Posted on Apr 17, 2026
I recently introduced my children to a game called "Buzz" (also known as "Fizz buzz"), in which players take turns reciting whole numbers in order. However, in one particular variant of the game, anytime a number is a multiple of 7 or at least one of its digits is a 7, the player must say "buzz" instead of that number.
For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.
How many times should "buzz" be said in the first 100 turns of the game?
Count the numbers from 1 through 100 that satisfy at least one of the two buzz rules.
| Condition | Count |
|---|---|
| Multiples of 7 | 14 |
| Contains the digit 7 | 19 |
| Counted twice | 3 |
The multiples of 7 are
\[ 7,14,21,28,35,42,49,56,63,70,77,84,91,98. \]
The numbers from 1 through 100 containing a 7 are
\[ 7,17,27,37,47,57,67,70,71,72,73,74,75,76,77,78,79,87,97. \]
The overlap is just \(7, 70,\) and \(77\), so by inclusion-exclusion the total number of buzzes is
\[ 14 + 19 - 3 = 30. \]
Answer: \( \boxed{30} \).
Enter any turn count \(N\) to see how many buzzes have occurred by then, along with the running percentage.
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As we just saw, in the first 20 turns of the game, 15 percent of the numbers were buzzed. But as the game proceeds, an increasing frequency of numbers get buzzed.
There is a certain minimum number \(N\) such that, for the \(N\)th turn in the game and for every turn thereafter, at least half the numbers up to that point have been buzzed. What is this value of \(N\)?
Let \(B(n)\) be the number of buzzes among the first \(n\) turns, and define
\[ D(n)=2B(n)-n. \]
Then "at least half the first \(n\) turns are buzzes" is exactly the condition \(D(n)\ge 0\).
So we want the smallest value \(N\) such that \(D(n)\ge 0\) for every \(n\ge N\).
The right way to organize the count is in blocks of 100000 consecutive numbers. Write
\[ I_q=\{100000q,100000q+1,\dots,100000q+99999\}. \]
If the prefix \(q\) contains a 7, then every number in \(I_q\) contains a 7, so every turn in that whole block is a buzz.
Now suppose \(q\) does not contain a 7. For a number \(n=100000q+r\) with \(0\le r<100000\), the only ways \(n\) can be buzzed are:
Since \(100000\equiv 5 \pmod 7\), the divisibility condition becomes
\[ r\equiv -5q \pmod 7. \]
So for a block whose prefix \(q\) has no 7, the entire buzz/non-buzz pattern depends only on \(q\bmod 7\). That means there are only 7 possible block types to check.
A finite check of those 7 cases gives two facts:
Now look at the first seven blocks, from 0 up to 699999. Their prefixes are \(0,1,2,3,4,5,6\), none of which contain a 7, so they are exactly one copy each of the 7 block types above. Counting buzzes in these seven blocks gives a total of 345706 buzzes from 0 through 699999, so from 1 through 699999 there are
\[ B(699999)=345705 \]
buzzes. Therefore
\[ D(699999)=2\cdot 345705-699999=-8589. \]
Next comes the block from 700000 through 799999. Every one of these numbers contains a 7, so every one of these 100000 turns is a buzz. That means \(D(n)\) increases by 1 at every step throughout this block.
Starting from \(D(699999)=-8589\), the running total first reaches 0 exactly 8589 steps later, at
\[ n=708588. \]
Indeed, at the previous turn we still have
\[ B(708587)=354293 \qquad\text{and}\qquad D(708587)=2\cdot 354293-708587=-1. \]
while at turn 708588 we have
\[ B(708588)=354294 \qquad\text{and}\qquad D(708588)=2\cdot 354294-708588=0. \]
So \(N\) cannot be smaller than 708588. It remains to show that once we reach that point, \(D(n)\) never becomes negative again.
By the end of the all-buzz block we are far above zero:
\[ D(799999)=91411, \qquad D(800000)=91410. \]
From then on, any block whose prefix contains a 7 is again all buzzes, so it only pushes \(D(n)\) upward.
For a block whose prefix does not contain a 7, we already know two things: the whole block decreases \(D\) by at most 1228, and even inside the block the temporary dip below the block's starting value is at most 9257.
Among any 10 consecutive block prefixes, one ends in 7, so there can never be more than 9 consecutive blocks whose prefixes avoid the digit 7. Therefore, starting at any block boundary, the worst possible drop before the next all-buzz block arrives is at most
\[ 9\cdot 1228 + 9257 = 20309. \]
But at \(n=800000\) we already have \(D(800000)=91410\), which is much larger than 20309. So from that point onward, even the worst possible stretch of non-all-buzz blocks cannot drive \(D(n)\) below 0.
Since \(D(708587)=-1\) but \(D(n)\ge 0\) for every \(n\ge 708588\), the minimum value is
Extra credit answer: \( \boxed{708588} \).