Find two sets of twin primes that are between 25 & 45 other then 5 & 7.?
98237
1) 29 and 31\n2) 41 and 43
888908
The answer is 23475.\n\nLet's look at how I got there:\n\n\nThe first digit is a prime number:  2,3,5,7\n\nOne is not a prime, so anything starting with 1 is not actually an option.  There are two valid numbers, but one of them has repeating digits.\n\nGood.  That narrows it down a lot.\n\nThe third digit is twice the first digit:  Following from the numbers possible for the first digit, it's 4,6,10,or 14.  It obviously cannot be the last 2 since they are two digits.  That means it is either 4 or 6, and the first digit is 2 or 3.\n\nSo far we have\n2x4xx or\n3x6xx\n\nThe fourth digit is the third digit plus 3.  That's easy.  We now have\n2x47x or\n3x69x\n\nThe fifth digit is the difference between the first digit and the fourth digit.  7-2=5 and  9-3=6  We now have\n\n2x475 or\n3x696\n\nWait!  The second number has a repeated digit, 6.  It thus cannot be the number.\n\nLet's go on with option 1.  The second digit is the fifth digit minus the first digit.  5-2=3\n\nThe number is 23475.\n\n\nHope this helps.
198100
7 / [8 2/5] = 7 / [42/5] = 7* 5/42= [7*5]/[7*6] = 5/6
523621
(10+15+23+37+45) / 5 = 26
875039
The first five are simple\nP(5,1) + P(5,2) + P(5,3) + P(5,4) + P(5,5) \n5 + 20 + 60 + 120 + 120 = 325\n\nSix and seven are a little more complicated\n6!/2! + 7!/2!2!\n360 + 1260 = 1620\n\nEight is special because you limit the maximum.\n\nList of bad values\n9x,xxx,xxx\n7x,xxx,xxx\n69,xxx,xxx\n67,xxx,xxx\n66,xxx,xxx\n65,9xx,xxx\n65,7xx,xxx\n65,6xx,xxx\n\n8!/3!2! - (2 * 7!/3!2!) - (3 * 6!/3!) - (3 * 5!/3!)\n3360 - 840 - 360 - 60 = 2100\n\nNow add the results.\n325 + 1620 + 2100 = 4045
465400
2, 3, 5, 7, 11\n28\n\nFYI- 1 is not a prime number! It is only divisible by 1. To be a prime number, it must be divisible by 2 numbers.\n\nprime numbers up to 20-\n2, 3, 5, 7, 11, 13, 17, 19\n\nto find a prime number, start at the number 2. cross out every other number (that would be 4, 6, 8, even numbers). Then cross out every 3rd number, every 4th number, and so on. The remaining numbers are prime!\n\nHope this helps!
231634
well i am trying to find out myself\n\n31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.\n\nthat makes 15\n\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29.\n\nthats 10\n\nU have all 25 prime less than 100 !\n\nAlways remember 1 is not a prime number ! cuz its divisible only by 1 and not any other number. and prime number requires divisibility by 1 and the number itself. i hope u got everything !\n\nmodified....i messed up somewhat. i wrote 65 and forgot 71. its clear now im sure watever i've writeen is 100% correct.
166167
you mean the prime factors. They are 2 and 7.\nActually 686 is equal to 2 * 7 * 7 * 7\n2 is the only even number considered prime.\n1 is not considered to be prime (for technical reasons)
635110
For doing squares of numbbers ending with 5, here it is\n\ntake the numbers before 5 and multiply it by its next number and write product followed by 25 e.g.\n65*65 = (6*(6+1)) and 25 = 6*7 and 25 = 4225
118857
Ahkens provided the correct answer and Aseem provided the correct answer and the formulaic method of determining the relevant prime numbers. If you are confused, an easier way of obtaining the prime numbers for low numbers like 255 is simply to start with the lowest primes and divide them into 255 until you exhaust the possibilities. Firstly, the definition of a prime number is a number that can be divided, without a remainder, only by itself and by 1. (The number 1 is itself, however, NOT considered to be a prime - notice that Aseem correctly excluded it from his answer despite it appearing in the results of his formulae). \n\nUsing this definition, the prime numbers between 0 and 100 are:\n\n2,3,5,7,11,13,17,19,23,29,\n31,37,41,43,47,53,59,61,\n67,71,73,79,83,89,97\n\nSo 2 is the first prime that fits the above definition; but 2 doesnt divide into 255 so it doesn't fit into this answer.\n\nNext is 3; 255 / 3 = 85 so 3 is your first number.\nNext prime is 5; 85 / 5 = 17 so 5 is your next number.\n17 is a prime because it too fits the definition so 17 is your next number and as there are no remainders you have solved the problem. 3,5,17
899336
84, without even needing to check. That is 2*2*3*7, which gives 3 prime factors. You can't get 4, since 2*3*5*7 is too big.
499816
Factor each number into their prime factors.\nTake the greatest number of occurrences of each\nprime and use them and only them  as the factors of your answer.\n\n  For example 4 and 10. Primes for each are 2,2 and\n2,5. 2 occurs at most two times and 5 occurs at most\nonce. So the factors in your answer are 2,2,5\nand the answer is 20. \n  Concise but complete. Exhaustive but exclusive.\nThat's elegance.
571951
Start at 2, and eliminate every second number after that.\nThen start at 3, and eliminate every third number after that.\n4 will already have been eliminated in step 1.\nStart at 5, and eliminate every fifth number after that.\nContinue this way.  The numbers left uneliminated are prime.
1098310
There are lots of them.  The easy way is to multiply 5 times 6 times 7.  This number is divisible by 5, 6, and 7.  So -- just add two to it.\n\nThis is 212.\n\nYou want another?  How about 2102?  210000000002?\n422? 420002?\n\nI got a million of 'em.
544625
3 and 23/32nds\nor simply\n119/32
927457
Is there an infinite number of twin primes? (i.e. N and N+2 are both prime).\n\nFor other problems, see:\nhttp://mathworld.wolfram.com/UnsolvedProblems.html
425892
get the 5 pair  $14.49/5=2.89/pair\n$8.75/3=2.92/pair