What is a 3 digit number divisable by 7 but not 2-The sum of my digits is 4?
919009
Because your result is odd (not divisible by 2), the last digit can't be 0 or 2, so it must be 1 or 3...\n\nIf the last digit is 3, the other digits must add to 1, so you have only one choice:\n103\n\nIf the last digit is 1, the other digits must add to 3, so your choices are:\n121\n211\n301\n\nNow test each one to see if it is divisible by 7:\n103 / 7 = 14.714285...\n121 / 7 = 17.285714...\n211 / 7 = 30.142857...\n301 / 7 = 43\n\nSo the only number that fits your criteria is:\n301\n\n301 is 3 digits\n301 is divisible by 7 (7 x 43)\n301 is *not* divisible by 2\nThe sum of my digits (3 + 0 + 1) is 4.
90506
I'll take it as far as I can.\n\n3 - Add all the numbers and if the result is divisible by 3 then the entire number is 9+1+2+0+3+1+2=18 which is divisible by 3.\n4 - The last 2 numbers are divisible by 4. 12 is divisible by 4 so the entire number is.\n6 - It's divisible by 2 and 3.\n8 - If the last 3 numbers are divisible by 8, then the entire number is. 312/8=39. \n9 - Like for 3, except it has to be divisible by 9. The result, from above, is 18 which is divisible by 9.\n12 - If divisible by 3 and 4.\n18 - If divisible by 2 and 9.\n24 - If divisible by 8 and 3.\n36 - If divisible by 4 and 9.\n72 - If divisible by 8 and 9.
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.
534946
it is 136641(square each digit of previous number and put them in order)\ne.g. 7*7=49\n     49--(4*4=16,9*9=81) so no. is 1681\ntherefore next no. is (1*1,6*6,8*8,1*1)=(1,36,64,1)=136641
60067
Definition: A whole number is divisible by 4 if the number formed by the last two digits is divisible by 4.\n\nUsing that logic:  is 48 divisible by 4?  Yes, 4 * 12 = 48.  Therefore 2648 is divisible by 4.
97133
7,248,000,000.\n\nLook at the digit of the place you want to round to.\n\n7,248,193,675.\n------^\nLook to the immediate right. If this digit is 4 or less, keep the digit of the place you want to round to the same and make everything else a zero. If this digit is 5 or more,   make the digit of the place you are rouding to 1 greater, and then make everything else a zero. It is like seeing whicho ne the number is closer to.
1108874
Any number you like, the roots of (x-1)(x-153)(x-370)(x-y)...\n\nHowever, what you may be after is something like 371, 407 - these are the only positive numbers that are the sums of their respective digit's cubes.\n\nAre there any 4 digit numbers such that a^4+b^4+c^4+d^4 = abcd\n\netc..?
692386
There are 9 with one digit, 90 with two digits, 900 with 3 digits, and 1 with 4 digits.  So 9(1) + 90(2) + 900(3) + 1(4) = 2893.\n\nSorry about my earlier answer, my brain was off or something!
131067
last digit of 3^17=3 because 3^1=3,3^2=9,3^3=27,3^4=81.Then the last no.repeats four times.so 17/4=4&remainder 1.therefore 3^1=3 =last number.\nlast digit of  7^13=7 because 7^1=7,7^2=49,7^3=243,7^4=1701.Then the last no.repeats four times.so 13/4=3&remainder 1.therefore 7^1=7 =last number.\nTherfore last number=7+3=10.So last number =0
935133
Ok, take it back.\n\nFlip the 6, turn it into a 9.  Move it over to the right hand side, move the 2 to an exponent over the 7.\n\n7^2 = 49.\n\nThat's it.
1110740
I -1 digit\nLOVE - 4 digits\nYOU - 3 digits
175757
I can get you started. \n\nThe digit A can't be very large if ABCD * 4 is a 4 digit number. In fact 4  3 = 12 so if A = 3 then ABCD x 4 is at least five digits. \n\nA is either 1 or 2. If A is 1 then 1BCD x 4 = DCB1 But this is not possible since the left side is even and the right side is odd. Thus A = 2. \n\nso, we have the first number!\n\n2BCD x 4 = DCB2 \n\nLook at D on the right side. What can it be? 4 x 2 = 8 so D might be 8 or, if there is a carry, D might be 9. But if D on the left side is 9 we get 4 x 9 = 36 which doesn't work with the units digit of 2 on the right side. So, D = 8 and we get:\n\n2BC8 x 4 = 8CB2 \n\nThis should be enough to get you started
924460
It depends on whether you allow a leading zero or not. \nIf you are allowed a leading 0, its easy. The last digit can be 0,2,4, or 6 (4 possibilities), and after you've chosen that, theres 7 possibilities for the first digit and 6 possibilities for the second digit, which makes 4*7*6 = 168.\nHowever, usually 3 digit numbers can't have leading zeroes. Now we have to split it up into two cases:\nCase 1:\n_ _ 0\nThere are 7 choices for the first digit, and 6 choices for the second - 42 possibilities.\nCase 2:\nIt ends with a 2, 4, or 6.\nThere are now 3 choices for the last digit, 6 choices for the first digit (anything *but* 0), and 6 choices for the second digit - 3*6*6 = 108.\nSo the total is 150.\n\nNote that we dealt with the 0 case specially, because it made a difference whether we had a 0 to work with for the first digit or not.\n\n(Note - again, in response to Wal C below, you *have* to split it up into cases. He said 4*6*6; however, if you happened to choose the 0 for the last digit as part of that '4', there will be *7* possibilities for the first digit, not 6. So you can't do it all together like that.)
681272
If decimals are allowed, then (73+7)*.3 is a nice solution :)\n\nIf not, then the only solution I could find is (3+(3/7))*7\n\nEDIT : As for the initial question, absolutely! Math is my favorite subject, even in college. I won several contests in high-school, and love any sort of math puzzles. If you know of any more, please e-mail me :)\n\nSubmarinerdave : Heh. That's what I thought of as well, my sister has a 24 contest at her school two years ago :) A similar but more challenging game is the 2006 math game, where you try to make numbers from 1 to 100 using the digits 2,0,0, and 6. Link below.
1013929
t's actually pretty easy. When you take a number, jumble around it's digits and then subract the new number from the old, you are always going to get the sum of the digits of the difference is divisible by 9 (ignoring any negative). \n\nExample: 12345-54321 =-41976 ----> 4+1+9+7+6= 27=9•3\n\n\nWhen you give all but one of the digits (and knowing that the last one is not 0), the computer just takes the sum of the digits you gave it, and subtracts it from a multiple of 9 (so that the remainder is between 1 and 9). \n\nLet's take 1234 and 4321 for example:\n\n1234-4321=-3087\n\nNow choose a non-zero number:\nif the number was 3 then the sum of the rest would be 15, and 18-15=3\nIf the number was 8, then the sum of the rest would be 10 and 18-10=8\nIf the number was 7, then the sum of the rest would be 11 and 18-11=7\n\n(now, the number you subract rom will not always be 18, but it will always be a multiple of 9)\n\nThe number that you picked will always be the remainder.\n\n\nWhy can't you pick 0?\n\nLets look at this example:\n\nChoosing 0, we have the sum or the remaineder is 3+8+7=18.  Therefore, the computer doesn't know if it should subtract 18 from 18, and the number you picked was 0, or subract 18 from 27 and the number you picked was 9.
343578
you search for a common factor, 2s,3s,5s and 10s are the most apparent. Once you look to a number you can immediately decide if it is divisible by 2 ( if the number is even) or by 3 (if the total of the digits is divisble by 3 i.e 168 total = 15 divisible by 3), or by 5 ( the last digit is 0 or 5) or divisble by 10 ( the last digit is 0).. good luck
94104
u can do it by trial and error. ok the possible digits for ones are just 1,3,5,7,9 since the statement says it is an odd number. ok then try to add 2 and equate it to 12. if it is = that's is the answer. 7+5= 12. so and 7 is greater than 5 by 2. the answer is 75 degrees.