Beaver Math Olympiad - Revision history

Mxdys: BMO9 was solved

2026-09-24T14:35:54Z

BMO9 was solved

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 for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_n = \left\lfloor \frac{b_n}{2} \right\rfloor + 1</math>?
 === 10. {{TM|1RB0LB_0RC0LD_1LD1RE_0LA0RF_0RD1RF_0RC---|undecided}} ===
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 Link to Discord discussion: https://discord.com/channels/960643023006490684/1421782442213376000/1431483206208852001
 == Practice Problems ==

C1: Improved the consistency of the content.

2026-08-04T00:29:29Z

Improved the consistency of the content.

← Older revision		Revision as of 00:29, 4 August 2026
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	occurrence may be chosen arbitrarily; there is no prescribed reduction order.		occurrence may be chosen arbitrarily; there is no prescribed reduction order.

	At any stage, call a symbol <math>\#</math> active if it is the leftmost symbol of an occurrence of one of the ~~four~~ left-hand sides above. Equivalently, a <math>\#</math> is		At any stage, call a symbol <math>\#</math> active if it is the leftmost symbol of an occurrence of one of the five left-hand sides above. Equivalently, a <math>\#</math> is
	active if the word beginning at that <math>\#</math> starts with one of		active if the word beginning at that <math>\#</math> starts with one of

Mxdys: add BMO11

2026-08-03T16:41:01Z

add BMO11

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 :<math>
 \#0,\qquad \#100,\qquad \#110,\qquad \#1010.
 </math>

Mxdys: add BMO10

2026-08-01T09:27:05Z

add BMO10

← Older revision		Revision as of 09:27, 1 August 2026
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	where <math>\mathbf x^{(n+1)}=T(\mathbf x^{(n)})</math> whenever the right-hand side is not <math>\mathrm{Halt}</math>.		where <math>\mathbf x^{(n+1)}=T(\mathbf x^{(n)})</math> whenever the right-hand side is not <math>\mathrm{Halt}</math>.

			=== 10. {{TM\|1RB0LB_0RC0LD_1LD1RE_0LA0RF_0RD1RF_0RC---\|undecided}} ===

			Let <math>\Sigma=\{0,1,\#\}</math>. Consider two-sided infinite words over <math>\Sigma</math>. The initial word is

			:<math>
			w^{(0)}=\cdots 10\#10\#10\#10\#0000\cdots,
			</math>

			that is, a left-infinite repetition of <math>10\#</math>, followed by a right-infinite tail of <math>0</math>'s.

			The rewrite rules are

			:<math>
			\#0 \to 1,
			</math>

			:<math>
			\#100 \to 000\#,
			</math>

			:<math>
			\#110 \to 000\#,
			</math>

			:<math>
			\#1010 \to 1010\#\#.
			</math>

			A rewrite step consists of choosing any occurrence of the left-hand side of one of these rules in the current word and replacing it by the corresponding right-hand side. The
			occurrence may be chosen arbitrarily; there is no prescribed reduction order.

			At any stage, call a symbol <math>\#</math> active if it is the leftmost symbol of an occurrence of one of the four left-hand sides above. Equivalently, a <math>\#</math> is
			active if the word beginning at that <math>\#</math> starts with one of

			:<math>
			\#0,\qquad \#100,\qquad \#110,\qquad \#1010.
			</math>

			The question is whether there exists some finite sequence of rewrite steps, starting from <math>w^{(0)}</math>, after which the rightmost occurrence of <math>\#</math> is not
			active. In other words, does there exist a rewrite order such that the rightmost <math>\#</math> can no longer be rewritten?

	== Solved problems ==		== Solved problems ==

Mxdys: add BMO9

2026-08-01T08:56:39Z

add BMO9

← Older revision		Revision as of 08:56, 1 August 2026
Line 55:		Line 55:

	for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_n = \left\lfloor \frac{b_n}{2} \right\rfloor + 1</math>?		for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_n = \left\lfloor \frac{b_n}{2} \right\rfloor + 1</math>?

			=== 9. {{TM\|1RB1LA_1RC0RD_1LA---_1RE1RD_1LF0LA_---0LE\|undecided}} ===

			Let <math>\mathbb{N}=\{0,1,2,\ldots\}</math>. Consider configurations given by infinite sequences of non-negative integers,

			:<math>
			\mathbb{N}^{\omega}
			=
			\{(x_0,x_1,x_2,\ldots): x_i\in\mathbb{N}\}.
			</math>

			Define a partial transition map

			:<math>
			T:\mathbb{N}^{\omega}\to \mathbb{N}^{\omega}\cup\{\mathrm{Halt}\}
			</math>

			by the following rewrite rules. For all <math>a,b,c\in\mathbb{N}</math> and every infinite tail sequence <math>\mathbf r\in\mathbb{N}^{\omega}</math>,

			:<math>
			(0,a,c,\mathbf r)\mapsto (3+a+c,\mathbf r),
			</math>

			:<math>
			(1,0,\mathbf r)\mapsto \mathrm{Halt},
			</math>

			:<math>
			(1,1+a,c,\mathbf r)\mapsto (a,0,1,1+c,\mathbf r),
			</math>

			:<math>
			(2+a,b,c,\mathbf r)\mapsto (a,1+b,1+c,\mathbf r).
			</math>

			The initial configuration is

			:<math>
			\mathbf x^{(0)}=(0,0,0,0,\ldots).
			</math>

			The question is whether the orbit of <math>\mathbf x^{(0)}</math> under repeated application of <math>T</math> eventually halts; that is, whether there exists some
			<math>n\in\mathbb{N}</math> such that

			:<math>
			T(\mathbf x^{(n)})=\mathrm{Halt},
			</math>

			where <math>\mathbf x^{(n+1)}=T(\mathbf x^{(n)})</math> whenever the right-hand side is not <math>\mathrm{Halt}</math>.

	== Solved problems ==		== Solved problems ==

Notxxdog: Removed an extra ")" in an equation

2026-05-21T13:30:07Z

Removed an extra ")" in an equation

← Older revision		Revision as of 13:30, 21 May 2026
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	Let <math>f(n) = n+1+(v_2(n+1) \bmod 2)</math>.		Let <math>f(n) = n+1+(v_2(n+1) \bmod 2)</math>.

	Now consider the iterated application of the function <math>f^{n+1}(b) = f(f^n(b)))</math>, <math>f^0(b)=b</math>.		Now consider the iterated application of the function <math>f^{n+1}(b) = f(f^n(b))</math>, <math>f^0(b)=b</math>.

	Let <math>(a_n)_{n \ge 0}</math> be a sequence such that <math>a_0=1</math> and <math>a_{n+1} = f^{n+2}\left(\left\lfloor\frac{a_n}{2}\right\rfloor\right)</math> for all non-negative integers <math>n</math>.		Let <math>(a_n)_{n \ge 0}</math> be a sequence such that <math>a_0=1</math> and <math>a_{n+1} = f^{n+2}\left(\left\lfloor\frac{a_n}{2}\right\rfloor\right)</math> for all non-negative integers <math>n</math>.

ADucharme: /* 6. 1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD (bbch) */ rename section to Space Needle like how BMO 2 is "Hydra and Antihydra"

2026-04-28T17:14:10Z

6. 1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD (bbch): rename section to Space Needle like how BMO 2 is "Hydra and Antihydra"

← Older revision		Revision as of 17:14, 28 April 2026
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	Does there exist a positive integer <math>i</math> such that <math>b_i = f(a_i)-1</math>?		Does there exist a positive integer <math>i</math> such that <math>b_i = f(a_i)-1</math>?

	=== 6. ~~{{TM\|1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD\|undecided}}~~ ===		=== 6. [[Space Needle]] ===
	Let <math>f(b) = b + k + 3a</math>, where <math>k</math> and <math>a</math> are non-negative integers satisfying <math>b = (2a+1)\cdot 2^k</math>.		Let <math>f(b) = b + k + 3a</math>, where <math>k</math> and <math>a</math> are non-negative integers satisfying <math>b = (2a+1)\cdot 2^k</math>.

Sligocki: /* 8. 1RB0LD_0RC1RB_0RD0RA_1LE0RD_1LF---_0LA1LA (bbch) */ Improve floor bracket sizes

2026-04-28T17:11:58Z

8. 1RB0LD_0RC1RB_0RD0RA_1LE0RD_1LF---_0LA1LA (bbch): Improve floor bracket sizes

← Older revision		Revision as of 17:11, 28 April 2026
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	<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}		<math display="block">(a_{n+1}, b_{n+1}) = \begin{cases}
	(a_n-\lfloor\frac{b_n}{2}\rfloor-3, 3\lfloor\frac{b_n+1}{2}\rfloor+6) & \text{if }a_n > \lfloor\frac{b_n}{2}\rfloor \\		(a_n-\left\lfloor\frac{b_n}{2}\right\rfloor-3, 3\left\lfloor\frac{b_n+1}{2}\right\rfloor+6) & \text{if }a_n > \left\lfloor\frac{b_n}{2}\right\rfloor \\
	(3a_n+5, b_n-2a_n) & \text{if }a_n \le \lfloor\frac{b_n}{2}\rfloor		(3a_n+5, b_n-2a_n) & \text{if }a_n \le \left\lfloor\frac{b_n}{2}\right\rfloor
	\end{cases}</math>		\end{cases}</math>

	for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_n = \lfloor\frac{b_n}{2}\rfloor + 1</math>?		for all positive integers <math>n</math>. Does there exist a positive integer <math>i</math> such that <math>a_n = \left\lfloor \frac{b_n}{2} \right\rfloor + 1</math>?

	== Solved problems ==		== Solved problems ==

Coda: Add link to Lean formalizations

2026-04-11T13:57:09Z

Add link to Lean formalizations

← Older revision		Revision as of 13:57, 11 April 2026
Line 2:		Line 2:

	The purpose of the BMO is twofold. First, statements where non-essential details (related to tape encoding, number of steps, etc.) are discarded are more suitable to be shared with mathematicians who perhaps are able to help. Second, it's a way to jokingly highlight how a hard question could appear deceptively simple.		The purpose of the BMO is twofold. First, statements where non-essential details (related to tape encoding, number of steps, etc.) are discarded are more suitable to be shared with mathematicians who perhaps are able to help. Second, it's a way to jokingly highlight how a hard question could appear deceptively simple.

			BMO problems have been formalized in Lean and added to the DeepMind formal-conjectures database ([https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/Other/BeaverMathOlympiad.lean Github]).

	== Unsolved problems ==		== Unsolved problems ==

RobinCodes: Fix formatting

2026-04-10T09:30:16Z

Fix formatting

← Older revision		Revision as of 09:30, 10 April 2026
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	Now consider the iterated application of the function <math>f^{n+1}(b) = f(f^n(b))</math>, <math>f^0(b)=b</math>. Does there exist a non-negative integer <math>n</math> such that <math>f^n(6)</math> equals a power of 2?		Now consider the iterated application of the function <math>f^{n+1}(b) = f(f^n(b))</math>, <math>f^0(b)=b</math>. Does there exist a non-negative integer <math>n</math> such that <math>f^n(6)</math> equals a power of 2?

	=== 8. {{TM\|1RB0LD_0RC1RB_0RD0RA_1LE0RD_1LF---_0LA1LA		=== 8. {{TM\|1RB0LD_0RC1RB_0RD0RA_1LE0RD_1LF---_0LA1LA\|undecided}} ===
	\|undecided}} ===

	Let <math>(a_n)_{n \ge 1}</math> and <math>(b_n)_{n \ge 1}</math> be two sequences such that <math>(a_1, b_1) = (10, 12)</math> and		Let <math>(a_n)_{n \ge 1}</math> and <math>(b_n)_{n \ge 1}</math> be two sequences such that <math>(a_1, b_1) = (10, 12)</math> and