SOLUTIONS

Consider the number m = (b_k ... b₀) ₁₀ = b_k10^k +··· + b₂10² + b₁10 + b₀. Since b_k10^k +··· + b₂10² is divisible by 4, we have that m is divisible by 4 if and only if (b₁ b₀)₁₀ = b₁10 + b₀ is divisible by 4.

Consider the number m = (b_k ... b₀) ₁₀ = b_k10^k + ··· + b₂10² + b₁10 + b₀. Then m is divisible by 8 if and only if (b₂b₁b₀)₁₀ is divisible by 8.
Now consider the binary representation of an integer m: (b_k ... b₀) ₂ = b_k2^k + ··· + b₂2² + b₁2 + b₀. Since b_k2^k + ··· + b₂2³ is divisible by 8, we now have that m is divisible by 8 if and only if (b₂b₁b₀)₂ = (000)₂.

Consider the number m = (b_k ... b₀) _a = b_ka^k + ··· + b₂a² + b₁a + b₀ = b_k(a^k - 1) + ··· + b₂(a² - 1) + b₁(a-1) + (b₀ + ··· + b_k).

Since b_j(a^j - 1) is divisible by a - 1 for all j, we have that m is divisible by a - 1 if and only if b₀ + ··· + b_k is divisible by a - 1.

Consider the number (abcabc) ₁₀ = a10⁵ + b10⁴ + c10³ +a100 + b10 + c = 100100a + 10010b + 1001c = 1001(100a + 10 b + c). The prime factorization of 1001 is 1001 = 7 · 11 · 13, implying that 7, 11 and 13 divide (abcabc)₁₀.

If n = 0 mod 3, then n⁴ + n² + 1 = 0 + 0 + 1 = 1 mod 3. Thus n⁴ + n² + 1 is not divisible by 3.

If n = 1 mod 3, then n⁴ + n² + 1 = 1 + 1 + 1 = 0 mod 3. Thus n⁴ + n² + 1 is divisible by 3.

If n = 2 mod 3, then n⁴ + n² + 1 = 16 + 4 + 1 = 0 mod 3. Thus n⁴ + n² + 1 is divisible by 3.

If for all nonzero x we have x^{p - 1} = 1 mod p, then every nonzero element x has an inverse, namely x^{p - 2}. Thus p is a prime.

• 10⁶ - 1 = (-3)⁶ - 1 = 9³ - 1 = (-4)³ - 1 = -64 - 1 = -65 = 0 (mod 13).
• 10⁸ + 1 = (100)⁴ + 1 = 2⁴ + 1 = 17 = 0 (mod 17).
• n⁴ + 64 = n⁴ - 1 = (n² - 1)(n² + 1) (mod 5). Since a square is 1 or -1 mod 5 this is always 0 mod 5. Hence n⁴ + 64 is divisible by 5.
• 2¹⁰⁰⁰ + 5 = (-1)¹⁰⁰⁰ + 2 = 1 + 2 = 0 (mod 3).
• 2²ⁿ - 1 = (2ⁿ - 1)(2ⁿ + 1) = 0 (mod 3).

Modulo 37 the inverse of 3 is 25.

The inverse of 4 modulo 14 does not exist.

• x = 8 (mod 21).
• x = 3 (mod 5).
• No solution.

We have to show that 641 divides 2²⁵ + 1, or in other words that 2²⁵ + 1 = 0 mod 641.

2³² + 1 = (2⁷)⁴ · 2⁴ + 1 = (2⁷)⁴ · (- 5⁴) + 1 = - (2⁷ · 5)⁴ + 1 = - (-1)⁴ + 1 = 0 mod 641.

The binomial coefficient n choose k equals

Since k is less than p we see that there is no factor p in the denominator, and for k > 0, p is a factor of the numerator. Hence the quotient is divisible by p.

First writing (x + y)^p with the help of Newton's binomium and then reducing mod p yields that

The invertible elements of Z/4Z are 1 and 3.

The invertible elements of Z/6Z are 1 and 5.

The invertible elements of Z/12Z are 1, 5, 7 and 11.

The invertible elements of Z/p²Z are all elements except for the multiples of p.

Let x be an integer. The value of x³ mod 9 equals -1, 0 or 1. So x³ + y³ + z³ = -3, -2, -1, 0, 1, 2 or 3 mod 9. In particular, x³ + y³ + z³ = 5 (mod 9) is not possible.

Now let us examine the possible values below 30 that x³ + y³ + z³ can take. By the above, we only have to consider those values that are -3, -2, -1, 0, 1, 2 or 3 mod 9.

0 = 0³ + 0³ + 0³

1 = 1³ + 0³ + 0³

2 = 1³ + 1³ + 0³

3 = 1³ + 1³ + 1³

6 = 2³ + (-1)³ + (-1)³

7 = 2³ + (-1)³ + 0³

8 = 2³ + 0³ + 0³

9 = 2³ + 1³ + 0³

10 = 2³ + 1³ + 1³

• 377 = (101111001)₂ = (571)₈ = (179)₁₆.
• 301 = (610)₇.
• (3456142)₇ = 3 + 4 + 5 + 6 + 1 + 2 + 4 = 1 (mod 6).
• Suppose x = (b_k ... b₀)₂ is a number given in binary form. Then let
  a_i = b_3i + 2b_{3i + 1} + 4b_{3i + 2}.

  Then x = (a_k' ... a₀)₁₆, where k' is the maximal index with a_i not zero.

Suppose the celebration is for the car number x·1000. Then the number of minutes gone by is x·1000· 11.

The celebration should be on Friday between 3 and 5 pm. The number of minutes for Monday 0 am till Friday 3 pm is (96 + 15) · 60. The number of minutes for Monday 0 am till Friday 5 pm is (96 + 17) · 60. The number of minutes in a week is 168 · 60.

This yields the following equation: (96 + 15) · 60 - 1 < x · 1000 · 11 (mod 168 · 60) < (96 + 17) · 60 + 1. The first positive integer x for which this is true, x = 73, is now easily found.

1. If the decimal representation of 1/n is finite, then there is a power of 10, say 10^m, with 10^m/n being an integer. So the primes dividing n are 2 or 5. The converse is trivial.

2. The decimal representation of 1/n contains a repeating part if there exist m and m' with the remainder of 10^m divided by n being equal to the remainder of 10^m' divided by n. In other words if 10^m = 10^m' mod n.

   Since the powers of 10 can only take finitely many values mod n, there we will be such m and m'.

Multiplying the first equaltion with 3 yields

x = 1 mod 5.

Multiplying the second equaltion with -2 yields

x = 2 mod 7.

So x = 16 mod 35.

Add the two equation and find

3x = 3 mod 11. So x = 1 mod 11.

But then y = 5 mod 11.

Let d be the inverse of c. Then ca = cb implies dca = dcb and thus a = dca = dcb = b.

Factor 2623 to find the prime factors 43 and 61.  Then (p - 1)(q - 1)  =  2520. Next we look for a decoding number w satisfying vw = 1 mod (p - 1)(q - 1). In our case the equation is 37w = 1 mod 2520. A solution to this equation is w = 613. Now compute (mod 2623)

The encoded text then reads