If n is a positive integer, what is the remainder when n^7 – n is divi

Answer A
Take 2: 2^7-2= 126/42=3 and 0 as remainder
Take 3: 37-3= 2187-3/42=52 and 0 as remainder

First, we can factor the common n from each term:

n(n^6 - 1)

We now have a difference of squares, which can be factored:
n(n^3 - 1)(n^3 + 1)

We can factor (n^3 - 1) as a difference3 of cubes and (n^3 + 1) as a sum of cubes, obtaining:

n(n - 1)(n^2 + n + 1)(n + 1)(n^2 - n + 1)

Rearranging the factors, we have:

(n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1)

We see that (n - 1)(n)(n + 1) is a product of 3 consecutive integers, which is always divisible by 3! = 6. However, to be divisible by 42, it also needs to be divisible by 7 since 42 = 6 x 7. To see if the expression is divisible by 7, we can substitute n by all the remainders from division by 7 (i.e., 0, 1, 2, 3, 4, 5, 6):

If n = 0, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = 0, which is divisible by 7.

If n = 1, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = 0, which is divisible by 7.

If n = 2, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = (1)(2)(3)(7)(3), which is divisible by 7.

If n = 3, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = (2)(3)(4)(13)(7), which is divisible by 7.

If n = 4, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = (3)(4)(5)(21)(13), which is divisible by 7.

If n = 5, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = (4)(5)(6)(31)(21), which is divisible by 7.

If n = 6, (n - 1)(n)(n + 1)(n^2 + n + 1)(n^2 - n + 1) = (5)(6)(7)(43)(31), which is divisible by 7.

As we can see, since all the values of n from 0 to 6, inclusive, yield a value of the expression that is divisible by 7, we see that n^7 - n is divisible by both 6 and 7, which makes it divisible by 42. Thus, the remainder when the expression is divided by 42 is 0.

[Note: The reason we can just use the 7 remainders from division by 7 is that all other integers will “behave” like one of these 7 remainders. For example, if n = 15 and since 15/7 = 2 R 1, 15 “behaves” like 2.]

Answer: A

let n=1
1^7-1=0
0/42 leaves remainder of 0
A

I got to the part: n(n - 1)(n^2 + n + 1)(n + 1)(n^2 - n + 1)

And then here's my logic:

(n-1)n(n+1) is always divisible by 3 and 2. So we got T = n^7-n is divisible by 6.

With divisibility by 7: n can have 6 remainders when divided by 7: r(emainder) = {0; 1; 2; 3; 4; 5; 6}

r = 1. (n-1) is divisible by 7. T is divisible by 42
r = 2. n = 7k+2. So n^2 +n +1 = ... + 2^2 + ... + 2 + 1 = 7. So T is divisible by 42.
r = 3. n = 7k+3. So n^2 - n +1 = ... + 3^2 - ... - 3 + 1 = 7. So T is divisible by 42.
r =4. n = 7k+4. So n^2 + n +1 = ... + 4^2 + ... + 4 + 1 = 21. So T is divisible by 42.
r =5. n = 7k+5. So n^2 - n +1 = ... + 5^2 - ... - 5 +1 = 21. So T is divisible by 42.
r =6. (n+1) is divisible by 7. So T is divisible by 42.

IMO A.

Asked: If n is a positive integer, what is the remainder when \(n^7 – n\) is divided by 42?

n^7 - n = n (n^6-1) = n(n^3-1)(n^3+1) = n(n-1)(n^2+n+1)(n+1)(n^2-n+1)
n(n-1)(n+1) is divisible by 6.

n^7-n is also divisible by 7. Confirmed after checking for n=1,2,3,4,...

n^7-n is divisible by 6*7 = 42

The remainder when \(n^7 – n\) is divided by 42 = 0

IMO A