,456,486 +1 -87 -998\n= 99,002,455,415\nNote to other answerers:  If you hold your mouse over the string of characters, it will show everything.
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
527322
1+1-21+30-5\n2-21+30-5\n-19+30-5\n11-5\n6
1110740
I -1 digit\nLOVE - 4 digits\nYOU - 3 digits
285803
It is\n\n   888\n  +088\n  +008\n  +008\n  +008\n --------\n  = 1000\n --------
1038063
If there was only one digit, there would be 10 (0-9)\nTwo digits, 100 (00-99)\nThree digits, 1000 (000-999)\n\nHey, there might be a pattern here\n\nSo 10 digits, 10,000,000,000 (0000000000-9999999999)
58688
I can show that the answer is 25927 without using computer.\n\n\nFirst, we can ignore 1000000 since its sum of digits is not 18. Also, let's add 0's in front of numbers to have 6 digits if the number has less than 6 digits. So our range is going to be from 000001 to 999999.\n\nNumber of partition of 18 into 6 whole number is\n(18+6-1)C(6-1)=(23)C(5). Assign these 6 numbers to the 6 digits.\n\nNow, we need to remove the cases where one digit is 10 or bigger. It is very good thing that two digits cannot be 10 or bigger, as the sum would go over 18.\n\nLet one digit has x, where x>=10. The sum of the other 5 digits is 18-x. Since 18-x is partitioned into 5 whole numbers, we have (18-x+4)C(4) possible ways. This should be multiplied by 6 since the big digit can be the first digit, second digit, ..., sixth digit. x is from 10 to 18. So we need to calculate 6((12)C(4) + (11)C(4) + .... + (4)C(4)).\n\nSince (n)C(m)+(n)C(m+1)=(n+1)C(m+1), we get\n(4)C(4) + (5)C(4) + (6)C(4) + (7)C(4) + .... + (12)C(4)\n=((5)C(5) + (5)C(4)) + (6)C(4) + (7)C(4) + .... + (12)C(4)\n=((6)C(5) + (6)C(4)) + (7)C(4) + .... + (12)C(4)\n...\n=(13)C(5). (This shows beauty of mathematics, doesn't it?)\n\nTherefore, the answer is (23)C(5)-6*(13)C(5)=25927.
1022090
99.\n\n2-55 can be thought of a 2 and then a matching tens and ones place 3 digits higher.\n\n6-xx then, can be thought of as the x being 3 tens and 3 ones higher than 6, ant that' 6-99, making 99 the answer.
187318
500+-200=-300\nwhen you add integers you subtract
1072126
10000\n\nThat's counting 0000, 0001, 0002, all the way to 9999
914266
20,000,000  (7)\n\n250,000,000(7)\n\n300,000,000 (8)
741904
You have a billion numbers from 000,000,000 to 999,999,999.  These are all the possible combinations of digits 0 to 9 at each position (units, tens, hundreds, thousands, tens of thousands etc... to hundreds of millions).\n\nYou have 100,000,000 numbers starting with a 0.\n100,000,000 starting with a 1.\netc...\n100,000,000 starting with a 9.\n\nSince 0+1+2+3+4+5+6+7+8+9=45, this means the sum of all first digits is 45 x 100,000,000 = 4,500,000,000.\n\nSimilarly you will find that the sum of alll second digits (tens of millions) is also 4.5 billion.  Same for third digits, etc... and for ninth digits(units).\n\nSo the sum of all digits from 0 to 999,999,999 is 9 times 45 billion = 40,500,000,000.\n\nYou need to add 1 for the number 1,000,000,000, so the answer is 40.5 billion + 1 = 40,500,000,001.