What is the smallest COMPOSITE number not divisible by 2,3,5,7,11?
470071
169 is 13 times 13.  Any smaller positive integer is either prime or divisible by 2, 3, 5, 7, or 11.
869678
1 and 1\n9 and 3\n25 and 5\n49 and 7\n81 and 9\n121 and 11\n...
1165002
This is a wild guess.\n\n3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 = the number.\nThe answer will be the product of the first 12 primes that aren't 2. (If 2 were a factor, the answer would be even, not odd.)\n\nFor the rest, use a calculator. And check to make sure all those numbers are prime. Sorry this is such a whack answer.
4059
First you have to take the 7 11/12 and 3 5/8 and change them into unmixed fractions (no whole numbers).  To do that you multiply 7 times 12 and then add 11.  That is your new numerator (7*12 + 11).  The denominator is still 12. (so 95/12) \n\nDo the same thing with 3 5/8 (3*8 + 5). (so 29/8)\n\nNow  you need to find a least common denominator so that you can subtract 29/8 from 95/12.  8*3 = 24.  12*2 = 24.  So 24 is your least common denominator (note that 8*12 is also a common denominator but not the LEAST common denominator).\n\nChange each of the fractions so that you have 24 as the denominator.  Then do the subtraction.
645443
Primes less than 100:\n2\n3\n5\n7\n11\n13\n17\n19\n23\n29\n31\n37\n41\n43\n47\n53\n59\n61\n67\n71\n73\n79\n83\n89\n97\n\nFrom this list we keep only:\n3\n7\n13\n17\n23\n31\n37\n43\n47\n53\n67\n71\n73\n79\n83\n97\n\nGood luck...
788011
There aren't too many two digit numbers which are 3 less than a perfect squares. Lets list the perfect squares between (10+3) and (99+3):\n16, 25, 36, 49, 64, 81, 100.\nSubtract 3:\n13, 22, 33, 46, 61, 78, 97\nThe only one which is a prime and a square reversed is 61.\n\nSo the answer was 61.
129045
Even numbers are divisible by 2 (2,4,6,8,10, etc) whereas odd are not (1,3,5,7,9, etc)
948776
60 is the smallest number that has exactly 12 factors.\nThe next numbers under 100 are 72, 84, 90 and 96.\n\n72 : 1 2 3 4 6 8 9 12 18 24 36 72\n84 : 1 2 3 4 6 7 12 14 21 28 42 84\n90 : 1 2 3 5 6 9 10 15 18 30 45 90\n96 : 1 2 3 4 6 8 12 16 24 32 48 96
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
133517
the solutions is 7,8,3,2 or 7,6,5,2. The problem works with both sets of numbers.
686152
A common multiple of 2,3,4,5,6,9 and 10 is 64800. But that is not the lowest common multiple which would be a preferable answer. The lowest common multiple is the smallest of the common multiples of a given set of numbers,\n2 = 2 x 1\n3 = 3 x 1\n4 = 2 x 2\n5 = 5 x 1\n6 = 2 x 3\n9 = 3 x 3\n10= 2 x 5\ntheir lowest common multiple would be\n2 x 2 x 3 x 3 x 5 = 180\n180 would be the most ideal answer
930195
yes because any number ending in zero is even but is divisable by five
842088
111 is divisible by 3 (plus others) so is not a prime\n63  is divisible by 3 (plus others) so is not a prime\n15 is divisible by 3 (plus others) so is not a prime\n41 and 43 are twin primes
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.
580109
ans is 70\n(5*2+1) (11*2+2) (24*2+3)\nrule is there r two series 5 11 24 51...\n7 16 34 70\n(7*2+2) (16*2+2) (34*2+2)
1067663
Seven prime integers.\n\n7\n17\n37\n47\n67\n71\n73
376218
Not sure how many you need, but here are the first few:\n2,3,5,7,11,13,17,19,23,29,31, 37,41,43,47,53,59,61,67, 71,73,79,83,89,97,101