Algebraic Manipulation

1.Factoring and the standard identities

Factoring rewrites a sum or difference as a product, and on the AMC it is usually the first move. An expression that looks large often collapses once it is written as a product of small pieces. The basic identities are $a^2-b^2=(a-b)(a+b)$ and $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$. They are valuable because each one turns a long computation into a short one: $103^2-97^2=6\cdot 200$ is immediate, while squaring both numbers is not.

The sum-of-cubes and difference-of-cubes identities follow from one multiplication. Expand $(a-b)(a^2+ab+b^2)$: the $a^2b$ and $ab^2$ terms cancel in pairs, leaving $a^3-b^3$. The sum of cubes is the same identity with $b$ replaced by $-b$, which flips the sign of the middle term. The skill is reading the identities in reverse. A term such as $x^6-1$ is $(x^3)^2-1^2$, so it factors twice: $(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$.

The Sophie Germain identity $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$ is the one students rarely see coming. It comes from completing a square: $a^4+4b^4=(a^2+2b^2)^2-4a^2b^2$, which is a difference of squares. Ordinary real factoring cannot split a sum of two terms like this, so the identity is worth memorising. Taking $b=1$ and $a=n$ gives $n^4+4=(n^2+2n+2)(n^2-2n+2)$. The second factor is $(n-1)^2+1$, so for every integer $n\ge 2$ both factors exceed $1$ and $n^4+4$ is composite.

Factoring by grouping and recognising a hidden square are two more routes. For $x^3+x^2+x+1$, group as $x^2(x+1)+(x+1)=(x^2+1)(x+1)$. For $x^4+x^2+1$, add and subtract $x^2$: it becomes $(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$. The second trick generalises. When a quartic has no visible factorisation, try adding a square to make a difference of squares, and subtract the same square to keep the value unchanged.

A common error is to treat $a^2+b^2$ as if it factored over the reals in the same way as $a^2-b^2$. It does not, and writing $(a+b)^2$ for it is wrong by $2ab$. A second error is expanding $(a-b)^3$ as $a^3-b^3$. The correct expansion is $a^3-3a^2b+3ab^2-b^3$, so cubing a difference is not the same as subtracting cubes. Always check a factorisation by multiplying it back out, at least in a quick numerical case.

• $a^2-b^2=(a-b)(a+b)$ and $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$
• Sophie Germain: $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$
• $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ by adding and subtracting $x^2$
• Factor by grouping: pair up terms that share a common factor

• Writing $a^2+b^2=(a+b)^2$, which is off by $2ab$
• Expanding $(a-b)^3$ as $a^3-b^3$ and forgetting the middle terms

2.Completing the square

Completing the square rewrites a quadratic as a squared term plus a constant: $x^2+bx=\left(x+\frac b2\right)^2-\frac{b^2}{4}$. It is the engine behind the quadratic formula, but on contests its most useful role is in equations in two variables where the expression is required to be zero. A sum of squares equal to zero forces every square to vanish, which pins down the variables exactly.

The derivation is one line: $\left(x+\frac b2\right)^2=x^2+bx+\frac{b^2}{4}$, so subtract $\frac{b^2}{4}$ to correct the constant. In a quadratic in two variables, complete the square in each variable separately. For $x^2-6x+y^2+4y+13$, write $x^2-6x=(x-3)^2-9$ and $y^2+4y=(y+2)^2-4$. The whole expression is then $(x-3)^2+(y+2)^2+13-9-4=(x-3)^2+(y+2)^2$, a sum of two squares with no constant left over.

The same expression shows how the method is used. If $x^2+y^2-6x+4y+13=0$, the equation says $(x-3)^2+(y+2)^2=0$. Because squares are nonnegative, both must be zero, so $x=3$ and $y=-2$ are the only real solutions. The question then reduces to reading off the answer. The method is most often needed when a problem gives an equation with no visible solution and asks for a single value such as $x+y$ or $xy$.

Completing the square also explains which quadratics fail to factor over the integers. The expression $x^2+6x+7$ has no integer factorisation, but $(x+3)^2-2$ shows that it is a difference of squares over the reals: $(x+3-\sqrt2)(x+3+\sqrt2)$. Recognising this form tells you the roots are irrational, which is a fact worth checking before trying to find them.

The common mistake is to complete the square on one variable and forget the constant that was added. The resulting sum of squares is then off by that constant, and the equation has no real solutions when it should have one. Always subtract back what you added, or expand your final form to check it against the original expression.

• $x^2+bx=\left(x+\frac b2\right)^2-\frac{b^2}{4}$
• A sum of squares equal to $0$ forces each square to be $0$
• $x^2+y^2-6x+4y+13=(x-3)^2+(y+2)^2$

• Forgetting the constant subtracted when completing a square
• Concluding that a sum of squares has many solutions when it equals zero

3.Simon's Favourite Factoring Trick

Many counting problems on the AMC reduce to an integer equation of the form $xy+ax+by=c$. Such an equation has no visible structure, and a direct search is slow and error-prone. Simon's Favourite Factoring Trick, usually abbreviated SFFT, forces the left side to factor, so that the solutions become divisor pairs of a fixed number. It is named after a contributor to the AoPS forum and is one of the most reliable tools for integer counting.

The key identity is $(x+b)(y+a)=xy+ax+by+ab$. The left side of the original equation differs from this expansion only by the constant $ab$. So add $ab$ to both sides of $xy+ax+by=c$ to obtain $(x+b)(y+a)=c+ab$. The two equations are equivalent, so no solutions are gained or lost: the trick is a rewriting of the same condition, not an approximation.

If the coefficient of $xy$ is not $1$, multiply the equation by it first. For $axy-cx-by=d$, multiply by $a$ and add $bc$ to both sides. Then $(ax-b)(ay-c)=ad+bc$. For example, $2xy-5x-3y=45$ becomes $(2x-3)(2y-5)=90+15=105$. The process is always the same: multiply, add the right constant, and factor. The constant to add is whatever makes the left side a product of two linear expressions.

Once the equation is factored, the rest is bookkeeping, and bookkeeping is where contests are decided. A factor such as $x+2$ must be at least $3$ when $x$ is a positive integer, and a product that is positive forces both factors to share a sign. List the divisor pairs of the constant in both signs, then discard any pair that gives a non-positive $x$ or $y$. Counting without this step is the main source of error.

The classic mistake is to count divisor pairs of the constant without checking signs and bounds. This produces solutions with $x=0$, or with a negative $y$, that the problem excludes. The opposite mistake is to discard a negative-times-negative pair that the bounds in fact allow. Test each candidate in the original equation before you count it.

• $xy+ax+by=c$ becomes $(x+b)(y+a)=c+ab$
• For $axy-cx-by=d$: $(ax-b)(ay-c)=ad+bc$
• Apply the sign and size bounds on $x$ and $y$ after factoring, before counting

• Counting divisor pairs without checking that $x$ and $y$ are positive
• Forgetting to multiply by the coefficient of $xy$ before factoring

4.Symmetric systems and the sum-product substitution

A system is symmetric in $x$ and $y$ when swapping the two variables leaves both equations unchanged. Such systems resist elimination but yield to a change of variables: set $s=x+y$ and $p=xy$. Every symmetric polynomial in $x$ and $y$ can be written as a polynomial in $s$ and $p$, so the system becomes a smaller system in $s$ and $p$, usually one that is easy to solve.

The identities needed most often are $x^2+y^2=s^2-2p$ and $x^3+y^3=s^3-3ps$. The cube identity comes from expanding $(x+y)^3=x^3+y^3+3xy(x+y)$ and rearranging. A quantity like $x^4+y^4$ is also symmetric and reduces to $(s^2-2p)^2-2p^2$, but in practice you substitute whichever identity matches the given quantities and do not memorise every power.

Every pair $(s,p)$ produced by the substitution must be checked for reality. The numbers $x$ and $y$ are the roots of $t^2-st+p=0$, and those roots are real exactly when $s^2\ge 4p$. A pair with $s^2<4p$ corresponds to complex numbers, which a problem asking for real values must exclude. Forgetting this check is the most common way to lose a point on a symmetric system, and it is the trap in the harder problems of this topic.

As an example, let $x+y=4$ and $x^3+y^3=28$. With $s=4$ we get $28=64-12p$, so $p=3$. The discriminant is $s^2-4p=16-12=4>0$, so $x$ and $y$ are real: they are $1$ and $3$, and indeed $1^3+3^3=28$. The reduced equation was linear in $p$, which is typical. When the reduction produces a cubic in $s$, decide which roots are admissible by the reality condition, not by which root looks nicer.

Do not divide by a quantity that could be zero, and do not discard a root of the reduced equation before testing it against the original system. Also remember that one unordered pair $\{x,y\}$ produces two ordered pairs, which matters when the question counts solutions.

• $x^2+y^2=(x+y)^2-2xy$ and $x^3+y^3=(x+y)^3-3xy(x+y)$
• Real $x,y$ with sum $s$ and product $p$ require $s^2\ge 4p$
• Substitute $s=x+y$ and $p=xy$ to reduce a symmetric system to two unknowns

• Accepting an $(s,p)$ pair with $s^2<4p$, which gives non-real $x,y$
• Counting an unordered pair once when the question asks for ordered pairs

5.Absolute value and piecewise equations

The absolute value $|t|$ equals $t$ when $t\ge 0$ and $-t$ when $t<0$. An equation containing absolute values is therefore a piecewise problem. It splits into cases according to the signs of the expressions inside the bars. The method is always the same: find the points where each inner expression changes sign, solve the equation on each interval, and keep only the solutions that lie in the interval you assumed.

Two basic facts do most of the work. First, $|t|=c$ with $c\ge 0$ has exactly the solutions $t=c$ and $t=-c$, and it has none when $c<0$. Second, $|t|=|s|$ holds exactly when $t=s$ or $t=-s$. Nested absolute values are solved from the outside in: for $\left||x-2|-3\right|=1$, first solve $|x-2|-3=\pm 1$, which gives $|x-2|=4$ or $|x-2|=2$, and then solve each of those.

As an example, take $|x-1|+|x+2|=5$. The inner expressions change sign at $x=1$ and $x=-2$, which gives three intervals. For $x\ge 1$ the equation is $2x+1=5$, so $x=2$. For $-2\le x<1$ the left side is $3$ for every such $x$, so there are no solutions there. For $x<-2$ it is $-2x-1=5$, so $x=-3$. Both candidates lie in their intervals, so the solutions are $2$ and $-3$.

Geometrically, $|x-a|+|x-b|$ is the total distance from $x$ to two fixed points. Outside the interval between them the sum grows, and between them it is constant. This explains the middle case above: a whole interval either satisfies the equation or none of it does, which is why the case analysis can be done quickly once you see it. The same picture gives the minimum of such sums without calculus.

Check each candidate against its own case. A root of $-t=c$ is valid only when $t<0$. Also note that $\sqrt{t^2}=|t|$, so a square root of a square forces a case split that is easy to forget. Students who solve $t=\pm c$ and stop, without testing both branches against the original expression, often report extra or missing solutions.

• $|t|=c$ with $c\ge 0$ gives $t=c$ or $t=-c$; no solutions if $c<0$
• $|t|=|s|$ if and only if $t=s$ or $t=-s$
• $\sqrt{t^2}=|t|$
• The sign-change points of the inner expressions are the case boundaries

• Solving $t=\pm c$ and forgetting to check each branch against its case assumption
• Dropping the middle interval, where the expression can be constant and equal to the target

6.Clever substitutions

A substitution replaces a messy expression by one letter so that the equation becomes one you already know how to solve. The skill is recognising the inner expression. In $x^4-5x^2+4=0$ the quantity $x^2$ is the only thing that appears, so $u=x^2$ turns the equation into $u^2-5u+4=0$, with roots $u=1$ and $u=4$. Returning to $x$ gives $x=\pm1$ and $x=\pm2$, four solutions in all.

Reciprocal pairs are the most common hidden structure. If an expression is unchanged when $x$ is replaced by $\frac1x$, write $u=x+\frac1x$. Since $\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2$, we have $x^2+\frac1{x^2}=u^2-2$, and applying this again gives $x^4+\frac1{x^4}=(u^2-2)^2-2$. Given $u=3$, the value of $x^2+\frac1{x^2}$ is $7$ and that of $x^4+\frac1{x^4}$ is $47$. Notice that the substitution answered a question about $x^4$ using only $x+\frac1x$.

Shift substitutions handle quantities that appear in two places with a constant difference. If an equation involves $x^2-3x$ in two places, let $u=x^2-3x$ and solve the quadratic in $u$ first. Each root $u_0$ then gives a quadratic in $x$ that must be solved separately. Solutions must be gathered from every branch, and a value that arises from two branches must be counted once if the question asks for distinct solutions.

Ratio substitutions work the same way. If $\frac xy+\frac yx$ is given, set $t=\frac xy$. The condition becomes $t+\frac1t=c$, a quadratic in $t$, and any ordering condition on $x$ and $y$ selects the right root. Once you have $t$, the requested expression is often another rational function of $t$, which is easy to evaluate.

Two pitfalls are common. First, the substitution must be inverted at the end: students solve for $u$ and report $u$ when the question asks for $x$. Second, respect the domain. For real $x\ne 0$ the quantity $x+\frac1x$ satisfies $\left|x+\frac1x\right|\ge 2$, so any root $u$ with $|u|<2$ is not attainable and must be discarded before you go further.

• With $u=x+\frac1x$: $x^2+\frac1{x^2}=u^2-2$ and $x^4+\frac1{x^4}=(u^2-2)^2-2$
• For real $x\ne 0$: $\left|x+\frac1x\right|\ge 2$
• Replace a recurring expression such as $x^2$ or $x^2-3x$ by one letter, solve, then return to $x$

• Reporting the value of the new letter instead of the value of $x$
• Accepting a value of $u=x+\frac1x$ with $|u|<2$, which no real $x$ produces

7.Nested radicals and rationalising

A nested radical is a square root inside another square root, such as $\sqrt{5+2\sqrt6}$. The goal is to rewrite it without the inner root, a process called denesting. The idea is that a square root of an expression of the form $a+b\sqrt d$ is sometimes itself of the form $p+q\sqrt d$ with rational $p,q$, and squaring the guess reveals whether this is so.

Suppose $\sqrt{a+\sqrt b}=\sqrt p+\sqrt q$ with $p,q>0$. Squaring gives $a+\sqrt b=p+q+2\sqrt{pq}$. Matching the parts that do not and do involve the root gives $p+q=a$ and $4pq=b$. So $p$ and $q$ are the roots of $t^2-at+\frac b4=0$, and they are rational exactly when $a^2-b$ is a perfect square. For $\sqrt{5+\sqrt{24}}$ we need $p+q=5$ and $pq=6$, which gives $p=2$, $q=3$, so the value is $\sqrt2+\sqrt3$.

The sign needs care, because squaring cannot tell $\sqrt p+\sqrt q$ from $\sqrt p-\sqrt q$ in its result. Decide the sign by checking the numerical value of the expression, and remember that $\sqrt t$ always denotes the nonnegative root. A nested radical that is a difference, such as $\sqrt{A+B}-\sqrt{A-B}$, is handled by squaring too, and the cross term then has the opposite sign.

Rationalising the denominator uses conjugates: $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b$. So $\frac1{\sqrt5-2}=\frac{\sqrt5+2}{5-4}=\sqrt5+2$. The same identity applied to a sum of two adjacent roots gives $\frac1{\sqrt k+\sqrt{k+1}}=\sqrt{k+1}-\sqrt k$, which is the start of a telescoping sum.

For a conjugate pair, squaring the sum simplifies everything: $\left(\sqrt{A+B}+\sqrt{A-B}\right)^2=2A+2\sqrt{A^2-B^2}$. This turns a sum of two radicals into a single radical. The usual error is to write $\sqrt{a+b}=\sqrt a+\sqrt b$, which is false: with $a=b=1$ the left side is $\sqrt2$ and the right side is $2$.

• $\sqrt{a+\sqrt b}=\sqrt p+\sqrt q$ requires $p+q=a$ and $4pq=b$
• $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b$
• $\left(\sqrt{A+B}+\sqrt{A-B}\right)^2=2A+2\sqrt{A^2-B^2}$
• In general $\sqrt{a+b}\ne\sqrt a+\sqrt b$

• Writing $\sqrt{a+b}=\sqrt a+\sqrt b$
• Choosing the wrong sign after squaring, instead of checking the numerical value

8.Telescoping sums

A telescoping sum is one in which consecutive terms cancel. The engine is a difference identity: if each term can be written as $f(k)-f(k+1)$, then $\sum_{k=m}^{n}\bigl(f(k)-f(k+1)\bigr)=f(m)-f(n+1)$, because every interior value of $f$ appears once with each sign. Finding the function $f$ is the entire problem.

The classic example is $\frac1{k(k+1)}=\frac1k-\frac1{k+1}$, so $\sum_{k=1}^{n}\frac1{k(k+1)}=1-\frac1{n+1}$. The identity is checked by putting the right side over a common denominator: $\frac{(k+1)-k}{k(k+1)}=\frac1{k(k+1)}$. For $n=4$ the sum is $\frac45$. Notice that the answer depends only on the endpoints, which is why such sums are fast once the identity is spotted.

The same mechanism handles radicals. Rationalising gives $\frac1{\sqrt k+\sqrt{k+1}}=\sqrt{k+1}-\sqrt k$, so a sum of such terms collapses to a difference of square roots at the endpoints: from $k=1$ to $k=99$ the sum is $\sqrt{100}-\sqrt1=9$. The gap between the roots matters. Since $\frac1{\sqrt k+\sqrt{k+2}}=\frac{\sqrt{k+2}-\sqrt k}{2}$, a gap of two leaves a factor of $\frac12$ in front.

Products telescope as well. In $\prod_{k=2}^{n}\frac{k^2-1}{k^2}$, factor each numerator as $(k-1)(k+1)$. The factors $\frac{k-1}{k}$ and $\frac{k+1}{k}$ then telescope separately, giving $\frac{1}{n}\cdot\frac{n+1}{2}=\frac{n+1}{2n}$. Recognising that a product splits into two telescoping products is the key step.

Telescoping can also be hidden inside a factorisation. The Sophie Germain identity gives $\frac{4k}{k^4+4}=\frac1{k^2-2k+2}-\frac1{k^2+2k+2}$, and the two denominators are shifts of one another, so the sum collapses. A common error is forgetting the constant produced by the rationalising or the shift, so check the first term by hand before trusting the telescope.

• $\sum_{k=m}^{n}\bigl(f(k)-f(k+1)\bigr)=f(m)-f(n+1)$
• $\frac1{k(k+1)}=\frac1k-\frac1{k+1}$
• $\frac1{\sqrt k+\sqrt{k+1}}=\sqrt{k+1}-\sqrt k$ and $\frac1{\sqrt k+\sqrt{k+2}}=\frac{\sqrt{k+2}-\sqrt k}{2}$

• Writing $\frac1{\sqrt k+\sqrt{k+2}}$ as $\sqrt{k+2}-\sqrt k$ and dropping the factor $\frac12$
• Trusting a telescope without checking that the first term actually matches the claimed difference

9.Radical equations: conjugates and extraneous roots

Equations with square roots are solved by isolating a root and squaring, but squaring can introduce solutions. Squaring $\sqrt A=B$ gives $A=B^2$, which holds for both $B$ and $-B$. So every candidate must be substituted back into the original equation. Keep track of the domain as well: $\sqrt t$ is defined only for $t\ge 0$, and it is nonnegative, so a candidate that makes the right side negative is rejected at once.

When two roots appear, a conjugate trick is often faster. For $\sqrt{x+7}-\sqrt{x-1}=2$, multiply the left side by $\sqrt{x+7}+\sqrt{x-1}$ and use the difference of squares: $(x+7)-(x-1)=8$. The given equation then implies $\sqrt{x+7}+\sqrt{x-1}=4$. Adding this to the original equation and halving gives $\sqrt{x+7}=3$, so $x=2$, which checks since $\sqrt9-\sqrt1=2$.

The general shape is $\sqrt{x+a}-\sqrt{x+b}=c$. The difference of the squares is $a-b$, so the sum of the two roots equals $\frac{a-b}{c}$, and the two linear relations determine each root. This works whenever the constant $c$ is nonzero, and it explains why such equations usually have a single clean solution.

Extraneous solutions usually come from a sign issue or a domain issue. For $x=\sqrt{x+2}$, squaring gives $x^2=x+2$, with roots $2$ and $-1$. The root $-1$ is extraneous, because $\sqrt{x+2}\ge 0$ can never equal a negative $x$. Test every candidate in the unsquared equation, not in the squared one.

Students often square a sum of roots term by term and write $(\sqrt a+\sqrt b)^2=a+b$. The cross term $2\sqrt{ab}$ is essential and must be kept. Once it is included, the squared equation is usually a simple linear or quadratic equation in one variable.

• $\sqrt A=B$ squares to $A=B^2$, which also admits $B=-\sqrt A$; check each candidate
• $\left(\sqrt{x+a}-\sqrt{x+b}\right)\left(\sqrt{x+a}+\sqrt{x+b}\right)=a-b$
• $(\sqrt a+\sqrt b)^2=a+b+2\sqrt{ab}$

• Squaring a sum of roots without the cross term $2\sqrt{ab}$
• Reporting a solution from the squared equation without testing it in the original

10.Word problems that reduce to algebra

A word problem on the AMC is an algebra problem in disguise, and most of the work happens before any algebra. Name the unknown quantities, write one equation for each relationship the story states, and then solve. The usual errors are choosing variables badly, or writing a relation in the wrong direction, and both are fixed by slowing down at the setup.

For rates, the relation is distance equals rate times time. Work problems use the same idea with the rate being the fraction of a job done per hour, and rates of cooperating agents add. A tap fills a pool in $6$ hours, a second fills it in $12$ hours, and a drain empties it in $9$ hours. The combined rate is $\frac16+\frac1{12}-\frac19=\frac5{36}$ of the pool per hour, so it fills the pool in $\frac{36}5$ hours.

Moving-object problems compare two equal times or two equal distances. For a boat of still-water speed $b$ in a current of speed $c$, the downstream speed is $b+c$ and the upstream speed is $b-c$. If the boat covers $24$ km downstream in the same time it covers $16$ km upstream and $c=2$, then $\frac{24}{b+2}=\frac{16}{b-2}$, which gives $b=10$. The equation is the whole problem: the rest is arithmetic.

Mixture problems conserve a substance. If $40$ liters of a $30\%$ solution are diluted with $w$ liters of water to reach $20\%$, the amount of pure substance does not change: $0.3\cdot 40=0.2(40+w)$, so $w=20$. The same conservation principle handles money, distance travelled by two parties, and any quantity that is transferred rather than created.

Choose variables so that the relations are as simple as possible. Use one unknown for the quantity asked, or for a quantity that makes the others easy to express. When the story is symmetric in two quantities, the sum-product substitution from the systems concept often solves it in a line. Finally, check the answer against the story: a negative speed or a fractional number of people signals a setup error.

• Time equals distance divided by rate, and the rates of cooperating workers add
• Downstream speed is $b+c$ and upstream speed is $b-c$ for still-water speed $b$ and current $c$
• A conserved quantity, such as pure salt or money, gives an equation between two expressions

• Writing a rate relation in the wrong direction, such as multiplying distance by speed to get time
• Accepting a negative speed or a fractional count without checking the story

Example 1: Difference of squares

Compute $103^2-97^2$.

Example 2: Cubes from a sum and a product

If $a+b=7$ and $ab=10$, what is $a^3+b^3$?

Example 3: Counting with Simon's trick

How many ordered pairs of positive integers $(x,y)$ satisfy $xy-x-3y=2$?

Example 4: Simplifying a conjugate pair of nested roots

Simplify $\sqrt{5+2\sqrt6}+\sqrt{5-2\sqrt6}$.

Example 5: Sum of solutions with two absolute values

Real numbers $x$ satisfy $|x+3|+|x-5|=10$. What is the sum of all such $x$?

Example 6: Ratio substitution with an ordering condition

Real numbers $x>y>0$ satisfy $\frac xy+\frac yx=\frac{10}{3}$. What is $\frac{x+y}{x-y}$?