Can Anyone please tell me the divisibility test for the no 7  ?????????
538636
Believe me, the test is more difficult than the division itself.
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 ?
385685
Consider the numbers\nA1,\nA1+A2,\nA1+A2+A3\n..\nA1+A2+A3+...+An.\n\nDo the computations modulo n. If none of these is divisible by n, then we have n numbers and n-1 non-zero residue classes (mod n), so there are two of these numbers which are equivalent (mod n) by the pigeon hole principle. But then subtracting gives a sum of the required form that is 0 (mod n), in other words, so that n divides that sum.\n\nLooking at the proof, there is no need for the sequence to be monotonic.
287547
Condition: The other number is different from 7.\n\nLet n = the other number\n\nSolution:\n\n7 / n = n / 7\n\nn^2 = 49\n\nn^2 - 49 = 0\n\n(n + 7)(n - 7) = 0\n\nTherefore:\n\nn + 7 = 0...........or...........n - 7 = 0\nn = -7..............or...........n = 7 (reject)\n\nTherefore, the other number is -7 ANSWER
269746
Make 2 7/15 an improper fraction, flip it over to its reciprical and mulitply the two.\n\nBoy...I hope I got that.  *bits fingernails*
532611
That statement is not true.  However, for any number that is divisible by 9, you can add up its digits and get another number that is divisible by 9.\n\nFor example, for 999, 9+9+9 = 27, which is divisible by 9.
323646
again really easy:\n\nint divisibility = (m % 10) == 0) ? 1 : 0;\n\nBut I think it is better you learn it yourself. Try a book that would be better. Just search C language tutorials.
1011386
I tried and I found that 100! is perfectly divisible by 350 as 350=2*5*5*7=50*7.\n\n100!= 100*99*98*97!\nnow 100 is divisible by 50 and 98 is divisible by 7.\n\nSo the least number is 1.
986330
It's multiple choice, so all you really need to do is start plugging numbers in and see if they work:\ng(7) = 2(343) -11(49) + 8(7) + 21\ng(7) = 686 - 539 + 56 + 21 =/= 0\n\ng(1) = 2(1) - 11(1) + 8(1) + 21\ng(1) = 2 - 11 + 8 + 21 =/= 0\n\ng(7/2) = 2(343/8) - 11(49/4) + 8(7/2) + 21\ng(7/2) = 343/4 - 539/4 + 112/4 + 84/4 = 0\n\nThus, d is the right answer.
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.  ;)
18671
To show that if a natural number is divisible by three, then the sum of its digits is divisible by three, use inductive reasoning:\n\n(A)  Base case: 3 is divisible by 3, as any natural number is divisible by itself.  The sum of its digits is 3, which is, of course, divisible by 3.\n\n(B)  Assumption: Let k=d(n)d(n-1)...d(1) be a multiple of 3, where the d(i) is the ith digit from the right in k.  Assume that d(n)+d(n-1)+...+d(1) is divisible by 3.\n\n(C)  Induction: Now, consider k+3.  One of the following occurs:\n\n     (1)  3 is added to d(1) and no carries are involved.  The sum of the digits becomes d(n)+d(n-1)+...+d(1)+3, which is divisible by 3.\n\n     (2)  3 is added to d(1) and a carry to the tens place is involved.  After the carry, 10 is subtracted from d(1) and 1 is added to d(2).  Now, The sum of the digits becomes:\n\n  d(n)+d(n-1)+...+d(1)+3-10+1\n= d(n)+d(n-1)+...+d(1)-6,\n\nwhich is divisible by 3.\n\n     (3)  3 is added to d(1) and multiple carries are involved.  This occurs when a string of m-1 9s ending in the tens place exists.  After the carries, 10 is subtracted from d(1) and 1 is added to d(2) (making d(2) 10), 10 is subtracted from d(2) (making d(2) zero) and 1 is added to d(3), and so on, involving m carries.  Then, Now, The sum of the digits becomes\n\n  d(n)+d(n-1)+...+d(1)+3-10m+m\n= d(n)+d(n-1)+...+d(1)+9m-6,\n\n which is divisible by 3.\n\nHence, by induction, if a natural number is divisible by three, then the sum of its digits is divisible by three.  \n\nNow, you wanted the converse.  Similar proofs can show that for base cases 1 and 2, the sum of digits for k=3*p+1 and k=3*p+2 (where p is a natural number) will not be divisible by three, giving the proof that a natural number is divisible by three if and only if the sum of its digits is divisible by three.  QED
200952
Write the number in binary.  Now, from right to left, write the digit for the even powers of 2.  Then write the digits for the odd powers of 2.\n\nNow, in decimal, subtract these 2 numbers, and if this number is divisible by 3, then the binary number is divisible by 3.\n\nExamples\n\n24 = 11000\ne - 001 -- starting w/the first on the right, and every other one.\no - 01 -- starting w/the second from the right, and every other one.\n\nNow in decimal, 1-1 = 0, so 11000 is divisible by 3\n\n~*~*~*~*~*~*~*~\n45 = 101101\ne - 110\no - 011; in decimal, 110 - 11 = 99, which is divisible by 3, so 101101 is divisible by 3\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nTo Milkyway:  Thanks!  I'm thinking it's probably got something to do with the fact that 3 written in binary is 11 ;)
883360
if you mean: how I know that number N is dividable by number k with no residue?\n1)if the unit digit of N is even, then N is divisible by 2;\n2)if sum of digits of N is divisible by 3, then N is divisible by 3;\n3)if sum of digits of N is divisible by 9, then N is divisible by 9;\n4)if the unit digit of N is 0 or 5, then N is divisible by 5;\n5)if sum of odd digits is equal to sum of even digits, then N is divisible by 11;\n6)if the last two digits represent a number divisible by 4, then the whole N is divisible by 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.
452386
7*6*5*4 = 840 if order matters.\n\n7*6*5*4 / 4! = 35 if order does not matter
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
222795
It is an easy problem.\n\nPut it as the\n\nProposition. There does not exist a positive integer n such that 2^n+1 is divisible by 7.\nProof. Re-write the comparison 2^n+1=0 mod 7 as 2^n=-1 mod 7 or else 2^n=6 mod 7.\n\nFurther, it is clear that 2^0=1 mod 7, 2^1=2 mod 7, 2^2=4 mod 7, 2^3=1 mod 7, hence, one observes the periodicity of remainders (periodicity with respect to n of length 3). Consequently it suffices to consider only 3 types of powers of 2: namely, for\n\n1) n=3*k, 2) n=3*k+1, 3) n= 3*k+2 (k is a positive integer).\n\nThese cases imply respectively:\n\n1) 2^(3*k)=1 mod 7;\n2) 2^(3*k+1)=2 mod 7;\n3) 2^(3*k+2)=4 mod 7;\n\nThese cases exhaust all possibilities. In neither of the cases the remainder 6 is attained (or, what*s the same, -1). The proposition has thereby been proved.