Twenty two methods from the Academy's Year Two curriculum.

This is the directed angle form of the inscribed angle theorem, and its value is that it needs no case split for the position of \(D\) relative to arc \(AB\). Forward direction: if \(A,B,C,D\) lie on a circle \(\omega\), the inscribed angle theorem for chord \(AB\) says every point of \(\omega\) (other than \(A,B\)) sees \(AB\) under the same directed angle mod \(180^\circ\); in the classical (undirected) statement this splits into 'same arc' and 'opposite arcs give supplementary angles' cases, but working mod \(180^\circ\) collapses both cases into the single equality \(\angle(CA,CB)=\angle(DA,DB)\). Converse: suppose \(\angle(CA,CB)=\angle(DA,DB)\) with \(A,B,C,D\) not collinear. Let \(\omega\) be the circumcircle of \(A,B,C\), and suppose for contradiction \(D\notin\omega\). Let \(D'\) be the second intersection of line \(BD\) with \(\omega\) (other than \(B\)). By the forward direction applied to \(A,B,C,D'\), \(\angle(CA,CB)=\angle(D'A,D'B)\). Combined with the hypothesis, \(\angle(D'A,D'B)=\angle(DA,DB)\), and since \(D,D'\) both lie on line \(BD\), this forces \(D=D'\), so \(D\in\omega\) after all, a contradiction. Hence \(A,B,C,D\) are concyclic. This is exactly the tool used to prove Miquel's theorem: to show three circles meet at one point \(M\), one names \(M\) as the second intersection of two of them and shows the directed angle condition forces \(M\) onto the third.

Let \(M\) be the second intersection point of circles \((BXZ)\) and \((CXY)\) (they already share the point \(X\)). We show \(M\) also lies on \((AYZ)\). Since \(B,X,Z,M\) are concyclic, the directed angle form of the inscribed angle theorem gives \[\angle(MZ,ZB) = \angle(MX,XB).\] Since \(C,X,Y,M\) are concyclic, \[\angle(MX,XC) = \angle(MY,YC).\] Because \(X\) lies on line \(BC\), \(XB\) and \(XC\) are the same line, so \(\angle(MX,XB)=\angle(MX,XC)\). Chaining the two equalities: \[\angle(MZ,ZB) = \angle(MY,YC).\] Now \(Z\) lies on line \(AB\) and \(Y\) lies on line \(AC\), so \(ZB\) is the same line as \(ZA\) and \(YC\) is the same line as \(YA\); rewriting, \[\angle(MZ,ZA) = \angle(MY,YA).\] By the converse of the inscribed angle theorem (directed angle form), this equality of directed angles subtended by segment \(MA\) at \(Z\) and at \(Y\) means \(A, Y, Z, M\) are concyclic. Hence \(M\) lies on \((AYZ)\) as well, so all three circles pass through \(M\).

By the intersecting chords form of Power of a Point, using ordinary positive lengths, \[PA\cdot PB = PC\cdot PD.\] Substituting gives \[8\cdot 3 = 6\cdot PD \implies PD = 4.\] Both positive chord products equal \(24\). Under the signed convention, the actual power of the interior point is therefore \(-24\).

Every multiple of \(2\) up to \(12\) contributes one factor of \(2\). Multiples of \(4\) contribute a second factor, and the multiple of \(8\) contributes a third. Therefore \[\nu_2(12!)=\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{4}\right\rfloor+\left\lfloor\frac{12}{8}\right\rfloor=6+3+1=10.\] Direct factorisation gives the same total.

Factor each number: \(12=2^2\cdot3,\ 18=2\cdot3^2,\ 30=2\cdot3\cdot5\). For the gcd, take the minimum exponent of each prime across all three: \(\nu_2\) minimum is \(\min(2,1,1)=1\); \(\nu_3\) minimum is \(\min(1,2,1)=1\); \(\nu_5\) minimum is \(\min(0,0,1)=0\). So \(\gcd(12,18,30) = 2^1\cdot3^1 = 6\). Check the characterization: \(k=6\) divides all three of \(12,18,30\), and indeed \(6\mid\gcd(12,18,30)=6\); no larger \(k\) divides all three, consistent with \(6\) being exactly the gcd. For the lcm, take the maximum exponent of each prime: \(\nu_2\) maximum is \(\max(2,1,1)=2\); \(\nu_3\) maximum is \(\max(1,2,1)=2\); \(\nu_5\) maximum is \(\max(0,0,1)=1\). So \(\operatorname{lcm}(12,18,30) = 2^2\cdot3^2\cdot5 = 4\cdot9\cdot5 = 180\). Check: \(12,18,30\) all divide \(180\) (\(180/12=15\), \(180/18=10\), \(180/30=6\), all integers), and \(180\) is the smallest such common multiple, confirming the dual characterization.

Factor \(756=2^2\cdot3^3\cdot7\). Its smallest prime factor is \(2\), so the identity gives \[M(756)=\frac{756}{2}=378.\] To justify maximality, let \(d<756\) be any divisor and write \(756=dq\). Because \(d\) is proper, \(q>1\). Every prime factor of \(q\) is also a prime factor of \(756\), so \(q\ge2\). Hence \(d=756/q\le756/2=378\). Since \(378\mid756\), the bound is attained and \(378\) is the greatest proper divisor.

Write each pile size in binary: \(5 = 101_2\), \(7 = 111_2\), and \(3 = 011_2\). Their nim sum is \(101 \oplus 111 \oplus 011 = 001\), so the position is winning. To make the nim sum zero, reduce the pile of \(7\) to \(6\), because \(5 \oplus 3 = 6\). The new position \((5,6,3)\) has nim sum \(000_2\). Repeating this response after each opposing move leads to the final stone.

Classify small pile sizes by backward induction. Position \(0\) is losing. Positions \(1\) and \(2\) can reach \(0\), so they are winning. Position \(3\) reaches only winning positions, so it is losing. The pattern repeats every three values: losing positions satisfy \(n\equiv0\pmod3\). The first player wins when \(n\equiv1\) or \(2\pmod3\) by leaving a multiple of \(3\) after each move.

Number the rows and columns \(1,2,\dots,15\). Before the jumps, the rooks occupy every row and every column exactly once, so the sum of all row and column coordinates is \[\sum_{i=1}^{15}(r_i+c_i)=2(1+2+\cdots+15)=240,\] an even number. A knight style jump changes one coordinate by \(\pm1\) and the other by \(\pm2\), so it changes \(r_i+c_i\) by an odd integer. The total change produced by all 15 rooks is therefore a sum of 15 odd integers, hence odd. If the rooks were again in distinct rows and distinct columns after the simultaneous jumps, the final coordinate sum would again be \(240\), so the total change would be \(0\), an even number. This contradiction shows that the final placement cannot have one rook in every row and every column; at least two rooks must share a row or a column.

Let \(T=A+B\). Each transfer preserves \(T\), while each coin addition increases it by \(1\). Thus \(T\) is a nondecreasing monovariant. If \(T\le N\) and the starting total is \(T_0\), at most \(N-T_0\) coin additions can occur.

The starting position \((7,7)\) has equal piles, so by the two pile symmetry principle it is a Losing position for the player to move, here, the first player. The second player's strategy: whenever the first player reduces one pile from \(x\) to some smaller value \(x'<x\) (leaving piles \((x',x)\) with \(x'\ne x\)), the second player immediately reduces the other pile from \(x\) down to \(x'\) as well, restoring the equal position \((x',x')\). This is always a legal move, since the pile being reduced still has \(x>x'\) stones in it before the copycat move. Following this strategy, the position is exactly equal piles after every one of the second player's turns: \((7,7)\to(x_1',x_1)\to(x_1',x_1')\to(x_2',x_1')\to(x_2',x_2')\to\cdots\), with the common pile size strictly decreasing each round. Since the common size is a nonnegative integer that strictly decreases, it must eventually reach \((0,0)\) right after one of the second player's moves, meaning the second player takes the last stone and wins.

Apply CBS with sequences \(a_i=\sqrt{x_i}\) and \(b_i = 1/\sqrt{x_i}\) for \(x_i \in \{x,y,z\}\): take vectors \((\sqrt{x},\sqrt{y},\sqrt{z})\) and \((1/\sqrt{x}, 1/\sqrt{y}, 1/\sqrt{z})\). Then \[\left(\sqrt{x}\cdot\tfrac{1}{\sqrt{x}} + \sqrt{y}\cdot\tfrac{1}{\sqrt{y}} + \sqrt{z}\cdot\tfrac{1}{\sqrt{z}}\right)^2 \le (x+y+z)\left(\frac1x+\frac1y+\frac1z\right).\] The left side is \((1+1+1)^2 = 9\), so \[9 \le (x+y+z)\left(\frac1x+\frac1y+\frac1z\right).\] Equality in CBS requires \(\left(\sqrt{x},\sqrt{y},\sqrt{z}\right)\) proportional to \(\left(\tfrac{1}{\sqrt{x}},\tfrac{1}{\sqrt{y}},\tfrac{1}{\sqrt{z}}\right)\), i.e. \(x = y = z\) (up to the same positive constant of proportionality), so equality holds exactly when \(x=y=z\).

This is the two term case of Titu's lemma. Taking numerators \(x,y\) and positive denominators \(a,b\) gives immediately \[\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}.\] Equality in Titu's lemma holds exactly when \(x/a=y/b\), equivalently \(bx=ay\). This example isolates the lemma before combining it with other inequalities.

By AM GM, \(ab \le \left(\dfrac{a+b}{2}\right)^2 = 1\), so \(a^2+b^2 = (a+b)^2 - 2ab = 4-2ab \ge 4-2=2\). Now combine the two fractions over a common denominator: \[\frac{1}{a^2+1}+\frac{1}{b^2+1} = \frac{(a^2+1)+(b^2+1)}{(a^2+1)(b^2+1)} = \frac{a^2+b^2+2}{(a^2+1)(b^2+1)}.\] Expand the denominator: \((a^2+1)(b^2+1) = a^2b^2+a^2+b^2+1\). By AM GM on \(a^2,b^2\) (or directly since \(ab\le1\)), \(a^2b^2 = (ab)^2 \le 1\), so the denominator satisfies \((a^2+1)(b^2+1) \le 1 + (a^2+b^2) + 1 = a^2+b^2+2\). Therefore, since both the numerator \(a^2+b^2+2\) and denominator \((a^2+1)(b^2+1)\) are positive and the denominator is at most the numerator, \[\frac{1}{a^2+1}+\frac{1}{b^2+1} = \frac{a^2+b^2+2}{(a^2+1)(b^2+1)} \ge \frac{a^2+b^2+2}{a^2+b^2+2} = 1.\] Hence \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge1\), with equality when \(ab=1\) and \(a=b\) simultaneously, i.e. \(a=b=1\), which is consistent with \(a+b=2\).

Step 1 (Titu setup): write \(\dfrac{a}{b+c}=\dfrac{a^2}{a(b+c)}\), \(\dfrac{b}{c+a}=\dfrac{b^2}{b(c+a)}\), \(\dfrac{c}{a+b}=\dfrac{c^2}{c(a+b)}\). Step 2 (apply Titu's Lemma): \[\frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)} \ge \frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}.\] Step 3 (simplify the denominator): expand \(a(b+c)+b(c+a)+c(a+b) = ab+ac+bc+ab+ac+bc = 2(ab+bc+ca)\). So the bound becomes \(\dfrac{(a+b+c)^2}{2(ab+bc+ca)}\). Step 4 (finish with a quadratic fact): expand \((a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\), so \((a+b+c)^2 \ge 3(ab+bc+ca)\) is equivalent to \(a^2+b^2+c^2 \ge ab+bc+ca\), which follows from \(\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]\ge0\). Substituting, \[\frac{(a+b+c)^2}{2(ab+bc+ca)} \ge \frac{3(ab+bc+ca)}{2(ab+bc+ca)} = \frac{3}{2}.\] Chaining Steps 2 to 4 gives \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac32\) as required. Equality requires both \(a/(a(b+c)) = b/(b(c+a)) = c/(c(a+b))\) (Titu's equality condition, i.e. \(1/(b+c)=1/(c+a)=1/(a+b)\), forcing \(a=b=c\)) and \(a=b=c\) from Step 4, both conditions coincide exactly at \(a=b=c\), confirming equality holds there and only there (checked directly: at \(a=b=c\), each fraction equals \(1/2\), summing to \(3/2\)).

Both sides are homogeneous of degree \(1\), so the correct weights are sufficient. Apply Titu's Lemma with numerators \(a,b,c\) and denominators \(b,c,a\): \[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{(a+b+c)^2}{b+c+a}=a+b+c.\] Equality requires \(a/b=b/c=c/a\), which gives \(a=b=c\).

For \(p=2\) the expression is not an integer. Let \(p\) be odd and put \(m=(p-1)/2\). If \((2^{p-1}-1)/p=q^2\), then \[(2^m-1)(2^m+1)=pq^2.\] The two odd factors are coprime, so one is a square and the other is \(p\) times a square. If \(2^m-1=x^2\), then for \(m\ge3\) reduction modulo \(8\) gives \(x^2\equiv7\pmod8\), impossible; checking \(m=1,2\) leaves \(p=3\). If \(2^m+1=y^2\), then \((y-1)(y+1)=2^m\). The two factors are powers of two differing by \(2\), hence they are \(2\) and \(4\); thus \(y=3\), \(m=3\), and \(p=7\). Direct substitution confirms that \(p=3\) and \(p=7\) both work.

The last digit of \(7^{222}\) is \(7^{222} \bmod 10\). Since \(\gcd(7,10)=1\), Euler's theorem applies with \(n=10\): \(\varphi(10) = 10\left(1-\tfrac12\right)\left(1-\tfrac15\right) = 10\cdot\tfrac12\cdot\tfrac45 = 4\), so \(7^4 \equiv 1 \pmod{10}\). Reduce the exponent modulo \(\varphi(10)=4\): \(222 = 4\cdot55+2\), so \(222 \equiv 2 \pmod 4\), giving \[7^{222} \equiv 7^2 = 49 \equiv 9 \pmod{10}.\] Thus \(7^{222}\) ends in \(9\). The cycle \(7,9,3,1\) confirms the period of four.

Here \(a=6, b=1, p=5, n=100\). Check the hypotheses: \(p \mid a-b\) since \(5 \mid 6-1=5\), yes; and \(5 \nmid 6\), \(5\nmid1\), so both hold. By LTE, \[\nu_5(6^{100}-1^{100}) = \nu_5(6-1) + \nu_5(100) = \nu_5(5) + \nu_5(100).\] Now \(\nu_5(5)=1\), and \(100 = 4\cdot25 = 2^2\cdot5^2\), so \(\nu_5(100)=2\). Hence \[\nu_5(6^{100}-1) = 1+2 = 3,\] meaning \(5^3=125\) divides \(6^{100}-1\) but \(5^4=625\) does not.

Suppose \(n>1\) and \(n \mid 2^n-1\). Let \(p\) be the smallest prime divisor of \(n\). Since \(n \mid 2^n-1\), also \(p \mid 2^n - 1\), so \(2^n \equiv 1 \pmod p\). Note \(p\) is odd, since \(2^n-1\) is odd so \(p\ne2\). Let \(d = \operatorname{ord}_p(2)\); then \(d \mid n\) (since \(2^n\equiv1\pmod p\)). Also by Fermat's Little Theorem, \(2^{p-1}\equiv1\pmod p\), so \(d \mid p-1\). Since \(d \mid p-1\), we have \(d < p\) (as \(d\le p-1\)); but \(d\mid n\) and \(p\) is the smallest prime factor of \(n\), so \(d\) cannot have any prime factor smaller than \(p\) while still dividing \(n\), since every divisor of \(n\) greater than 1 has a prime factor at least \(p\), and \(d<p\), the only divisor of \(n\) that is less than \(p\) is \(1\). Hence \(d=1\), meaning \(2^1 \equiv 1 \pmod p\), i.e. \(p \mid 1\), which is impossible for a prime \(p\). This contradiction shows no \(n>1\) satisfies \(n\mid2^n-1\), so \(n=1\) is the only solution (and indeed \(1 \mid 2^1-1=1\) trivially holds).

Suppose \(n>1\) and \(n\mid2^n+1\). Let \(p\) be the smallest prime factor of \(n\); since \(2^n+1\) is odd, \(p\ne2\), so \(p\) is odd. From \(n\mid2^n+1\), also \(p\mid2^n+1\), so \(2^n\equiv-1\pmod p\), hence \(2^{2n}\equiv1\pmod p\). Let \(d=\operatorname{ord}_p(2)\). Then \(d\mid2n\) (from \(2^{2n}\equiv1\)), and \(d\mid p-1\) (Fermat). Also, since \(2^n\equiv-1\not\equiv1\pmod p\) (as \(p\) is odd, \(-1\ne1\) mod \(p\)), \(d\nmid n\); combined with \(d\mid 2n\), this forces \(d\) to contain a factor of \(2\) that \(n\) alone doesn't supply, i.e. \(d\mid2n\) but \(d\nmid n\) means \(\nu_2(d) = \nu_2(2n)\) exceeds \(\nu_2(n)\) by exactly one, so \(d = 2e\) for some \(e\mid n\) with \(d/\gcd(d,n) = 2\). Since \(p\) is the smallest prime factor of \(n\), any divisor \(e\mid n\) with \(e>1\) has \(e\ge p\); the smallest prime divisor squeeze on \(d\mid p-1<p\) forces \(e=1\), so \(d=2\). Then \(\operatorname{ord}_p(2)=2\) means \(2^2\equiv1\pmod p\), i.e. \(p\mid3\), so \(p=3\) (since \(p\) is an odd prime). As \(p\) is the smallest prime factor of \(n\) and \(p=3\), we conclude \(3\mid n\).

Compute successive powers of \(3\) mod \(7\): \(3^1=3,\ 3^2=9\equiv2,\ 3^3=3\cdot2=6,\ 3^4=3\cdot6=18\equiv4,\ 3^5=3\cdot4=12\equiv5,\ 3^6=3\cdot5=15\equiv1\pmod7\). The powers \(3,2,6,4,5,1\) are exactly the six nonzero residues \(\{1,2,3,4,5,6\}\) mod \(7\), each appearing exactly once, and the order of \(3\) is \(6=p-1\). So \(3\) is indeed a primitive root mod \(7\). By the counting fact, the primitive roots mod \(7\) are exactly \(3^k\) for \(\gcd(k,6)=1\), i.e. \(k\in\{1,5\}\) among \(\{1,\dots,6\}\): these give \(3^1=3\) and \(3^5=5\). So there are \(\varphi(6)=\varphi(2\cdot3)=6\cdot\tfrac12\cdot\tfrac23=2\) primitive roots mod \(7\), namely \(3\) and \(5\), and indeed neither \(2\) (order \(3\), since \(2^3=8\equiv1\)) nor \(4,6\) has order \(6\), confirming exactly two primitive roots exist mod \(7\).