hi!one math problem.I have number of divisors N for an unknown number K,how i can find it?
103412
If you have the number of divisors N, take K to be for example p^(N-1), for an odd prime p. Then your K would have the desired number N of divisors, given in the conditions. But, you have found one such way, and cannot guarantee that it is unique, there may be several such numbers. Therefore, you cannot uniquely determine the unknown number, there will always be several possibilities for that number, but you can always find at least one as follows from Euler's multiplicative function in number theory that counts the number of divisors.\n\nF(N)=F(P_1^k1*...*P_n^Kn)=F(p_1^k1)*...F(p_n^kn).\nHope that helps.
885353
Each zero at the end requires a 5 and a 2 somewhere in the factors.  Since there are more twos than 5s, all you need to do is count the number of 5s.  That's still an awfully big task.  Each number that is divisible by 5 has at least one 5 as a factor (e.g., 5, 10, 15...)  Of course there are 2000/5 = 400 of these.\n\nEach number divisible by 25 has at least 2 factors of 5.  There are 2000/25 of these or 80.  We already accounted for one of these 5s in the first count so this adds 80 more.\n\nEach number divisible by 5^3 (125) add another factor of 5 for 2000/125 = 16.\n\nNow, on to 5^4 this time the division is uneven: 2000/625 = 3.2 This means there are 3 numbers below 2000 divisible by 5^4\n\nOnce we get to 5^5 = 3125, it is bigger than 2000 so there are no more factors of 5.\n\nAll in all this comes to:  400 + 80 + 16 + 3 = 499 zeros.\n\nYou also asked about 1000!.  Using the same approach:\n\n200 + 40 + 8 + 1 = 249 zeros
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.
275586
3k - (k-2) = (k-2) - (k+7)\n3k = (k-2) - (k+7) + (k-2)\n3k = 2(k-2) - (k+7)\n3k = 2k - 4 - k - 7\n3k = k - 11\n2k = -11\n\nk = -5.5
153978
For any number i (values for 1573 in parentheses): \ndefine i=10j+k (j=157, k=3)\nif j+4k (=157+12=169) is divisible by 13, then i is\n\nif you don't know if j+4k is divisible by 13, repeat the test on it, i.e.,\n(j=16, k=9, j+4k=16+36=52, is divisible by 13)\n\nSee the question, "number divisibility test?" for rationale and divisibility tests for other divisors.\n\nIn this case just dividing by 13 is quicker!  I don't how you can just look at it and tell.
828966
This is exactly what I am doing in math right now:\nso with the remainder theorem you plug d+4 into the equation so it becomes -4\nso\n(-4)^4+2(-4)^2+5(-4)-10\n256+32-20-10=258
736040
Looks like it's 50 numbers. For example , let the numbers be\n51 thru 100 inclusive. Since even the lowest one can't divide\ninto the highest, none are divisible by the other.  Now lets' suppose there is another number less than 51 in this set. Then\nthat number times 2 will be some number less than or equal to\n100 so by contradiction, there is no other number less than 51\nin the set.
823097
Can't prove it as n cannot = + or -1 and k be an interger.\nIt appears to work with other odd integers. Change your question then it may be possible to prove it.
427860
There is an infinite number of solutions, \nthe set of solutions is :\n{(k,-k+12):k belongs to the set of real numbers}
303516
Each element x of S_n can be written as a product of disjoint cycles, e.g. x=(132)(45). Since the order of a cycle is the number of elements it permutes, the order of x is the least common multiple of the orders of each of its cycles.\nNow, the order of x should be 3000 = 2^3*3*5^3. To make the number n of elements permuted by x minimal, we have to find the set of coprime numbers d_i, i=1..k with d_1*...*d_k = 3000 that makes n=d_1+d_2+d_3 minimal. All possible choices are listed below :\n1. (k=3) = 3, 2^3, 5^3   \n2. (k=2) : 3*2^3, 5^3    \n3. (k=2) : 3, 2^3*5^3    \n4. (k=2) : 2^3, 3*5^3    \n5. (k=1) : 3*2^3*5^3     \nObviously, the first choice where k is maximal (= the number of different prime factors of 3000) minimizes n to the value 3+2^3+5^3=136.\n\njust in case you want to know... :\nQuite generally, the minimal n for an element x with order d is given by n = p_1^e_1+...+p_k^e_k where the p_i's are different prime numbers with d = p_1^e_1*...*p_k^e_k.
689300
Q98 = 1,1,2,3,5 \n\nP.S: Looks like the fibonacci series
899137
let us evaluate this upto n terms\n\nwe get 2^(n) -1\n\n\nnow this to be divisible by 9\n\n2^n mod 9 = 1\n\n2^1 mod 9 = 2\n2^2 mod 9 = 4\n2^ 3 mod 9 = 8 or -1 \n\nso 2^6 mod 9 = 1\n\nso (2^6n)\nso 2^6-1 mod 9 = 0\n\nso 6k  terms sum is dvisible by 0\n\nso n = 6, 12 , 18 so on of the form 6k
5300
(n+k-1) over (k) = (n+k-1) / (k)\n\n=> (n+k-1) / (k) = n/k +1 -1/k
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
186591
solve it by induction...\nlet n=1, then 5^2n-1 +1 =6 so divisible by 6\nlet n=k, then 5^2n-1 +1=5^2k-1 +1 and let it be divisible by 6\nlet n=k+1 then 5^2n-1 +1=5^(2(k+1)-1)) +1\nie 5^(2k+2-2) +1=(5^2k-1)(5^1)\nbut we know that 5^2k-1 is a multiple of 6,so (5^2n-1)*5 is also a multiple od 6. since it satisfies this for n=0, it should satisfy for all n
853794
Evaluate the expression after the ∑ for each of k = 1, 2, ..., up to 6, and add them:\n\n[ 1 + (-1)^1 ] 2^-1 +\n[ 1 + (-1)^2 ] 2^-2 +\n[ 1 + (-1)^3 ] 2^-3 +\n[ 1 + (-1)^4 ] 2^-4 +\n[ 1 + (-1)^5 ] 2^-5 +\n[ 1 + (-1)^6 ] 2^-6\n\nNow use (-1)^1 = -1, (-1)^2 = (-1) × (-1) = 1, ... to get\n\n[ 1 + (-1) ] 2^-1 +\n[ 1 + (1) ] 2^-2 +\n[ 1 + (-1) ] 2^-3 +\n[ 1 + (1) ] 2^-4 +\n[ 1 + (-1) ] 2^-5 +\n[ 1 + (1) ] 2^-6\n\n=\n\n[ 0 ] 2^-1 +\n[ 2 ] 2^-2 +\n[ 0 ] 2^-3 +\n[ 2 ] 2^-4 +\n[ 0 ] 2^-5 +\n[ 2 ] 2^-6\n\n= 2 × 2^-2 + 2 × 2^-4 + 2 × 2^-6\n\nUsing a^m × a^n = a^(m + n), simplify the exponents, to get\n\n2^-1 + 2^-3 + 2^-5 \n= 1 / 2 + 1 / 8 + 1 / 32\n= 16 / 32 + 4 / 32 + 1 / 32 \n= 21 / 32
262222
the answer is 96012000\n\nuse the equation \ns= n( 2*a+(n-1)d)/2