Sunday Times Teaser 3011 – Optical Illusion

by Danny Roth

Published Sunday June 07 2020 (link)

George and Martha are studying optics. If you place an object a specific distance from a lens, an image will appear at a distance from that lens according the following formula:

The reciprocal of the object distance plus the reciprocal of the image distance is equal to the reciprocal of the focal length of the lens.

The object distance was a two-digit whole number of cm (ab). The image distance and the focal length of the lens were also two-digit whole numbers (cd and ef respectively). The six digits were all different and non-zero. Also, the object distance and the focal length were of the same parity and b was an exact multiple of d. Martha pointed out that the sum of those three two-digit numbers was a prime number.

What was that prime number?

4 Comments Leave one →

Brian Gladman permalink


from itertools import combinations, permutations

# small primes 
prms = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

# the tens digits of the object distance and the focal length
for e, a in combinations(range(1, 10), 2):

  s8 = set(range(1, 10)).difference([e, a])
  # their units digits
  for b, f in permutations(s8, 2):
    
    # compute the image distance
    ab, ef = 10 * a + b, 10 * e + f
    cd, r = divmod(ab * ef, ab - ef)
    
    # apply the constraints
    if not r and b & 1 == ef & 1 and b % (cd % 10) == 0:

      # the sum of the three lengths is a prime
      pr = ab + cd + ef
      if all(pr % x for x in prms):
        print(f"The prime is {pr} (ab = {ab}, cd = {cd}, ef = {ef}).")

from itertools import combinations, permutations

# small primes

prms = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

# the tens digits of the object distance and the focal length

for e, a in combinations(range(1, 10), 2):

s8 = set(range(1, 10)).difference([e, a])

# their units digits

for b, f in permutations(s8, 2):

# compute the image distance

ab, ef = 10 * a + b, 10 * e + f

cd, r = divmod(ab * ef, ab - ef)

# apply the constraints

if not r and b & 1 == ef & 1 and b % (cd % 10) == 0:

# the sum of the three lengths is a prime

pr = ab + cd + ef

if all(pr % x for x in prms):

print(f"The prime is {pr} (ab = {ab}, cd = {cd}, ef = {ef}).")

GeoffR permalink


from number_theory import is_prime
from itertools import permutations

for p1 in permutations('123456789', 6):
  a, b, c, d, e, f = p1

  # check b is a multiple of d
  if int(b) % int(d) == 0:
    ab, cd, ef = int(a+b), int(c+d), int(e+f)
    # check parity is same for ab and ef
    if ab % 2 != ef % 2:
        continue
    
    # sum of the three two-digit numbers is a prime number
    if is_prime(ab + cd + ef):
      # optical formula: 1/ab + 1/cd = 1/ef (in integers)
      if (cd * ef) + (ab * ef) == (ab * cd):
        print(f"Prime number was {ab+cd+ef}")
        print(f"Three two-digit numbers were {ab},{cd},{ef}")

from number_theory import is_prime

from itertools import permutations

for p1 in permutations('123456789', 6):

a, b, c, d, e, f = p1

# check b is a multiple of d

if int(b) % int(d) == 0:

ab, cd, ef = int(a+b), int(c+d), int(e+f)

# check parity is same for ab and ef

if ab % 2 != ef % 2:

continue

# sum of the three two-digit numbers is a prime number

if is_prime(ab + cd + ef):

# optical formula: 1/ab + 1/cd = 1/ef (in integers)

if (cd * ef) + (ab * ef) == (ab * cd):

print(f"Prime number was {ab+cd+ef}")

print(f"Three two-digit numbers were {ab},{cd},{ef}")

Tony Smith permalink

The lens equation 1/U+1/V=1/F has general solution in positive integers U=Am(m+n), V=An(m+n), F=Amn where A,m,n are integers and hcf (m,n)=1.

Since U+V+F is prime A=1 and since U and F have the same parity, m is even and n is odd.

A very small number of values of (m,n) can give 2-digit numbers with non-zero digits and a few lines of analysis and a small amount of calculation (or, I assume, a very simple program) give the solution and show its uniqueness.

U+V is square if and only if A is square.

It is possible to find the solution without using the primality and parity tests but it would need a lot of calculation (or, I assume, a more complex program) to show uniqueness.

Brian Gladman permalink

Thanks for your input Tony. As you suggest, this offers another path to the solution and one that is easy to code in Python:


# small primes 
prms = {2, 3, 5, 7, 11, 13, 17}

# The solution of 1 / U + 1 / V = 1 / W in integers is 
# given by:
#
#    U = k.m.(m + n), V = k.n.(m + n), W = k.m.n
#
# Since U + V + W is prime k = 1; Since U and W have
# the same parity m is even and n is odd

for m in range(2, 10, 2):
  for n in range(m + 1, 99 // m + 1, 2):
    
    u, v, w = m * (m + n), n * (m + n), m * n
    
    if not all(10 <= x < 100 for x in (u, v, w)):
      continue
    
    (a, b), (c, d), (e, f) = divmod(u, 10), divmod(v, 10), divmod(w, 10)
    
    if len({a, b, c, d, e, f}) == 6 and b % d == 0:

      sm = u + v + w
      if all(sm % p for p in prms):
        print(f"The prime is {sm} (ab = {u}, cd = {v}, ef = {w}).")

# small primes

prms = {2, 3, 5, 7, 11, 13, 17}

# The solution of 1 / U + 1 / V = 1 / W in integers is

# given by:

# U = k.m.(m + n), V = k.n.(m + n), W = k.m.n

# Since U + V + W is prime k = 1; Since U and W have

# the same parity m is even and n is odd

for m in range(2, 10, 2):

for n in range(m + 1, 99 // m + 1, 2):

u, v, w = m * (m + n), n * (m + n), m * n

if not all(10 <= x < 100 for x in (u, v, w)):

continue

(a, b), (c, d), (e, f) = divmod(u, 10), divmod(v, 10), divmod(w, 10)

if len({a, b, c, d, e, f}) == 6 and b % d == 0:

sm = u + v + w

if all(sm % p for p in prms):

print(f"The prime is {sm} (ab = {u}, cd = {v}, ef = {w}).")

Sunday Times Teaser 3011 – Optical Illusion

by Danny Roth

Published Sunday June 07 2020 (link)

Leave a comment to GeoffR Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta