My daughter is in 3rd grade and her homework instructions confused me.  It states "Write the value of the digit 9 in standard form" Example 1: 495,703\nI would think the value in standard form would be 90,000\nAm I right, or am I digging into this too far?\nAny help would be appreciated.
215356
In the number: 495,703 the value of 9 is 9 X 10,000 = 90,000.\n\nActually IF I recall correctly (been a long time since 3rd grade math) the value is 10,000.\n\nI think this lesson is probably about learning the value placement in the positioning of numbers so that understanding big numbers becomes easier. But the wording of this question to a 3rd grader (AND some other people too) IS not easy to comprehend, so hopefully, I am telling this answer correctly.\n\nSounds like it's time for a parent/teacher conference.
443122
There's a better way of looking at it.\n\nFor every real number x in [0,1) you define D(x, n) as the nth decimal of x.  This is defined for every natural number n.\n\nFor x=1/3, we have D(1/3, n) = 3 for every n.  The number 0.3333... (the dots are important, they indicate that the nth decimal is a 3 for every n) is not "very close" to 1/3, it does not "tend toward" 1/3.  It is EXACTLY the number 1/3.  Show me a number other than 1/3 and I will show you that its decimal representation is not 0.33333...\n\nFor 0, we have D(0,n) = 0 for every n.  There is never a '1' decimal.  If you give me an arbitrarily small number 0.00000...1, the digit 1 is at a finite position p.  I can then reply with a number such that D(x,n)=0 for all n<=p, and D(x,p+1) = 1.  That number is greater than zero and smaller than the number that you gave me.  \n\nThe 0.5 = 0.49999... thing is just a quirk of decimal notation.  We can formally prove the equality as follows:\n\nLet a = 0.5, i.e. D(a, 1) = 5 and D(a,n) = 0 for all n>2.\nLet b = 0.4999..., i.e. D(b, 1) = 4 and D(b,n) = 9 for all n>2.\n\nProof by contradiction that a=b: Let's suppose on the contrary that a > b.  This means there must be a number x between a and b.\n\nLet's figure out the digits of x.  Since x is less that a=0.5, its first decimal must be a 4.  D(x,1) = 4.\n\nThe second decimal of x must be 9 since x > b `> 0.49.\nThe third decimal of x must be 9 since x > b >0.499.\nAnd so on, the nth decimal of x must be a 9 because x > b > the number y such that D(y,1)=4, D(y,i)=9 for i=2 to n and D(y,i) = 0 for i>n.\n\nTherefore x has exactly the same digits as b=0.499999... (I just showed that for every n, the nth digit of x is the same as the nth digit of b), and so x=b.  Contradiction, ergo a = b.\n\nThat's the difference between 0.499999... vs. 0.5 and 0.0000...1 vs. 0.  In the latter case, it is possible to find a number that is between the two.
640634
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.
20804
You need to round off the number to 4 digits and use the proper exponent to reflect the entire number:\n\n1.496 x 10^11  Your first answer is correct.
540057
Start by picking the units digit, you have 4 choices {1, 3, 5, 7} since 9 is not a choice.\n\nNext pick the ten-thousands digit (one of the digits is removed already, but you can't have zero or nine as the first digit either.)  So there are 7 choices for the ten-thousands digit.\n\nFor the thousands digit you have 7 choices (take away the two choices from before, and ignore the 9, but now you can use zero).\n\nContinuing down, you have 6 choices for the hundreds digit and 5 choices for the tens digit.\n\nThat comes out to 7 x 7 x 6 x 5 x 4 ways or 5,880 ways for an odd five-digit number with distinct digits (no nine).\n\nFor the decreasing digits, start with:\n87654321 (it can't include 9 nor can it end in 0).\n\nYou need to remove 3 digits leaving the last digit as a 1 or 3 (to be odd)\n\nIn the first case (ending in 3), you will have:\n876543 and one more digit to remove (5 choose 1, or 5 choices)\nThey are:\n87653, 87643, 87543, 86543, 76543\n\nIn the second case (ending in 1), you start with:\n87654321\n\nYou must leave the 1, so you have 7 digits choose 3, or 35 ways to remove 3 digits.\n87651, 87641, 87631, 87621, 87541,\n87531, 87521, 87431, 87421, 87321,\n86541, 86531, 86521, 86431, 86421,\n86321, 85431, 85421, 85321, 84321,\n76541, 76531, 76521, 76431, 76421,\n76321, 75431, 75421, 75321, 74321,\n65431, 65421, 65321, 64321, 54321\n\nAdding it up there are only 40 ways to create a 5 digit odd number, with distinct digits in descending order (no nine).
81222
Maybe because we're in base 10 and 9 is the highest base10 one-digit number? Try other bases, and you might get the same conclusion^_^(E.g. 1 in base2, 2 in base3, 3 in base4, 4 in base5, etc.) Just remember that when you sum them up you are still in that base!!!^_^
206321
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
293289
0123456789\n6210001000\n\nHere you can see there are 6 zeros, 2 ones, 1 two, and 1 six.\n\nI suppose you were looking for something more like this though.\n1234567890\n2100010006\n\nFor those of you with the 1000.0008 solution, where is the 1 8?  I suspect that solution is incorrect.
670616
It's so that the person you are talking to knows what you are talking about.\n\nWhat would the alternative be? "Twelve and seven hundred and eighty-nine thousandths?" That's a bit cumbersome.\n\nThe decimal point *creates* a division between the numbers, so we need to replicate that division by saying "point." It just makes sense.
37744
I rember doing something like this: you had to take the int you store the input in, and divide it by ten with the mod operator (%), and store it in a new int.  Since the mod operator divides the two numbers, and then returns the remainder, that will separate your last number.  For example:\n\nSay the input is 4321, and you have it stored in int input;\n\nint first_num = input % 10;\n\nthis takes your value in input, divides it by 10, then takes the remainder and stores it in first_num.  In this case, first_num should now equal 1.  \n\nAfter that, you have to shorten your input from 4321 to 432 by dividing (regular dividing this time) it by ten, and storing it in another new int.  (Since int can't have any decimals, dividing by ten just shortens it by the first number).  Then you do the whole thing over until you have all four numbers in separate ints, and then output them to separate lines.  \n\nHope that's clear enough--it's hard to explain this stuff so it's easy to understand.  Good Luck!
123220
such a no. cannot exist but hey dude a great question lets look at it this way\nthe last digit of the given no. will be 6\nand the resulting no will be 4 times that no\nlast digit of 4*6 is 4 so the resulting nos last digit digit has to 4 \nthat would mean the first digit of the original no will  be 4\nso just as an eg let us consider the no. 46\n46*4 will be 184\nthat is the first digit will be 1(or 2 or 3 or 4 but never 6)\nthis will be followed by every no \nso u c such a no just canno exist
82972
It is true that the ones digit for 5! and up is 0.  The reason for this is that when the 5 is included, the number would have to end in either a 0 or 5.  Then when you multiply by the 2, it has to be even, so it's the 0.  So you would only need to add up 4! + 3! + 2! + 1! + 0!\n\nDon't forget that 0! is 1.\n\n24+6+2+1+1 = 34, so the one's digit will be 4.
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
465138
i would guess this is 3rd or 4th grade?\n\nyou could use a calculator of course . . .\n\nto show your work, you need to do it on paper\n\ngood luck!
206223
Create a method called numDigits(long). Then you can use code something like this:\n\nint num;\ncin >> num;\n\nint numDigs = numDigits( num );\nint digits[ numDigs ];\nint temp = num;\nfor ( int i = numDigs - 1; i >= 0; i-- )\n{\n digits[ i ] = temp % 10;\n temp = (int)( temp / 10 );\n}\n\nThe digits array holds each of the digits in the number.\nThis is the code I used for numDigits():\n\nint numDigits( long num )\n{\nint digits;\nfor ( digits = 0; num > 0; digits++ )\n{\nnum /= 10;\n}\n\nreturn digits;\n}
373214
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.
170982
I believe it is 9599...because the digits of the right side add up to the left side.\n\n3+9+8+3 = 23\n9+5+9+9 = 32