Have You Heard the Buzz?

You count aloud, starting from $1$ and working upward through the whole numbers. For any number that is a multiple of $7$, or that contains the digit $7$, you say “buzz” instead of the number. So the game begins $$1,2,3,4,5,6,\text{buzz},8,9,\dots,13,\text{buzz},15,16,\text{buzz},18,19,20,\dots$$ where “buzz” has replaced $7$, $14$, and $17$. How many times do you say “buzz” in the first $100$ turns?

The Fiddler, Zach Wissner-Gross, April 17, 2026(original post)

Solution

A number is buzzed when it is divisible by $7$ or contains the digit $7$, so count each set and remove the double-count. Among $1,\dots,100$: multiples of $7$ number $\lfloor 100/7\rfloor = 14$; numbers containing a $7$ number $10+10-1=19$ (units-digit $7$, plus the seventies, with $77$ shared); numbers in both ($7,70,77$) number $3$. By inclusion–exclusion, $$14 + 19 - 3 = \boxed{30}.$$

The computation

Apply the rule itself to each of the first hundred numbers, buzzing on a multiple of $7$ or a digit $7$, and count.

buzz = lambda n: n % 7 == 0 or '7' in str(n)
print(sum(buzz(n) for n in range(1, 101)))  # 30

Extra Credit

In the first $20$ turns only $15\%$ of the numbers were buzzed, but the buzzed share climbs as the count goes on. Find the smallest $N$ such that, from the $N$th turn onward, at least half of the numbers counted so far have been buzzed.

Solution

The pressure comes from the digit rule. Among the $d$-digit numbers, the fraction containing no $7$ is $(9/10)^d$, so the fraction containing a $7$ is $1-(9/10)^d$. This first passes $\tfrac12$ at $d=7$, since $(9/10)^6 = 0.5314$ but $(9/10)^7 = 0.4783$. Multiples of $7$ add a steady $\tfrac17$ on top, so the running buzz fraction climbs through $\tfrac12$ partway into the six-digit numbers. Tracking the cumulative count, it falls just short of half for the last time at $n = 708{,}587$, so from $$N = \boxed{708{,}588}$$ onward at least half the numbers so far have always been buzzed. (The source’s value is behind its paywall; this is my own.)

The computation

Count up through the millions, tracking the cumulative buzzes, and record the last turn at which fewer than half the numbers so far were buzzed. The answer is one past it.

cum, last = 0, 0
for n in range(1, 5_000_001):
    if n % 7 == 0 or '7' in str(n): cum += 1
    if 2 * cum < n: last = n        # fewer than half buzzed so far
print(last + 1)                     # 708588 (no relapse out to 5,000,000)