Another Prime number riddle. Can you an even number that is the sum of two primes three ways?
598439
There are quite a few, 22 is the smallest such that there are exactly 3 ways. The ones less than one hundred are:\n\n22= 19+3 = 17+5 = 11+11\n24= 19+5 = 17+7 = 13+11\n26= 23+3 = 19+7 = 13+13\n30= 23+7 = 19+11 = 17+13\n40= 37+3 = 29+11 = 23+17\n44= 41+3 = 37+7 = 31+13\n52= 47+5 = 41+11 = 29+23\n56= 53+3 = 43+13 = 37+19\n62= 59+3 = 43+19 = 31+31\n98= 79+19 = 67+31 = 61+37\n\nOf course the more interesting question is Goldbach's Conjecture: Can every even number greater than 2 be expressed as the sum of 2 primes in at least one way? An interesting open problem.
301463
You r wrong, Not for every positive number... Such as your referance is not true for 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53...its a long list... you cant sum them up using 2 prims...\nbut its true for all even number(because no such even number has found yet) and Goldbach's Conjecture says so..\n**Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.\n\noh now I see u also ment even Integer.. I am sorry..\n\nand 2nd part...\nif all even number greater than 3 is sum of 2 primes..\nthen all even number greater then 6(not 5)is sum of 3 primes..\nthat means n = 5+p;\nbut 5 = 2+3;\nso, n = 2+3+p
861916
I can't see the rest of your question, but even numbers can be multiples of three. Like 6 (2x3). They are not mutually exclusive.
678736
the sum of the two smallest numbers must be higher than the third one so your answer is d)
1050937
go with #2 ... 253 is not prime (253 / 11 = 23)
751938
if two consecutive numbers are multiplied the products smallest prime number is always 2 whichever two consecutive numbers you choose. this is because the product of any two consecutive numbers is always even.
898910
If you mean a number with three prime factors, it's a 3-almost prime.  If you mean a number with three integral factors (prime, composite, or 1), it's the square of a prime.\n\nList of 3-almost primes: http://www.research.att.com/~njas/sequences/A014612\nList of squares of primes: http://www.research.att.com/~njas/sequences/A001248
236479
Any number can be broken down into it's prime factors.  So find all the prime factors of a number, and see what there are 3 of.\n\n8 is an easy example - the first prime number that goes evenly into 8 is 2 so\n\n2*4 = 8; but 4 is not a prime number.  The prime number 2 also goes into 4 so\n\n2*(2*2) = 8; 2*2*2 = 8; 2^3 = 8, so 2 is the cube root of 8.\n\nA slightly trickier one - 3,375.  Five is a prime number that clearly goes into this, so\n\n5*675 = 3,375; now 5 will also go into 675\n\n5*5*135 = 3,375; again, you can factor out another 5\n5*5*5*27 = 3,375; no more fives, but the prime number 3 will go into 27\n\n5*5*5*3*9 = 3,375; and 9 is still not a prime number, but 3 goes again - \n\n5*5*5*3*3*3 = 3,375\n\nOk, now it's broken down into prime numbers, so see what there are 3 of - there are 3 5's and 3 3's so you could regroup like this:\n\n5*3*5*3*5*3 = 3,375 or\n15*15*15 = 15^3 = 3,375\n\nSo the cube root of 3,375 is 15\n\n~~Now~~  If there had been leftover numbers that there weren't three of, say the number had been 48 - first pull out the prime factors -\n\n2*24 = 48\n2*2*12 = 48\n2*2*2*6 = 48\n2*2*2*2*3 = 48\n\nYou would group 3 of the 2's and then group the leftovers\n\n2^3 * 2*3 = 48\n2^3 * 6 = 48\n\nThen the cube root of 48 = cuberoot(2^3) * cuberoot(6), or cuberoot48 = 2*(cuberoot6)
853554
Hello Rajeash\nYour additional question "\nProve that when x is even then it is impossible to write the above equation For n>2.\n"\nIsnt this what thye good o'l  Sir Fermat already has stated even for odd x ?\n\n\na^(n) - b^(n) = x^(n) \nsuppose that n is not prime, thus n = p*q, with p a prime\nthen we can write\n(a^q)^p -(b^q)^p = (x^q)^p \nso we can limit our problem to n a prime. Fermat stated his statement for p>2 thus we only have to investigate the odd primes.I do not agrede with You Rashesj that we have to investigate 4 as a separate case, because 4 = 2^2 we can write the equation as (a^2)^2 - etc , Do you agree with this ?\n\nYou also say that we can assume that if we have a solution then x will be an odd prime. If a,b,x have a common factor then we can divide that factor out. So we can assumet that they dont have a common factor. If x is even then a and b should be odd.
728630
Prime factor each\n4 = 2 x 2\n7 = 7\n9 = 3 x 3\n\nTake union of all factors, removing duplicates (none here)\nAnswer = 2 x 2 x 3 x 3 x 7 = 252
1139439
Prime numbers are numbers that are divisible by itself and produce only two factors..\n\nThe two factors are 1 and the number itself(other than 1)...\n\nBetween 1 to 10.. There are only 4 prime numbers..\n\n2 = 2 x 1\n3 = 3 x 1\n5 = 5 x 1\n7 = 7 x 1\n\nThe remaining numbers are not prime numbers as they do not fit the criteria...\n\n1 = 1 x 1\n4 = 2 x 2 = 4 x 1\n6 = 3 x 2 = 6 x 1\n8 = 4 x 2 = 8 x 1\n9 = 3 x 3 = 9 x 1\n10 = 5 x 2 = 10 x 1\n\n\nHope it helps to clear your suspicions...\n\n\nCheers.... (",)
892074
Because 3 and 23 are prime, the number will be a multiple of (3 x 23) = 69.\n\nNext, the two digit number in the thousand and hundreds place must be *just before* the two digit number in my tens and ones place.\n\nThe answer is 2829.\n\nYou can confirm this by checking the list below:\n2001\n2070\n2139\n2208\n2277\n2346\n2415\n2484\n2553\n2622\n2691\n2760\n2829 <--\n2898\n2967\n3036\n3105\n3174\n3243\n3312\n3381\n3450\n3519\n3588\n3657\n3726\n3795\n3864\n3933\n4002\n4071\n4140\n4209\n4278\n4347\n4416\n4485\n4554\n4623\n4692\n4761\n4830\n4899\n4968\n5037\n5106\n5175\n5244\n5313\n5382\n5451\n5520\n5589\n5658\n5727\n5796\n5865\n5934\n6003\n6072\n6141\n6210\n6279\n6348\n6417\n6486\n6555\n6624\n6693\n6762\n6831\n6900\n6969\n7038\n7107\n7176\n7245\n7314\n7383\n7452\n7521\n7590\n7659\n7728\n7797\n7866\n7935\n8004\n8073\n8142\n8211\n8280\n8349\n8418\n8487\n8556\n8625\n8694\n8763\n8832\n8901\n8970\n9039\n9108\n9177\n9246\n9315\n9384\n9453\n9522\n9591\n9660\n9729\n9798\n9867\n9936
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.
402740
if you mean 3 , 2 ,  6 then it is the answer :\n(3 * 6) +2 = 20 .
216982
2, 3, 5, 7, 11, 13, 17, 19\n1 is NOT prime
899443
39 can be devided by 3 people.  Scary.  \n\nBut you get the jist of what is a prime and a square from the other posts.
136558
none of those numbers are prime numbers. a prime number is a number which nothing besides itself and 1 go into.\n\ni.e. 13. 13 is a prime number because the onyl way to get 13 is 13 x 1. 7 is also a prime number.\n\nBut 6, 12, 24, 36 are not prime because 2, 3, 6, and more #'s go into all of them.