Science Olympiad coding events help for 2027

Source: Electric Vehicle C event page and its 2027 Kit Modification Guidance PDF, read 31 October 2026. Full rules are in the official 2027 rules, free on the 2027 Rules page.

1. Number the alphabet. A is 0, B is 1, and so on to Z, which is 25.
2. Write two equations. Our message MEET AT THE LAB, encrypted with our key a = 5 and b = 8, begins QC. So M (12) became Q (16), and E (4) became C (2). That gives 12a + b = 16 and 4a + b = 2, both mod 26.
3. Subtract to remove b. 8a = 14, mod 26. Dividing everything by 2 gives 4a = 7, mod 13.
4. Solve mod 13. The inverse of 4 mod 13 is 10, because 4 times 10 is 40, which is 1 more than 39. So a = 7 times 10 = 70, which is 5 mod 13. Back in mod 26 that allows a = 5 or a = 18.
5. Reject the impossible key. 18 shares a factor of 2 with 26, so it has no inverse and can never be an affine key. Only a = 5 is left, and b = 16 minus 60, which is 8 mod 26.
6. Decrypt. The inverse of 5 mod 26 is 21. Each cipher letter y becomes 21 times (y minus 8), mod 26. Q (16) gives 21 times 8, which is 168, and 168 mod 26 is 12, which is M.

# Our practice tool for affine ciphers: encrypt, decrypt and crack from a crib.
# For practice at home only: tournament rules bar calculator apps on laptops and phones.
from math import gcd

ABC = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

def encrypt(text, a, b):
    return "".join(ABC[(a * ABC.index(c) + b) % 26] if c in ABC else c for c in text)

def inverse(a):
    # the number that undoes multiplying by a, working mod 26
    return next(x for x in range(26) if a * x % 26 == 1)

def decrypt(text, a, b):
    ai = inverse(a)
    return "".join(ABC[ai * (ABC.index(c) - b) % 26] if c in ABC else c for c in text)

def keys_from_crib(p1, c1, p2, c2):
    """Every legal key that sends plain letter p1 to c1 and p2 to c2."""
    found = []
    for a in range(26):
        if gcd(a, 26) != 1:
            continue          # no inverse, so not a legal key
        b = (ABC.index(c1) - a * ABC.index(p1)) % 26
        if (a * ABC.index(p2) + b) % 26 == ABC.index(c2):
            found.append((a, b))
    return found

secret = encrypt("MEET AT THE LAB", 5, 8)
print(secret)                              # QCCZ IZ ZRC LIN
print(keys_from_crib("M", "Q", "E", "C"))  # [(5, 8)]
print(decrypt(secret, 5, 8))               # MEET AT THE LAB

1. Encrypt one letter. With a = 3 and b = 5, what does H become? Answer: H is 7, and 3 times 7 plus 5 is 26, which is 0 mod 26, so H becomes A.
2. Find an inverse. What is the inverse of 7 mod 26? Answer: 15, because 7 times 15 is 105, and 105 is 4 times 26 plus 1.
3. Spot a bad key. Why can a = 13 never be an affine key? Answer: 13 shares a factor with 26, so 13 times any even letter number is 0 mod 26 and many letters collide. With no inverse, decryption is impossible.
4. A special case. Show that a = 25, b = 25 sends A to Z and Z to A. Answer: A is 0, and 25 times 0 plus 25 is 25, which is Z. Z is 25, and 25 times 25 plus 25 is 650, which is exactly 25 times 26, so 0, which is A. That is AtBash, as the overview says.
5. Count the work. A brute-force program tries every legal affine key on a 40-letter message. How many letter decryptions is that at most? Answer: 312 keys times 40 letters, or 12,480, a fraction of a second for Python and an afternoon for a pencil.