Q: The Perfect Square Puzzle

Let n = a² and n + 1 = b², where a and b are positive integers. Then: b² - a² = 1 (b - a)(b + a) = 1 Since b - a and b + a are positive integers whose product is 1, we must have: b - a = 1 and b + a = 1 But this gives: 2a = 0, so a = 0, which is not positive. Wait, let me reconsider the problem. Actually, consecutive perfect squares differ by at least 2a + 1, which is at least 3 for a ≥ 1. So it's impossible to have two consecutive perfect squares that differ by 1! But if we allow n = 0: 0² = 0 and 1² = 1, so 0 + 1 = 1 works! Answer: n = 0 (the only solution, but 0 is not positive) Actually, there are no positive integer solutions to this puzzle!

Q: The Modular Inverse

We need to find x such that 7x ≡ 1 (mod 26). This means: 7x + 26y = 1 for some integer y. Using Extended Euclidean Algorithm: 26 = 3 × 7 + 5 7 = 1 × 5 + 2 5 = 2 × 2 + 1 2 = 2 × 1 + 0 Working backwards: 1 = 5 - 2 × 2 1 = 5 - 2 × (7 - 1 × 5) 1 = 3 × 5 - 2 × 7 1 = 3 × (26 - 3 × 7) - 2 × 7 1 = 3 × 26 - 11 × 7 So: -11 × 7 + 3 × 26 = 1 Therefore: 7 × (-11) ≡ 1 (mod 26) Since -11 mod 26 = 15, we have: 7 × 15 ≡ 1 (mod 26) Answer: x = 15 Verification: 7 × 15 = 105 = 4 × 26 + 1 ≡ 1 (mod 26) ✓

Question 1

The Missing Digit

Accepted Answer

The sum of the known digits is 4 + 5 + 7 + 2 = 18.

Since 18 is already divisible by 9, the missing digit must be 0 or 9.

However, if we check:
- 45072: 4 + 5 + 0 + 7 + 2 = 18 ✓
- 45972: 4 + 5 + 9 + 7 + 2 = 27 ✓

Both work! But typically, we'd use 0 to get the smallest valid number.

Answer: 0 (or 9)

Question 2

The Prime Sum

Accepted Answer

Since 50 is even, and the only even prime is 2, one of the primes must be 2.

So we need: 2 + p = 50, where p is prime.

This gives us p = 48, but 48 is not prime (48 = 2 × 24).

Wait, let me reconsider... Actually, if both primes are odd, their sum would be even. But 2 is the only even prime.

Let's check: 2 + 3 = 5 (too small)
2 + 47 = 49 (not 50)
2 + 43 = 45 (not 50)

Actually, 3 + 47 = 50! And 7 + 43 = 50! And 13 + 37 = 50! And 19 + 31 = 50!

Answer: There are 4 pairs: (3, 47), (7, 43), (13, 37), and (19, 31)

Question 3

The Clock Arithmetic

Accepted Answer

We need to find: 3 + 100 mod 12

First, calculate 100 mod 12:
100 ÷ 12 = 8 remainder 4
So 100 mod 12 = 4

Therefore: 3 + 4 = 7

Answer: It will be 7 o'clock

Question 4

The GCD Challenge

Accepted Answer

Since gcd(a, b) = 6, we can write:
a = 6x, b = 6y where gcd(x, y) = 1

And a + b = 30, so:
6x + 6y = 30
x + y = 5

Since x < y and gcd(x, y) = 1, the possible pairs are:
- x = 1, y = 4 → a = 6, b = 24
- x = 2, y = 3 → a = 12, b = 18

Let's verify:
gcd(6, 24) = 6 ✓
gcd(12, 18) = 6 ✓

Answer: (a, b) = (6, 24) or (12, 18)

Question 5

The Perfect Square Puzzle

Accepted Answer

Let n = a² and n + 1 = b², where a and b are positive integers.

Then: b² - a² = 1
(b - a)(b + a) = 1

Since b - a and b + a are positive integers whose product is 1, we must have:
b - a = 1 and b + a = 1

But this gives: 2a = 0, so a = 0, which is not positive.

Wait, let me reconsider the problem. Actually, consecutive perfect squares differ by at least 2a + 1, which is at least 3 for a ≥ 1.

So it's impossible to have two consecutive perfect squares that differ by 1!

But if we allow n = 0: 0² = 0 and 1² = 1, so 0 + 1 = 1 works!

Answer: n = 0 (the only solution, but 0 is not positive)

Actually, there are no positive integer solutions to this puzzle!

Question 6

The Modular Inverse

Accepted Answer

We need to find x such that 7x ≡ 1 (mod 26).

This means: 7x + 26y = 1 for some integer y.

Using Extended Euclidean Algorithm:
26 = 3 × 7 + 5
7 = 1 × 5 + 2
5 = 2 × 2 + 1
2 = 2 × 1 + 0

Working backwards:
1 = 5 - 2 × 2
1 = 5 - 2 × (7 - 1 × 5)
1 = 3 × 5 - 2 × 7
1 = 3 × (26 - 3 × 7) - 2 × 7
1 = 3 × 26 - 11 × 7

So: -11 × 7 + 3 × 26 = 1

Therefore: 7 × (-11) ≡ 1 (mod 26)

Since -11 mod 26 = 15, we have:
7 × 15 ≡ 1 (mod 26)

Answer: x = 15

Verification: 7 × 15 = 105 = 4 × 26 + 1 ≡ 1 (mod 26) ✓