what is the lowest number with 17 factors?
963675
65536 = 2^16 has exactly 17 (unique) factors. (2^0 through 2^16)\n1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, and 65526 all divide evenly into 65536 and thus are its factors.\n\nIf a number is written in its prime factorization:\nx = p1^r1 * p2^r2 * p3^r3 * ... *pn^rn , then the number of factors of x is  (r1+1)(r2+1)(r3+1)...(rn+1)\nIf this product = 17, then r1=16 is the only solution.\nThe smallest prime is 2, so x = 2^16 = 65536 IS THE SMALLEST NUMBER WITH EXACTLY 17 FACTORS.\n\nAll those who are multiplying the first 17 primes together or the first 17 natural numbers together are incorrect.  In the primes example, If 2 and 3 are factors, then so is 6.  In the natural numbers example, if 16 and 17 are factors, then so is 16*17.\nAnd, 1 is a factor of every number.  The question did not ask for prime factors.\n\nAdditionally, if the question meant what is the smallest number with AT LEAST 17 factors, the answer is 180.  [Find a number with 18 factors:  18 = 3*3*2.  so r1=2, r2=2, r3=1 \nthis leads to the number 2^2 * 3^2 * 5^1 = 180]\nThey are: 1,  2,  3,  4,  5,  6,  9,  10,  12,  15,  18,  20,  30,  36,  45,  60,  90,  180. (18 factors)  HOWEVER, if the question was really looking for this, it was misleadingly phrased, as it should have used the term "at least" or it could have been more exclusively written as "What is the smallest number with 18 factors?"  [By the way, numbers with exactly an odd number of factors must be perfect squares.]
248920
here is my answers i think they are right 16-A , 17-D , 18-B , 19-A , and the last one is 20-C.
723554
b, c, d and e\n\nonly 17 can be....(8+9)
1108542
actually, i think it's 15 because however much you add or subtract in the first two, it always adds up to 6.\nAND\nyou always add once and subtract once\nTHUS\nyou have to subtract 3 from 18
455486
20 (9,19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99)
470071
169 is 13 times 13.  Any smaller positive integer is either prime or divisible by 2, 3, 5, 7, or 11.
707816
factors of 81 are 1,3,9,27,81\nfactors of 87 are 1,3,29,87\n\nif you mean factorize 87= 3*29 and 81 = 3*3*3*3
513981
ok, let's start with the biggest number:\n\n18: 18 can only go with 7 because the nearest perfect square is 25 and other number can put it to 36.\n\nsimilarly, for 17 and 16. they can only go with 8 and 9.\n\nso, so far, we have\n\n18 + 7\n17 + 8\n16 + 9\n\nnow, 2 can only go with either 7 or 14, but since 7 is taken, 2 must go with 14\n\nnow, 11 can only go with 5 or 14, but since 14 is taken, 11 must 5.\n\nnow, 4 can only go with 5 or 12, but since 5 is taken, 4 must 12.\n\nnow, 13 can only go with 3 or 12, but since 12 is taken, 13 must 3.\n\nnow, 6 can only go with 3 or 10, but since 3 is taken, 6 must 10.\n\nnow, 15 can only go with 1 or 10, but since 10 is taken, 15 must 1.\n\nnow, 10 can only go with 6 or 15, but since 15 is taken, 10 must 6.\n\nfor 3, it can go with either 1 or 6 or 13, but 6 and 13 is taken, we only have 1 left.\n\nso....sally goes with 3\n\nin summary\n\n18+7\n17+8\n16+9\n15+1\n14+2\n13+3\n12+4\n11+5\n10+6\n\n\nso...lets summarize what we have\n\n18 + 7\n17 + 8\n16 + 9\n2 + 14\n13 + 12\n11 + 5\n14 + 7
147903
15,17,43,44,48,and PB29
7818
There is no smallest number.\n\nEvery number (real number, for example) can be halved.\n\nThe set of real numbers is closed under the operation of multiplication.  (See item (II)-(5) at the linked reference.)\n\nSo, since 0.5 is a real number. if X is a real number, X*0.5 is also a real number.
777032
Since there are too many numbers to list, here's a program to do it for you:\nhttp://wims.unice.fr/wims/en_tool~algebra~factor.html
709001
Here are 3.  Asking for the 3 lowest common multiples is alot more interesting...  \n\n198,  396,  1188
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
730130
Domain = x... should be all real numbers\n\nRange = y... 2 and up... x can be a negative number, but will always end up being postive when added to two, therefore the lowest number is 2
199084
factors of 27 are 1 and 27, 3 and 9\nfactors of 54 are 1 and 54, 2 and 27, 6 and 9
844111
1 and 2 are the most important.  Obviously the two most outdated would be 18 and 21, since they cancel each other out.  But besides those, 16 and 17 are probably ones we could do without.
714266
f = 30 x 25 x 20 x 15 x 10 x 5 x the other 24 integers. \n\nSo f contains 7 factors of 5 (25 supplies 2 of them) and more than 7 factors of 2. \n\n(The 15 even numbers supply a total of more than 15 such factors as there are three powers of 2 and 3 of the other even numbers (12, 20 & 28) are also divisble by 4 and 24 is divisible by 8)\n\nBut 7 factors of 2 is all we need for our purposes and even if we had a multitude of surplus 2s, they cannot give us another 10\n\ni.e. f contains 7 factors of 10. And ONLY 7 such factors as each 10 can only be made up of a 5 and a 2 \n\nso d = 7 is the only possible answer, given that we are told that d >6\n\nGENERALISING:\n\n50, 75, 100, also supply two 5s to a factorial they are part of, \n\nso for f = 55! and d> 12, d must = 13\nso for f = 80! and d> 18, d must = 19\nso for f = 105! and d> 24, d must = 25\n\nhowever\n\nfor f = 130! and d> 30, d is not 31 but 32\nCAN YOU SEE WHY?