Can someone helps me test 365 for divisibility by 2,3,5,6, and 10?
814787
365/2 = 182.5\n\n365/3 = 121.666\n\n365/5 = 73\n\n365/6= 60.833\n\n365//10 = 36.5\n\nthe numer 365 is only divisible by 5 becos it does not leave a remainder unlike the other numbers\n\n=)
210139
the only way to make 1991_1991 divisible by nine is to choose _ the missing number by 5\n\n199151991 is divisible by nine\n\nso 199151991 / nine =221 2 7999 \n\nSo there is 3 numbers not 4 in the first so i say\n0221 2 7999 x nine = 1991 5 1991
930195
yes because any number ending in zero is even but is divisable by five
681251
1. If any number has an even number at its end, it is divisible by two. So 23000 is divisible.\n\n2. If the sum of digits of a number equal a number divisible by three, then the first number is divisible by three. In 23000, 2+3+0+0+0=5, which is not divisible by three. So 23000 is not divisible.\n\n3. Sorry, but I dont remember the divisiblity test for 4.\n\n4. If any number has 5 or 0 as its last digit, it is divisible by 5. So 23000 is.
149171
1,365\n\nor 15!/(11! * 4!)\n\nie n!/[k! * (n-k)!]
757625
1) \n\n1+2+3+4+5+....n +(n-1)+(n-2)+...+3+2+1...regrouping\n\n=2(1+2+3+4+5+....(n-1)) +n...regrouping the term in parenthesis\n\n=2( (1+n-1)+(2+n-2)+(3+n-3)...)+n\n=2(n+n+n+...(n-1) terms)+n\n=2(n)(n-1)/2 +n\n=n^2-n+n\n=n^2\n\n\n2)\n 1=10^0 and 11=(10^1+10^0) and 111=(10^2+10^1+10^0)...\n\n(10^n+10^(n-1)+....10^2+10^1+10^0) times itself\n=(10^0+10^2+10^3+....10^n)^2\n=10^2n+2*10^(2n-1)+.....\n\n\n3)\n\nn^2-n=n(n-1)...one of them must be even and is therefore divisible by two.\n\nn^3-n=n(n^2-1)=n(n+1)(n-1)=(n-1)(n)(n+1) or three consecutive integers.\n\nIf the product is three consecutive integers, then one of them must be divisible by three and therefore the product is divisible by three.
770675
38894 0's between 1-100000\n\nWrite EVERY number (except 100000) as a 5-digit number with leading 0's.  We will subtract out the leading 0's later.\n\nIf the first digit is zero, there are 10*10*10*10=10000 numbers.\nif the second is zero, there are also 10*10*10*10=10000 \nsame with if the third, fourth or fifth is zero.\nSo now, we have 5*10000 or 50000 zeroes.\nNow we subtract out the number of 0's in numbers with leading 0's.\n\nIf the first is 0, there are 10000 numbers, so we added 10000 too many 0's, so subtract 10000\nIf the first two are 0, there are 1000 numbers, so we added 1000 too many more 0's\nIf the first three are 0, we added another 100 too many.\nIf the first four are 0, we added another 10 too many.\nIf all five are zero, we add 1 more too many.\n\nSo N = 50000 - 10000 - 1000 - 100 - 10 - 1 = 38889 for the numbers 1 through 9999.\nThen we add 5 more for the number 100000, so\nN = 38894\n\n*****\nThe equation for the number of 0's between 1 and 10^n inclusive is:\nn10^(n-1) - 10^(n-1) - 10^(n-2) - 10^(n-3) -  ... - 10 - 1 + n\n= (n-10/9)*10^(n-1) + n + 1/9
216703
Yes, 1000003 happens to be prime but unless I'm missing something your reasoning is not convincing. How do we know that r + r' is not divisible by n ?
1154166
To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number. \nIf you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again. \n Check to see if 203 is divisible by 7. \n\ndouble the last digit: 3 x 2 = 6 \nsubtract that from the rest of the number: 20 - 6 = 14. \ncheck to see if the difference is divisible by 7: 14 is divisible by 7, therefore 203 is also divisible by 7. \n\nFor 11:\n\nTake the first digit, subtract the second digit from it, add the third digit, subtract the fourth digit and so on. If the number is divisible by 11, the answer will be zero.\nconsider the number is 12354.\nDo this calculation:\n1-2+3-5+4\nAnswer is 1. This is not equal to 0 thus the number is not divisible by 11.\n\nConsider 108416:\n\n1-0+8-4+1-6=0\nThus the number is divisible by 11!
790673
count the number of decimal places\nin your example, its 7\nyour example becomes the numerator (remove the decimal point)\nand 1+ seven 0s be the denominator\nso its\n1285714/10000000
1131356
U cannot divide by zero\n4000+2000 is 6000 and definitely not zero.
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.
578566
To check if a number is divisible by 36, use the divisibility rules for both 4 and 9.  If a number is divisible by both 4 and 9, it must be divisible by 36.\n\nIf you need to know the rules for 4 and 9 (or other numbers up to 13, as well as methods for larger numbers), visit the source site.
915778
1) 6.4/100\n2) 100|6.400..since 100 cannot divide into 6, nor 64, you have to move the decimal point over until you reach a whole number that is divisible by the denominator (100). \n3) 0.0640\n\nI hope this clarified the problem for you!
483173
((6p - 18)/(9p)) / ((3p - 9)/(p^2 + 2p))\n((6(p - 3))/3(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) / ((3(p - 3))/(p(p + 2)))\n((2(p - 3))/(3p)) * ((p(p + 2))/(3(p - 3))\n(2p(p - 3)(p + 2))/((9p)(p - 3))\n(2p(p + 2))/(9p)\n(2(p + 2))/9\n(2p + 4)/9
732682
1110\n\nTo be divisible by 3, the total of the digits of a number has to be divisble by 3, so 1+1+1+0 = 3 (so you need at least 3 "1" digits).\nBut you need it to be divisble by 5 as well, and only numbers that end in 5 or 0 are divisble by five.\nSo, you take the 111 and add a 0, and that is your number.
779807
Yiu would first decide how many decimal places you are going to consider.  The usual is 3.  Then work it from there...;