I am a 3 -digit # divisable by 3-My tens digit is 3 times as great as my hundreds digit and the sum is 15?
930433
The answer is 393.\n\n Start with 333.\nIt's an easy 3-digit number that is divisible by 3\n Tens digit id 3 times more than hundreds.\nSo 393. \n Does it add up to 15? \nYes it does!\n\n:) :) :)
164857
1010020\n1_1_0_0 summing digits results in 2.\n_0_0_2_ summing digits results in 2.\nSo it is divisible by 11.\n\nAnother way:\n1 - 0 + 1 - 0 + 0 - 2 + 0 = 0\nSo 1010020 is divisible by 11.
146728
no, it is divisible by 1, 3, 23, 29, and itself (with the numbers being reduced as far as possible)
149171
1,365\n\nor 15!/(11! * 4!)\n\nie n!/[k! * (n-k)!]
593943
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
744222
Let the number be X\nThen X = 1000a + 100b + 10c + d, where a,b,c,d are the digits of X.\nLet the 'flipped' number be Y.\nThen Y = 1000d + 100c + 10b + a\nSo, X - Y = 1000a + 100b + 10c + d - (1000d + 100c + 10b + a)\nor\nX - Y = 999a + 90b - 90c - 999d\n= 9 * (111a + 10b - 10c - 111d) .\nwhich is divisible by 9.\nThe sum of the digits of any number divisible by 9 is itself divisible by 9.*  So if you sum the digits of X-Y, you will get a number divisible by 9.  Then if you sum the digits of that number, you will get another (smaller) number divisible by 9.  If you keep doing that, you will eventually get to 9. [I assume that's what you meant.  If you sum the digits just once you wont necessarily get 9, as in your example. 6354-4536 = 1818 which sums to 18.]\n\n* You can prove that the sum of the digits of any number divisible by 9 is itself divisible by 9 by induction.\nThe first 2-digit multiple of 9 is 18, and 1+8 = 9.\nAssume it is true for k*9 where k is a natural number (1,2,3,...)\nProof for (k+1)*9:\n(k+1)*9 = k*9 + 9\nIf the last digit of k*9 is 0, then adding 9 just adds 9 to the sum of the digits of k*9.  Since the sum of the digits of k*9 is divisible by 9, so is this sum + 9.  We are done.\nIf the last digit is anything else and the tens digit is <9, adding 9 is the same as adding 1 to the tens digit and subtracting one from the one's digit.  So the sum is the same as the sum of k*9 + 1 - 1 = k*9 and we are done\nIf the last digit is anything else and the tens digit=9 then adding 1 to the tens digit means subtracting 9 from the tens digit and adding 1 to the hundred digit, which means the sum is the same as the sum of k*9 + 1 - 9 + 1 = the sum of k*9 - 9 which is divisible by 9.  (This can be extended to carrying a 1 to the other places as well.)
175248
Unfortunately I don't know of a quick method but I can help you a little here. \n\nFirst, as you probably already know, prime numbers can only be evenly divided by 1 and the number itself.\n\nSecond, since all numbers are a product of prime numbers you can save time in by not testing division by non-prime numbers like 4 and 6 and 9 and so on.\n\nThird, you only have to test those prime number that are smaller than the square root of the number.\n\nIn your example, 117, you need only test the prime numbers less than 10.81.... since that is the square root of 117. So do a trial divide of 117 by 2, 3, 5, and 7.\n\nOf course there are a few other rules that can help you eliminate the first three choices. If the number is odd, it can't be evenly divided by 2. If the sum of digits isn't divisible by 3 then the number isn't either. And all numbers that are divisible by 5 end in 5 or zero.\n\nFor 117, the number is odd so that eliminates 2. The sum of 1 and 1 and 7 is 9 which is divisible by 3 so 117 is divisible by three and therefore not prime.\n\nSo now you may notice that 117 is divisible by 3 and 39 but I said you only need to test numbers smaller than the square root. What about 39? Well, since the number is divisible by three that alone proves the number is not prime. If you want the full prime factorization, you need to keep dividing.\n\nGood luck.
808949
500007 is divisible by 3!\n\n166669 / 3333333
271046
look at it this way:\n\n1/3 = .3333...\n2/3 = .6666...\n\nadd the two and you get \n\n3/3 = .9999...\nor, equivilently\n\n1 = .9999...\n\nso, looking at it that way, yes they are equivalent\n\nthe only reasone this works is because we could never add the .000....1 to the .999... to make it 1 because we never know where the last 9 occurs, infact there is no last 9, so we must conceed that they are equivilent.\n\nor, one could look at it as .999... converges to 1, no matter how close you say .999... has to be to 1 in order to say they are equivilent, .999... could be taken that many digits out so that the next digit would round it up to 1\n\nhope this was helpful\n\nmatttlocke
406228
I could be wrong but is it 15t.   They each have one t in common.  60 and 15 are both divisible by 15.\n\nTake the time to really think about these problems before posting them all on the internet.  It's better to have all the answers than to find someone else that does.\n\nGood luck with your math homework.  ;)
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.
306730
well we know the 3 space must be even, in order to avoid the last space resulting in another 5\n\nhmmm... i'l work on it, provided you're working as well!\n\n..and #3 mustn't be 0, to avoid another 0\n..lol i'd have had it if i had been home...i made a program to do it, and then left before i finished/ran it lol\n\nbut that, is of course, the correct asnwer, 396*45=17820\n\nhaha thanx for the challenge!
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.)
1067
ummmm... i cant count that high. let me grab my calculator. i calculate 3 as the answer. ;) lol
170800
A number is divisible by 11 if the sum of the digits in the odd places minus the sum of the digits in the even places is 0 or a number divisible by 11. (What the hell does that mean?) Another way: a number is divisible by 11 if the difference of the two sums of alternate digits is divisible by 11. For example, 6,523 is divisible by 11 since the sum of 6 and 2 is 8; and the sum of 5 + 3 = 8 and the difference is 8 – 8 = 0. What about 4,092 – is it divisible by 11? The sum of 4 + 9 = 13 and the sum of 0 + 2 = 2 and 13 – 2 = 11 which is divisible by 11, so 4,092 is divisible by 11.\n\n\nWhat this means is you add first digit and third digit and minus the second and fourth digit and if you get 0 or something divisible by 11\n1+5-8-9=-11 which is divisible by 11
747071
Look at each clue one at a time...\n\n●  "a 4 digit number number is divisible by 5,it is not divisible by 10" -- this tells you that the number ends with a 5, since numbers divisible by 5 end in 5 and 0 only, and numbers divisible by 10 end in 0 only.\n\n●  "the 10s digit is greater than 4 & power of 3" -- powers of three are 1 (1³) and 8 (2³).  You wouldn't use 3³ since that equals 27.  Therefore our 10s digit is 8, since it's greater than 4 it can't be 1.\n\n● "the hundreds digit is greater than 4 and is a power of 2" -- do this the same way we did the 10s digit.  Single-digit powers of 2 are 1, 4, and 9.  Since it's greater than 4, our only choice is 9 for our 100s digit.\n\n● "the thousands digit is less than 4 and is a power of 2&3" -- single-digit powers of 2 are 1, 4 and 9; single-digit powers of 3 are 1 and 8.  The only one less than 4 that's both a power of 2 and 3 is 1, therefore the 1000s digit is 1.\n\nPutting this altogether we get 1985 as our answer.\n\nANSWER:  1985\n\n~ ♥ ~
1110740
I -1 digit\nLOVE - 4 digits\nYOU - 3 digits