what are the factors of 43?

Answer 1

43 is a prime number
so the only factors would be 1 and 43

I hope you tried to work this out on your own first before you asked for help if this was your homework, if not, then I hope this answers your question. ^^

btw, here's a list of prime numbers just in case, hehe: http://www.math.utah.edu/~pa/math/primelist.html

Answer 2

Factors Of 43

Answer 3

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For the first question, recall the formula for factoring sums of the form x^(2n + 1) + y^(2n + 1): x^(2n + 1) + y^(2n + 1) = (x + y)(∑((-1)^k * x^k * y^(2n - k))), where the sum is taken from k=0 to 2n. (This is just like the formula for factoring the sum of two cubes, except the second term goes on longer.) Thus: (43^43 + 47^43) = (43 + 47)(43^42 - 43^41 * 47 + ... - 43 * 47^41 + 47^42) (43^47 + 47^47) = (43 + 47)(43^46 - 43^45 * 47 + ... - 43 * 47^45 + 47^46). So 90 is a common factor. It happens that it is the greatest common factor, but I am not sure how to prove this without using a computer. For the second question: We can write 7^126 = (7^2)^63 = (49)^63. Notice that, for any n, 49n = 48n + n. This means that 49n differs from n by a multiple of 48. Thus, if we divide 49n by 48, we get the same remainder as if we were to divide n by 48. Thus, if we divide 49^63 by 48, we get the same remainder as if we divided 49^62 by 48; which is the same remainder as if we divided 49^61 by 48; and so on down, until we get the same remainder as if we divided 49 by 48. Therefore, the remainder when dividing 7^126 by 48 is 1. (If you know group theory: Work out 7^126 within Z/48Z, the cyclic group of order 48. The argument is easier.)

Answer 4

1,43 because its prime number

Answer 5

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