Sunday Times Teaser 3095 – Diddums!

by BRG on January 14, 2022

by Stephen Hogg

Published Sunday January 16 2022 (link)

“Diddums” was Didi’s dada’s name for numbers where the number of positive DIvisors (including 1 and the number), the number of Digits, and the Digit sUM are equal.

Didi enjoyed mental arithmetic, doing it speedily and accurately. For numbers of digits from 1 to 9 she listed in ascending order all possible numbers whose digit sum equalled the number of digits. Working through her list in order, she quickly found the first two diddums (1 and 11) and the next three diddums. After many further calculations, Didi confirmed that the sixth diddum (which is even) was listed.

“Now I’m bored,” she moaned. “Diddums!” said Didi’s dada.

What is the sixth diddum’s digit sum?

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Brian Gladman permalink

This runs in under 40 milliseconds.


from number_theory import tau

# generate <nd>-digit integers with digit sums of <nd>
def gen(nd, dsum, nbr=0):  
  if not nd:
    if not dsum:
      yield nbr
  else:
    for d in range(0 if nbr else 1, min(10, dsum + 1)):
      yield from gen(nd - 1, dsum - d, 10 * nbr + d)

nd, cnt = 0, 0
while True:
  nd += 1
  # consider integers with both digit sums and ...
  for nbr in gen(nd, nd):
    # ... numbers of divisors equal to their length
    if tau(nbr) == nd:      
      cnt += 1
      sfx = f"(Answer = {nd})" if cnt == 6 else ""
      print(f"D[{cnt}] = {nbr} {sfx}")
      if cnt == 6:
        exit()

from number_theory import tau

# generate <nd>-digit integers with digit sums of <nd>

def gen(nd, dsum, nbr=0):

if not nd:

if not dsum:

yield nbr

else:

for d in range(0 if nbr else 1, min(10, dsum + 1)):

yield from gen(nd - 1, dsum - d, 10 * nbr + d)

nd, cnt = 0, 0

while True:

nd += 1

# consider integers with both digit sums and ...

for nbr in gen(nd, nd):

# ... numbers of divisors equal to their length

if tau(nbr) == nd:

cnt += 1

sfx = f"(Answer = {nd})" if cnt == 6 else ""

print(f"D[{cnt}] = {nbr} {sfx}")

if cnt == 6:

exit()

Here is a variation that allows “diddums” with a range of lengths to be output:


from sys import argv
from number_theory import factor, tau

# generate <nd>-digit integers with digit sums of <nd>
def gen1(nd, dsum, nbr=0):  
  if not nd:
    if not dsum:
      yield nbr
  else:
    for d in range(0 if nbr else 1, min(10, dsum + 1)):
      yield from gen1(nd - 1, dsum - d, 10 * nbr + d)

# generate integers of length <nd> with both digit sums
# and numbers of divisors equal to their length
gen2 = lambda nd: (nbr for nbr in gen1(nd, nd) if tau(nbr) == nd)

# allow 'diddums' to be output if there are input parameters
# giving the minimum and maximum number of digits to output
min_nd, max_nd, cnt_max = 1, 8, 6
if (ln:= len(argv)) >= 2:
  cnt_max = 0
  min_nd, *r = (int(x) for x in argv[1:])
  max_nd = min_nd if not r else r[0]

cnt = 0
for nd in range(min_nd, max_nd + 1):
  for nbr in gen2(nd):
    cnt += 1
    sfx = f" -> Answer = {nd}" if cnt == cnt_max else ""
    fct = '.'.join(f"{p}" if e == 1 else f"{p}^{e}" for p, e in factor(nbr))
    print(f"D[{cnt:3}] = {nbr} = {fct}  {sfx}")
    if cnt_max and cnt == cnt_max:
      break
  else:
    continue
  break

from sys import argv

from number_theory import factor, tau

# generate <nd>-digit integers with digit sums of <nd>

def gen1(nd, dsum, nbr=0):

if not nd:

if not dsum:

yield nbr

else:

for d in range(0 if nbr else 1, min(10, dsum + 1)):

yield from gen1(nd - 1, dsum - d, 10 * nbr + d)

# generate integers of length <nd> with both digit sums

# and numbers of divisors equal to their length

gen2 = lambda nd: (nbr for nbr in gen1(nd, nd) if tau(nbr) == nd)

# allow 'diddums' to be output if there are input parameters

# giving the minimum and maximum number of digits to output

min_nd, max_nd, cnt_max = 1, 8, 6

if (ln:= len(argv)) >= 2:

cnt_max = 0

min_nd, *r = (int(x) for x in argv[1:])

max_nd = min_nd if not r else r[0]

cnt = 0

for nd in range(min_nd, max_nd + 1):

for nbr in gen2(nd):

cnt += 1

sfx = f" -> Answer = {nd}" if cnt == cnt_max else ""

fct = '.'.join(f"{p}" if e == 1 else f"{p}^{e}" for p, e in factor(nbr))

print(f"D[{cnt:3}] = {nbr} = {fct} {sfx}")

if cnt_max and cnt == cnt_max:

break

else:

continue

break

Sunday Times Teaser 3095 – Diddums!

by Stephen Hogg

Published Sunday January 16 2022 (link)

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