I have three questions:

1. What is the remainder when 27 to the power of 1001 is divided by 13?

2. What is the remainder of when 38 to the power of 101 is divided by 13?

3. How do you show that

70 x 27 (to the power of 1001) + 31 x 38(to the power of 101)

can be divisible by 13?

Thanks

1. 27/13 = 2 with a remainder of 1

27^2/13 = 56 with a remainder of 1
27^3/13 = 1514 with a remainder of 1
27^4/13 = 40,880 with a remainder of 1
...
So 27 to any integer power, divided by 13, has a remainder of 1.

Define x Mod(13) as the remainder of x if you divide x by 13. You can iterchange taking Mod with taking powers.

Since:

27 Mod(13) = 1 --------->

27^n Mod(13) = 1^n = 1

2)
Let's omit writing Mod(13) every time, equality simply means both sides are the same Mod (13).

We can use that 38 = 12 = -1

So 38^101 = (-1)^101 = -1 = 12

70 x 27^(1001) + 31 x 38^(101) =

70 - 31 = 39 = 0

1. To find the remainder when 27 to the power of 1001 is divided by 13, we can use the remainder theorem: if a number is divided by 13, the remainder will be the same as when the number is divided by the positive integer remainder of 13.

We can simplify the problem by first finding the remainder for smaller powers of 27 when divided by 13.

27^1 (mod 13) = 1
27^2 (mod 13) = 1 * 27 (mod 13) = 14 (mod 13) = 1
27^3 (mod 13) = 1 * 27^2 (mod 13) = 1 * 1 (mod 13) = 1
27^4 (mod 13) = 1 * 27^3 (mod 13) = 1 * 1 (mod 13) = 1

We see that the remainder of powers of 27 repeats with a period of 1, so we can find the remainder for any power of 27 by dividing the exponent by 1 and taking the remainder.

Since 1001 divided by 1 leaves a remainder of 0, we can conclude that:

27^1001 (mod 13) = 27^0 (mod 13) = 1.

Therefore, the remainder when 27 to the power of 1001 is divided by 13 is 1.

2. Using a similar method, we can find the remainder when 38 to the power of 101 is divided by 13.

38^1 (mod 13) = 12
38^2 (mod 13) = 12 * 38 (mod 13) = 456 (mod 13) = 6
38^3 (mod 13) = 6 * 38 (mod 13) = 228 (mod 13) = 11
38^4 (mod 13) = 11 * 38 (mod 13) = 418 (mod 13) = 2
38^5 (mod 13) = 2 * 38 (mod 13) = 76 (mod 13) = 2

Again, we see that the remainder of powers of 38 repeats with a period of 4. Since 101 divided by 4 leaves a remainder of 1, we can conclude that:

38^101 (mod 13) = 38^1 (mod 13) = 12.

Therefore, the remainder when 38 to the power of 101 is divided by 13 is 12.

3. To show that the expression 70 * 27^1001 + 31 * 38^101 can be divisible by 13, we need to show that the sum of the remainders of each term, when divided by 13, is 0.

We have already determined the remainders for each term:
70 * 27^1001 (mod 13) = 70 * 1 (mod 13) = 70 (mod 13) = 5
31 * 38^101 (mod 13) = 31 * 12 (mod 13) = 372 (mod 13) = 4

Now, let's add the remainders:
5 + 4 = 9

Since 9 is not divisible by 13, we can conclude that the expression 70 * 27^1001 + 31 * 38^101 is not divisible by 13.

1. To find the remainder when 27 to the power of 1001 is divided by 13, we can use the concept of modulus. Begin by noticing a pattern in the remainders of powers of 27 when divided by 13:

27^1 ≡ 1 (mod 13)
27^2 ≡ 3 (mod 13)
27^3 ≡ 9 (mod 13)
27^4 ≡ 10 (mod 13)
27^5 ≡ 12 (mod 13)

We can observe that the remainders repeat after every 6th power. This is known as Fermat's Little Theorem, which states that if p is a prime number and a is any positive integer not divisible by p, then a^(p-1) is congruent to 1 modulo p.

Since 27 is not divisible by 13, we can use this theorem to simplify the calculation. In this case, 27^6 ≡ 1 (mod 13). Therefore, we can rewrite 27^1001 as:

27^1001 = (27^6)^166 * 27^5 ≡ 1^166 * 12 ≡ 12 (mod 13).

Hence, the remainder when 27^1001 is divided by 13 is 12.

2. Similarly, to find the remainder when 38^101 is divided by 13, we can use the same concept. Find the pattern in the remainders of the powers of 38 when divided by 13:

38^1 ≡ 12 (mod 13)
38^2 ≡ 3 (mod 13)
38^3 ≡ 1 (mod 13)

Once again, we notice that the remainders repeat after every 3rd power. Applying Fermat's Little Theorem, we can simplify the calculation:

38^101 = (38^3)^33 * 38^2 ≡ 1^33 * 3 ≡ 3 (mod 13).

Hence, the remainder when 38^101 is divided by 13 is 3.

3. To show that 70 * 27^1001 + 31 * 38^101 is divisible by 13, we can evaluate both terms separately and then add them together.

Using the results from questions 1 and 2, we know that 27^1001 leaves a remainder of 12 when divided by 13, and 38^101 leaves a remainder of 3 when divided by 13.

Now, substitute these values into the expression:

70 * 27^1001 + 31 * 38^101 ≡ 70 * 12 + 31 * 3 ≡ 840 + 93 ≡ 933 ≡ 0 (mod 13).

Since 933 is divisible by 13 (it leaves no remainder), we can conclude that 70 * 27^1001 + 31 * 38^101 is divisible by 13.