use fermat's little theorem to calculate 5 raised to the power of 74 mod 7?
406361
well the theorem says that a^p ≡ a(modp) where p is prime.\nequivalent to this is a^(p-1) ≡ 1(modp) by just taking the a to the other side. \n\nso we know 5^6≡1(mod7)\n\nso\n5^74 = (5^6)^12 x 5^2 ≡ 1 x 5^2 ≡ 25 ≡ 4 (mod 7)
676773
Since your examples are of the form "x^y mod N",\nI would recommend that you look at the page:\n\nhttp://en.wikipedia.org/wiki/Exponentiation_by_squaring\n\nand, in particular, the section titled:\n\nFurther applications\n\nwhere it discusses the most efficient method for\n"fast computation of large exponents modulo a number".\nFor your examples, it isn't as critical, but if you're\nexploring cryptography, you'll end up having to\ndeal with much larger exponents.\n\nFor your specific examples, you could use a "bignum"\npackage and just brute-force compute 123^7 and take\nthe remainder after division by 187.
1007006
add or subtract 1\n\nor else subtract the S at the begining of Seven
919015
(8X^6/7) * (56/24)\n8X^6/3\n\n(Hint:  X^m/X^n = X^(m-n)
1089449
It's as simple as plugging in numbers.  The number 15 becomes the letter "a", the number 7 is "b", and the number 55 is "c".  Just substitute the number for the letter in the formula and solve.
545123
12 to 1  less than 2 .....12 to 5 for number over a 7
822221
Using mods...\n\nn^3 + n - 4 = 0 mod 7\n\nn^3 + n = 4 mod 7\n\nn(n^2+1) = 4 mod 7\n\nIf n=0 mod 7 then n(n^2+1) = 0 mod 7, doesn't work.\nIf n=1 mod 7 then n(n^2+1) = 1(1^2+1) mod 7 = 2 mod 7, doesn't work.\nIf n=2 mod 7 then n(n^2+1) = 2(4+1) mod 7 = 3 mod 7, doesn't work.\nIf n=3 mod 7 then n(n^2+1) = 3(9+1) mod 7 = 2 mod 7, doesn't work.\nIf n=4 mod 7 then n(n^2+1) = 4(16+1) mod 7 = 5 mod 7, doesn't work.\nIf n=5 mod 7 then n(n^2+1) = 5(25+1) mod 7 = 130 mod 7 = 4 mod 7 - YESSSS!\nIf n=6 mod 7 then n(n^2+1) = 6(36+1) mod 7 = 222 mod 7 = 5 mod 7, doesn't work.\n\nSo, n must be equal to 7m+5 for some integer m.  This gives us:\nn^3 + n - 4 =\n(7m+5)^3 + 7m+5 - 4 =\n7^3m^3 + 3*7^2m^2*5 + 3*7m*5^2 + 5^3 + 7m + 1 = \na whole bunch of multiples of 7 + 5^3 + 1 = \nmultiples of 7, plus 126 =\nmultiples of 7, plus 18*7.
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
796039
(5 * 5 * 5 - 5) / 5 + 5 * 5
804536
1 + 7 + 7(6) + 42(5) + 210(4) + 840(3) + 2520(2) + 5040(1) = 13,700
703752
simply subtract 7 from both sides. This leave you with\n\nx<_3
605852
multiply n-m by 0.737562
484601
The previous poster didn't see the whole problem. This is a question about modular arithmetic, not standard algebra. The equations are to be done modulo n and the solutions are to be integers.\n\nBy adding 3 times the first to the second, we need 5x=8. By subtracting twice the first from the second, we get -5y=-2, so 5y=2. So all that is required is that 5 be invertible modulo n, i.e., that 5 does not divide n. For example, if n=4, x=0 and y=2 works.
769578
I don't follow this formula.  I tried it for 7.\n\n((7-1)/2)! (mod 7) is 3! mod 7 = 6 mod 7 = 6.\n\nNow,6 mod 4 = 2 and is not =1.\n\nAlso, it doesn't work for p=2.  I think it should be odd primes.
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.
198100
7 / [8 2/5] = 7 / [42/5] = 7* 5/42= [7*5]/[7*6] = 5/6
810912
105 or any number that equals 105 time any multiple of 3,5 or 7