Can You Take the Heat?

On the show Hot Ones you face $10$ hot sauces ranked $1$ to $10$. Rather than buy all ten, you buy a few sauce numbers and combine sauces (adding their numbers, each used at most once) to hit any value you are missing. With the set $\{1,2,3,4\}$ you can already make every value from $1$ to $10$. Counting that set, how many sets of four sauce numbers let you generate all of $1$ through $10$?

The Fiddler, Zach Wissner-Gross, November 28, 2025(original post)

Solution

A set works when every target $1,\dots,10$ is a subset sum. To make $1$ you need sauce $1$; to make $2$ you then need sauce $2$ (nothing else is small enough). Sorting the four numbers $1<2<a<b$, the subset sums fill $1,2,\dots$ contiguously as long as each new number is at most one more than the running total, and the four must total at least $10$. So $a\le 1+(1+2)=4$, giving $a\in\{3,4\}$, and $b\le 1+(1+2+a)=4+a$ with $b\ge 10-(3+a)$. Counting: $$a=3:\ b\in\{4,5,6,7\};\qquad a=4:\ b\in\{5,6,7,8\}.$$ That is $4+4 = \boxed{8}$ sets.

The computation

Test every four-sauce set directly: form all its subset sums and check that $1$ through $10$ are all reachable. Eight sets pass.

import itertools as it
def covers(S, top):
    sums = {0}
    for x in S: sums |= {s + x for s in sums}
    return all(t in sums for t in range(1, top + 1))
print(sum(covers(S, 10) for S in it.combinations(range(1, 11), 4)))   # 8

Extra Credit

For “Hotter Ones,” the sauces run $1$ to $100$. Let $N$ be the fewest sauces needed to generate every value from $1$ to $100$. For how many sets of $N$ sauce numbers is it possible?

Solution

With $N$ sauces there are only $2^N-1$ nonempty subset sums, so covering $100$ values needs $2^N-1\ge100$, forcing $N\ge7$; and $\{1,2,4,8,16,32,64\}$ shows $N=7$ suffices. Counting all $7$-element subsets of $\{1,\dots,100\}$ whose subset sums cover $1$ to $100$ (each must start at $1$, grow contiguously until the running total reaches $100$, after which the remaining sauces are free) gives $$N = 7,\qquad \#\text{ sets} = \boxed{1014}.$$ (The source’s count is behind its paywall; this is my own enumeration.)

The computation

Count valid $7$-sets by the contiguous-coverage rule: choose sauces in increasing order, each at most one past the running total (so subset sums stay gap-free), until coverage reaches $100$, after which any larger sauces complete the set.

from math import comb
from functools import lru_cache
N = 7
@lru_cache(None)
def cnt(k, last, reach):           # k chosen, last value used, coverage [0, reach]
    if k == N: return 1 if reach >= 100 else 0
    if reach >= 100: return comb(100 - last, N - k)     # remaining sauces are free
    return sum(cnt(k + 1, v, reach + v) for v in range(last + 1, min(100, reach + 1) + 1))
print(cnt(1, 1, 1))                # 1014   (every set must start with sauce 1)