Teaser 3251: Number test
From The Sunday Times, 12th January 2025 [link] [link]
I asked my daughter to find as many three-digit numbers as she could, each of which had the property that the number equalled the sum of the cubes of its individual digits. If she only found one such number, I asked her to give me this number — otherwise, if she had found more than one, to give me the sum of all such numbers found.
She gave me a prime number, from which I could see that not all answers had been found.
Which number or numbers did she not find?
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Jim Randell 7:07 am on 12 January 2025 Permalink |
It is straightforward to test all 3-digit numbers for the required property, and then look for (proper) subsets of the numbers found, to find a subset with a prime sum.
This Python program runs in 65ms. (Internal runtime is 1.1ms).
from enigma import (irange, nsplit, subsets, is_prime, diff, printf) # find 3-digit numbers that are the sum of the cube of their digits cb = list(d**3 for d in irange(0, 9)) ns = list(n for n in irange(100, 999) if n == sum(cb[d] for d in nsplit(n))) printf("numbers = {ns}") # find (proper) subsets that sum to a prime for ss in subsets(ns, min_size=1, max_size=len(ns) - 1): t = sum(ss) if not is_prime(t): continue # output solution missing = diff(ns, ss) printf("sum{ss} = {t}; missing = {missing}")Solution: The numbers not found are: 371, 407.
There are only four such numbers (if leading zeros are disallowed):
And the only strict subset of these with a prime sum is:
If all four numbers had been found, the sum would also have been prime:
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Frits 2:56 pm on 12 January 2025 Permalink |
from itertools import combinations def is_prime(n): if n < 4: return n > 1 if n % 2 == 0 or n % 3 == 0: return False # all primes > 3 are of the form 6n +/- 1 # start with f=5 (which is prime) and test f, f+2 for being prime f, r = 5, int(n**0.5) while f <= r: if n % f == 0: return False if n % (f + 2) == 0: return False f += 6 return True nums = [] for a in range(1, 10): a3 = a * a * a for b in range(10): a3b3 = a3 + b * b * b for c in range(10): t = a3b3 + c * c * c if t // 100 != a: continue if t % 10 != c: continue if (t % 100) // 10 != b: continue nums += [t] sum_nums = sum(nums) print("answer(s):") for k in range(1, len(nums)): # pick <k> numbers from <ns> she has missed for s in combinations(nums, k): # she gave me a prime number if is_prime(sum_nums - sum(s)): print(s)LikeLike
ruudvanderham 12:09 pm on 13 January 2025 Permalink |
Rather straightforward:
import istr istr.repr_mode("str") solutions = {i for i in istr(range(100, 1000)) if sum(digit**3 for digit in i) == i} print(*list(f"missing: {solutions - set(x)}" for r in range(1, len(solutions)) for x in istr.combinations(solutions, r=r) if sum(x).is_prime()))LikeLike