two consecutive numbers are multiplied,what is the product smallest prime factor?
826780
With two consecutive numbers, one is always even, so the smallest prime factor of the product is 2.
738875
To see what conjecture they are interested in, it really is necessary to consider a number of examples, exactly as they said. So lets set up a table. We will call the product in question N.\nRemember that we are only interested in prime numbers for p.\n\np               N                N/p\n---------------------------------\n2                    2               1\n3                    3               1\n5                  25               5\n7                721           103\n11      3628801     329891\n13  479001601 36846277\n\nNotice that the numbers N so far have all been divisible by p. So, if we represented  them on a p-clock, they would all be p o'clock (or 0 o'clock if you prefer). That is the conjecture they are interested in. You are supposed to conjecture that the value of 1*2*3*...*(p-1)+1 in p-hour arithmetic is always p o'clock.
614283
the first step is basically to factor 120175.  the obvious factor to do first is 5.  this gives us 5*24035.  then factoring another 5 gives us 5*5*4807.  since 4807 obviously isn't divisible by 5 or 2, try 3.  doesn't work - so keep on trying the next biggest prime number (1,2,3,5,7,11,13,17...)  turns out that 4807 is divisible by 11.  this gives us 5*5*11*437.  keep on going and you get 5*5*11*19*23.  just checking, yes - 5*5*11*19*23=120175.\n\nnow converting these factors into letters, you get the available letters EEKSW.  this can be rearranged to be WEEKS.  however, remember that 5*5*11*19*23 can also be written as 25*11*19*23, since this also equals 120175.  this means the word could also have KSWY.  but i don't think you can rearrange those letters to make a word.\n\nso therefore the english word encoded as 120175 is WEEKS.  basically, you factored the number, converted them into letters, and solved the anagram.\n\n[[edit]]\n\ninuyasha's right - there could be an infinite number of "a"s in there too since 5*5*11*19*23 = 5*5*11*19*23*1*1*1*...\n\nhowever, this gives us WEEKS+AAA... so using the online anagram solver, found at http://wordsmith.org/anagram/ , there are no words that have "WEEKS" in them and "A"s.  therefore, the answer is still WEEKS.
352698
7 & 103\n\nYou can go here to find the prime factors for any number:\nhttp://www.acme.com/software/factor/
890978
To get prime factors, keep dividing by numbers and divide those until you can't divide them no more!  For example,\n\n34 \n17 * 2\n\n(Can't divide 17 any more and can't divide 2 -- both of these are prime numbers)\n\nAnother example,\n\n120\n12 * 10\n2 * 6 * 2 * 5\n2 * 2 * 3 * 2 * 5 (<-- all prime numbers here)\n\nTry to get the rest yourself and post the answers, we'll tell you if you got them right!  ;-)
935700
two consecutive numbers could be called x, and (x+1). Add these two together to find the sum. x+(x+1) = 85.\nSo 2x+1 = 85.\nsubtract 1 from both sides and we have 2x = 84.\nDivide both sides by two to get x by itself so x = 42.\nSo if x=42, the consexutive numbers from the beginning of the problem would be 42, and 43.
754767
Assuming that the two numbers are positive, the best you can do is take double and triple 850.  Both numbers will be divisible by 850, and since neither can be divisible by the other they need each to be multiplied by numbers which are not divisible by the other.  The smallest numbers that work are 2 and 3, so 850 * 2 and 850 * 3 are the smallest solutions to your problem.
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.
283645
if p is a factor of both, than p should also be a factor of the difference.\n\nn + 10 - (n + 3)\n\nThe resulting integer only has one factor greater than 1.  That's your answer!\n\n(I'll let you figure out the rest, don't want to spoil it for you.)
594189
34,\n\nYou are having the series as 2 times <prime numbers>\n2*2, 2*3, 2*5, 2*7, 2*11, 2*13, 2*17, 2*19, 2*23...\n = \n4, 6, 10, 14, 22, 26, 34, 38, 46 ...
726076
Yep, it's 202.  Factor out the x and add the resulting cofactors\nin pairs whose sums are 100.\n\nx[(1+99)+(2+98)+...+\n(49+51)+50+100] =\nx[5050]  Now factor 5050 to its primes.\n\n  you get  2,5,5,101 so all you need are two factors\nof 2 and 101 to get all primes in pairs. Thus the answer is 202.\n\n202X4950=1020100 = (1010)^2\n\n Pretty clean , eh?
812323
The prime factors of 1992 are 2 x 2 x 2 x 3 x 83.\n\nBy including 2 * 4 * 6, you have already covered 2 x 2 x 2 x 3, but you need to cover 83.  83 is prime, so the first time you have an even number that has 83 as a factor will be 166 (2 * 83).\n\nThat means you must figure the product of 2 * 4 * 6 * ... * 164 * 166 for it to be divisible by 1992.\n\nThe answer is largest even number in the product is 166.
906131
Having researched this topic as far as possible online, I found very few helpful resources.\n\nWhile the result you mention, using the subsequence 6(5)^n, is intriguing, I suspect that other sequences will yield vastly different bounds on the index.  In particular, is it not true that any prime number p always has a corresponding Fibonacci number F(p+1) of which it is a factor?\n\nGood luck with your own research!
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.
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.
571480
A mathematical statement can be proven in several ways. Your method of proof will depend on the statement. For the statement above (assuming you mean integers by numbers) the proof would be somthing like this\n\nlet P = {x: x is a prime number}\n\nlet M1 = min(P)\nlet M2 = min(P - M1)\nlet M2 = min(P - M1 - M2)\n\nthis means that M1, M2, M3 are the 3 smallest prime numbers. We know that \nM1 = 2, M2 = 3, M3 = 5\nthis implies M1 + M2 + M3 = 2 + 3 + 5 = 10 > 6\ntherefore for every integer N > 6,\nN = P1 + P2 + P3, P1,P2,P3 are prime numbers
1014042
The range of consecutive prime numbers for a "strictly" 9-digit product begins at 463 (* 467 * 479 = 103569859) and ends with 991 (* 997 * 1009 = 996919243).  Within this numerical range, there are NO consecutive primes whose product results in unique digits.\n\nEven if you expand the definition of a 9-digit product to include those with a leading zero as a digit, you can also check all 8-digit products and consider the first digit to be a zero.  The range then begins at 211 (* 223 * 227 = 010681031).  Unfortunately, even with this expanded definition, there are no products with unique digits.  (The solution 233*239*241=13420567 is incorrect because if expressed as nine digits, the product becomes 013420567 which repeats the 0 digit.)\n\nFor the morbidly curious, the full solution set follows:\n\n211 x 223 x 227 = 10681031 -- 0x3 1x3 repeat\n223 x 227 x 229 = 11592209 -- 0x2 1x2 2x2 9x2 repeat\n227 x 229 x 233 = 12112039 -- 0x2 1x3 2x2 repeat\n229 x 233 x 239 = 12752323 -- 2x3 3x2 repeat\n233 x 239 x 241 = 13420567 -- 0x2 repeat\n239 x 241 x 251 = 14457349 -- 4x3 repeat\n241 x 251 x 257 = 15546187 -- 1x2 5x2 repeat\n251 x 257 x 263 = 16965341 -- 1x2 6x2 repeat\n257 x 263 x 269 = 18181979 -- 1x3 8x2 9x2 repeat\n263 x 269 x 271 = 19172437 -- 1x2 7x2 repeat\n269 x 271 x 277 = 20193023 -- 0x3 2x2 3x2 repeat\n271 x 277 x 281 = 21093827 -- 0x2 2x2 repeat\n277 x 281 x 283 = 22027871 -- 0x2 2x3 7x2 repeat\n281 x 283 x 293 = 23300239 -- 0x3 2x2 3x3 repeat\n283 x 293 x 307 = 25456133 -- 3x2 5x2 repeat\n293 x 307 x 311 = 27974761 -- 7x3 repeat\n307 x 311 x 313 = 29884301 -- 0x2 8x2 repeat\n311 x 313 x 317 = 30857731 -- 0x2 3x2 7x2 repeat\n313 x 317 x 331 = 32842151 -- 1x2 2x2 repeat\n317 x 331 x 337 = 35360399 -- 0x2 3x3 9x2 repeat\n331 x 337 x 347 = 38706809 -- 0x3 8x2 repeat\n337 x 347 x 349 = 40811711 -- 0x2 1x4 repeat\n347 x 349 x 353 = 42749359 -- 4x2 9x2 repeat\n349 x 353 x 359 = 44227723 -- 2x3 4x2 7x2 repeat\n353 x 359 x 367 = 46508809 -- 0x3 8x2 repeat\n359 x 367 x 373 = 49143869 -- 4x2 9x2 repeat\n367 x 373 x 379 = 51881689 -- 1x2 8x3 repeat\n373 x 379 x 383 = 54143561 -- 1x2 4x2 5x2 repeat\n379 x 383 x 389 = 56466073 -- 0x2 6x3 repeat\n383 x 389 x 397 = 59147839 -- 9x2 repeat\n389 x 397 x 401 = 61927633 -- 3x2 6x2 repeat\n397 x 401 x 409 = 65111573 -- 1x3 5x2 repeat\n401 x 409 x 419 = 68719771 -- 1x2 7x3 repeat\n409 x 419 x 421 = 72147191 -- 1x3 7x2 repeat\n419 x 421 x 431 = 76027969 -- 0x2 6x2 7x2 9x2 repeat\n421 x 431 x 433 = 78568283 -- 8x3 repeat\n431 x 433 x 439 = 81927497 -- 7x2 9x2 repeat\n433 x 439 x 443 = 84208541 -- 0x2 4x2 8x2 repeat\n439 x 443 x 449 = 87320173 -- 0x2 3x2 7x2 repeat\n443 x 449 x 457 = 90900499 -- 0x4 9x4 repeat\n449 x 457 x 461 = 94593973 -- 3x2 9x3 repeat\n457 x 461 x 463 = 97543451 -- 4x2 5x2 repeat\n461 x 463 x 467 = 99677881 -- 7x2 8x2 9x2 repeat\n463 x 467 x 479 = 103569859 -- 5x2 9x2 repeat\n467 x 479 x 487 = 108938491 -- 1x2 8x2 9x2 repeat\n479 x 487 x 491 = 114537043 -- 1x2 3x2 4x2 repeat\n487 x 491 x 499 = 119319383 -- 1x3 3x3 9x2 repeat\n491 x 499 x 503 = 123239527 -- 2x3 3x2 repeat\n499 x 503 x 509 = 127757473 -- 7x4 repeat\n503 x 509 x 521 = 133390067 -- 0x2 3x3 repeat\n509 x 521 x 523 = 138693847 -- 3x2 8x2 repeat\n521 x 523 x 541 = 147413303 -- 1x2 3x3 4x2 repeat\n523 x 541 x 547 = 154769821 -- 1x2 repeat\n541 x 547 x 557 = 164831339 -- 1x2 3x3 repeat\n547 x 557 x 563 = 171534277 -- 1x2 7x3 repeat\n557 x 563 x 569 = 178433279 -- 3x2 7x2 repeat\n563 x 569 x 571 = 182918137 -- 1x3 8x2 repeat\n569 x 571 x 577 = 187466723 -- 6x2 7x2 repeat\n571 x 577 x 587 = 193397129 -- 1x2 3x2 9x3 repeat\n577 x 587 x 593 = 200848507 -- 0x3 8x2 repeat\n587 x 593 x 599 = 208506509 -- 0x3 5x2 repeat\n593 x 599 x 601 = 213479407 -- 4x2 7x2 repeat\n599 x 601 x 607 = 218519393 -- 1x2 3x2 9x2 repeat\n601 x 607 x 613 = 223626691 -- 2x3 6x3 repeat\n607 x 613 x 617 = 229580147 -- 2x2 repeat\n613 x 617 x 619 = 234118799 -- 1x2 9x2 repeat\n617 x 619 x 631 = 240993413 -- 3x2 4x2 9x2 repeat\n619 x 631 x 641