Sunday Times Teaser 2681 – Inconsequential

by Victor Bryant

A sequence of integers in which the difference between consecutive terms is constant is known as an arithmetic sequence. For example, 10, 29, 48, … is an arithmetic sequence with a ninth term (162) that is divisible by 9.

I have formed another arithmetic sequence that starts with two three digit numbers and has a ninth term that is also divisible by 9. When I consistently replace different digits with different letters, this sequence starts ONE, TWO, … and its ninth term is NINE.

What nunber does WIN represent?

4 Comments Leave one →

brian gladman permalink


from itertools import permutations

for I, E in permutations(range(10), 2):
  # NINE % 9 = 0 ==> 2.N + I + E == 0 % 9 ==>
  N = (4 * (I + E)) % 9
  set_ien = set((I, E, N))
  if N and len(set_ien) == 3:
    for O in set(range(10)) - set_ien:
      # compute the sequence values
      ONE, NINE = 100 * O + 10 * N + E, 1010 * N + 100 * I + E
      TWO, rem = divmod(NINE + 7 * ONE, 8)
      # TWO must be a three digit integer ending in O
      if rem == 0 and TWO < 1000 and TWO % 10 == O:
        T, W = divmod(TWO // 10, 10)
        # check that we have six different digits
        if len(set((O, T, W)) | set_ien) == 6:
          fs = 'WIN = {} (ONE = {}, TWO = {}, NINE = {})'
          print(fs.format(100 * W + 10 * I + N, ONE, TWO, NINE))

from itertools import permutations

for I, E in permutations(range(10), 2):

# NINE % 9 = 0 ==> 2.N + I + E == 0 % 9 ==>

N = (4 * (I + E)) % 9

set_ien = set((I, E, N))

if N and len(set_ien) == 3:

for O in set(range(10)) - set_ien:

# compute the sequence values

ONE, NINE = 100 * O + 10 * N + E, 1010 * N + 100 * I + E

TWO, rem = divmod(NINE + 7 * ONE, 8)

# TWO must be a three digit integer ending in O

if rem == 0 and TWO < 1000 and TWO % 10 == O:

T, W = divmod(TWO // 10, 10)

# check that we have six different digits

if len(set((O, T, W)) | set_ien) == 6:

fs = 'WIN = {} (ONE = {}, TWO = {}, NINE = {})'

print(fs.format(100 * W + 10 * I + N, ONE, TWO, NINE))

ahmet cetinbudaklar permalink

Set ONE = abc, TWO = dea = abc + k and NINE = bfbc = abc + 8k where a .. f are the six
different digits involved and k is the difference between terms. We then have:

NINE = 1000b + 100f + 10b + c = ONE + 8k = 100a + 10b + c + 8k

giving k = 125b + (25/2)(f – a) with f – a even. Hence we have:

b = 1 -> k = 125 +/- (50,100)
b = 2 -> k= 250 +/- (25, 75)
b = 3 -> k = 375 +/- (50,100)

From here we can reach the values of b, k, ONE, TWO, NINE and hence WIN by trial and error.

Reply

geoffrounce permalink

Here is a permutation solution:


from itertools import permutations

for p in permutations (range(10),6):
    o,n,e,t,w,i = p
    if o !=0 and t != 0:
        one = e + n*10 + o*100
        two = o + w*10 + t*100
        if two > one:
            diff = two - one
            nine = e + n*10 + i*100 + n*1000
            if nine % 9 == 0 and one + diff*8 == nine:
                win = n + 10*i + 100*w
                print('WIN = ',win)
                print('ONE, TWO, NINE = ',one,two,nine)

from itertools import permutations

for p in permutations (range(10),6):

o,n,e,t,w,i = p

if o !=0 and t != 0:

one = e + n*10 + o*100

two = o + w*10 + t*100

if two > one:

diff = two - one

nine = e + n*10 + i*100 + n*1000

if nine % 9 == 0 and one + diff*8 == nine:

win = n + 10*i + 100*w

print('WIN = ',win)

print('ONE, TWO, NINE = ',one,two,nine)

Jim Randell permalink

This solution uses some handy routines from my enigma.py library.


from itertools import permutations
from enigma import irange, nconcat, split, printf

# possible digits
ds0 = set(irange(0, 9))

# choose O and T such that T > O > 0
for O in irange(1, 8):
  for T in irange(O + 1, 9):
    # allocate the remaining digits
    for (N, E, W) in permutations(ds0.difference((O, T)), 3):
      ONE = nconcat(O, N, E)
      TWO = nconcat(T, W, O)
      # compute NINE
      NINE = 8 * TWO - 7 * ONE
      # NINE must be divisible by 9
      if NINE % 9 > 0: continue
      # and be 4 digit
      if not(999 < NINE < 10000): continue
      # check the letters match
      (N1, I, N2, E1) = split(NINE, int)
      if not(N1 == N2 == N and E1 == E): continue
      # and are all different
      if I in (O, T, N, E, W): continue
  
      printf("WIN={W}{I}{N} [ONE={ONE} TWO={TWO} NINE={NINE}]")

from itertools import permutations

from enigma import irange, nconcat, split, printf

# possible digits

ds0 = set(irange(0, 9))

# choose O and T such that T > O > 0

for O in irange(1, 8):

for T in irange(O + 1, 9):

# allocate the remaining digits

for (N, E, W) in permutations(ds0.difference((O, T)), 3):

ONE = nconcat(O, N, E)

TWO = nconcat(T, W, O)

# compute NINE

NINE = 8 * TWO - 7 * ONE

# NINE must be divisible by 9

if NINE % 9 > 0: continue

# and be 4 digit

if not(999 < NINE < 10000): continue

# check the letters match

(N1, I, N2, E1) = split(NINE, int)

if not(N1 == N2 == N and E1 == E): continue

# and are all different

if I in (O, T, N, E, W): continue

printf("WIN={W}{I}{N} [ONE={ONE} TWO={TWO} NINE={NINE}]")

Sunday Times Teaser 2681 – Inconsequential

by Victor Bryant

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