Elementary Mathematics 2018 (I)

Step-by-step solution

1. 1
   $5^{17}+5^{18}+5^{19}+5^{20}=5^{17}\left(1+5+5^2+5^3\right)$

   Why: Every term shares the factor $5^{17}$, and taking it out reduces a sum of four huge powers to $5^{17}$ times a small integer.

   Power of a power, and the move to a common base
2. 2
   $1+5+5^2+5^3=1+5+25+125=156$

   Why: The bracket is a short sum that can be added directly.

3. 3
   $156=2^2\times 3\times 13$, so the whole expression is $5^{17}\times 2^2\times 3\times 13$

   Why: Prime-factorising $156$ lists every prime divisor; $5^{17}$ contributes only the prime $5$, so the primes available are $2$, $3$, $5$ and $13$.

   Prime factorisation
4. 4
   $13$ divides the expression; $7$, $9$ and $11$ do not

   Why: Only $13$ appears among the prime powers, and $9=3^2$ fails because $3$ occurs to the first power only.

Common mistakes

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Step-by-step solution

1. 1
   $a+b=2c\Rightarrow b=2c-a$

   Why: The single given relation lets $b$ be removed from the expression entirely.

   An evenly spaced run averages to its middle term
2. 2
   $b-c=(2c-a)-c=c-a=-(a-c)$

   Why: Because $c$ is the mean of $a$ and $b$, the two numbers sit symmetrically about $c$: whatever $a$ is above $c$, $b$ is below it.

3. 3
   $\frac{a}{a-c}+\frac{c}{b-c}=\frac{a}{a-c}-\frac{c}{a-c}$

   Why: Substituting the previous line turns the second denominator into the first one, up to a sign.

   Clear the denominators of an equation
4. 4
   $\frac{a-c}{a-c}=1$

   Why: One common denominator, and the numerator is exactly that denominator. Note $a\ne c$ is forced by the expression being defined at all.

Common mistakes

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Step-by-step solution

1. 1
   $x=y^{1/a}={\left(z^{1/b}\right)}^{1/a}=z^{1/(ab)}$

   Why: Substituting $y=z^{1/b}$ and multiplying the indices brings $x$ onto the base $z$.

   Power of a power, and the move to a common base
2. 2
   $x={\left(x^{1/c}\right)}^{1/(ab)}=x^{1/(abc)}$

   Why: Substituting $z=x^{1/c}$ closes the cycle and puts both sides on the base $x$.

3. 3
   $x^1=x^{1/(abc)}$ with $x\ne 1$, so $1=\frac{1}{abc}$

   Why: For a positive base other than $1$, equal powers force equal indices.

4. 4
   $abc=1$

   Why: Invert the previous line.

Common mistakes

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Step-by-step solution

1. 1
   $2b=a+c\Rightarrow b-a=c-b=d$, so $b=a+d$ and $c=a+2d$

   Why: A middle term equal to the mean of its neighbours is exactly the condition for three terms in arithmetic progression, so one common difference $d$ describes all three.

   An evenly spaced run averages to its middle term
2. 2
   $b-c=-d,c-a=2d,a-b=-d$

   Why: Each index in the expression is now a multiple of the single unknown $d$.

3. 3
   $x^{b-c}{y}^{c-a}{z}^{a-b}=x^{-d}{y}^{2d}{z}^{-d}={\left(\frac{y^2}{xz}\right)}^d$

   Why: Collecting the common index $d$ turns the product into a single power of $\frac{y^2}{xz}$, which is the quantity the second given controls.

   Power of a power, and the move to a common base
4. 4
   $y^2=xz\Rightarrow \frac{y^2}{xz}=1$, so the value is $1^d=1$

   Why: A negative index is a reciprocal, which is what lets $x^{-d}{z}^{-d}$ become the denominator ${\left(xz\right)}^d$.

   Negative exponent is a reciprocal

Common mistakes

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Step-by-step solution

1. 1
   A rational $\frac{p}{q}$ in lowest terms terminates exactly when $q=2^m{5}^n$; otherwise it is non-terminating but recurring.

   Why: This is the full description of the rational case, and it shows at once that neither 'always terminating' nor 'always non-terminating' is true of rationals.

   When a rational number's decimal expansion terminates
2. 2
   $\frac{1}{2}=0.5$ terminates while $\frac{1}{3}=0.\overline{3}$ does not, so options A and B each fail on the other's example.

   Why: One counterexample apiece is enough to reject both halves of the rational claim.

3. 3
   Any terminating decimal equals $\frac{\text{digits}}{10^k}$, a rational, so an irrational number's expansion cannot terminate. Option C is false.

   Why: A terminating expansion is itself a proof of rationality, which an irrational number cannot have.

4. 4
   A recurring expansion also converts to a fraction, so an irrational expansion can be neither terminating nor recurring: it is non-terminating and non-repeating.

   Why: Rationals occupy exactly the terminating and the recurring expansions, so the irrationals are precisely what is left.

   Convert a recurring decimal to a fraction

Common mistakes

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Step-by-step solution

1. 1
   Let the roots be $\alpha$ and $\frac{1}{\alpha }$.

   Why: 'Reciprocal to each other' is exactly the statement that one root is the reciprocal of the other.

   Sum and product of roots (Vieta)
2. 2
   Product of roots $=\alpha \cdot \frac{1}{\alpha }=1$

   Why: Whatever $\alpha$ is (and it cannot be $0$, since $0$ has no reciprocal), the product is $1$.

3. 3
   For $p{x}^2+x+r=0$, product of roots $=\frac{r}{p}$, so $\frac{r}{p}=1$

   Why: Vieta's product relation is $\frac{c}{a}$ with $a=p$ and $c=r$.

4. 4
   $r=p$, i.e. $p=r$

   Why: Cross-multiplying, valid because $p\ne 0$ for the equation to be a quadratic at all.

Common mistakes

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Step-by-step solution

1. 1
   The coefficient ratios differ, $\frac{65}{33}\ne \frac{-33}{-65}$, so the pair has exactly one solution.

   Why: Confirms before any work that a unique $(x,y)$, and hence a unique $xy$, exists.

   When a pair of linear equations meets, is parallel, or coincides
2. 2
   Adding: $(65x-33y)+(33x-65y)=97+1\Rightarrow 98x-98y=98\Rightarrow x-y=1$

   Why: The $x$ coefficients and the $y$ coefficients each sum to $98$, so one addition gives a clean equation with unit coefficients.

   Add and subtract when the coefficients are reversed
3. 3
   Subtracting: $(65x-33y)-(33x-65y)=97-1\Rightarrow 32x+32y=96\Rightarrow x+y=3$

   Why: The same near-symmetry makes the difference collapse to $32(x+y)$.

4. 4
   $x-y=1$ and $x+y=3$ give $x=2$, $y=1$, so $xy=2$

   Why: Adding and halving gives $x$; subtracting and halving gives $y$.

Common mistakes

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Step-by-step solution

1. 1
   From $\frac{b}{y}+\frac{z}{c}=1$: $\frac{z}{c}=1-\frac{b}{y}=\frac{y-b}{y}$

   Why: Isolating the term the two equations share is what lets them be joined.

   Clear the denominators of an equation
2. 2
   Taking reciprocals, $\frac{c}{z}=\frac{y}{y-b}$

   Why: The second equation contains $\frac{c}{z}$, not $\frac{z}{c}$, so invert. This needs $y\ne b$, which holds because $\frac{z}{c}$ is defined and non-zero.

3. 3
   Substituting into $\frac{c}{z}+\frac{x}{a}=1$: $\frac{x}{a}=1-\frac{y}{y-b}=\frac{-b}{y-b}$

   Why: With $\frac{c}{z}$ known in terms of $y$ and $b$, the second equation becomes a relation between $x$, $a$, $y$ and $b$ alone — the four letters the question asks about.

4. 4
   $x(y-b)=-ab\Rightarrow xy-bx=-ab\Rightarrow ab+xy=bx$

   Why: Cross-multiplying and collecting puts the required numerator $ab+xy$ against the required denominator $bx$.

5. 5
   $\frac{ab+xy}{bx}=\frac{bx}{bx}=1$

   Why: The numerator is literally the denominator.

Common mistakes

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Step-by-step solution

1. 1
   $\frac{a^2-1}{a}=a-\frac{1}{a}=5$

   Why: Splitting the fraction term by term shows the given is the classic 'number minus its reciprocal'.

2. 2
   $\frac{a^6-1}{a^3}=a^3-\frac{1}{a^3}$

   Why: The same split applied to the target: it is the cube version of the given.

3. 3
   $a^2+\frac{1}{a^2}={\left(a-\frac{1}{a}\right)}^2+2=25+2=27$

   Why: Squaring $a-\frac{1}{a}$ produces $a^2-2+\frac{1}{a^2}$, since the cross term is $-2\cdot a\cdot \frac{1}{a}=-2$.

   Square of a sum and of a difference
4. 4
   $a^3-\frac{1}{a^3}=\left(a-\frac{1}{a}\right)\left(a^2+a\cdot \frac{1}{a}+\frac{1}{a^2}\right)=5\left(27+1\right)=140$

   Why: Difference of cubes with $b=\frac{1}{a}$: the middle term of the quadratic factor is $ab=1$.

   Sum and difference of two cubes

Common mistakes

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Step-by-step solution

1. 1
   $y+z=-x$, so $y+z-x=-x-x=-2x$; likewise $z+x-y=-2y$ and $x+y-z=-2z$

   Why: The condition $x+y+z=0$ replaces each pair sum by minus the remaining variable, which is what makes every bracket collapse.

   $a^3+b^3+c^3-3abc$, and the case $a+b+c=0$
2. 2
   $(y+z-x{)}^3+(z+x-y{)}^3+(x+y-z{)}^3=(-2x{)}^3+(-2y{)}^3+(-2z{)}^3=-8\left(x^3+y^3+z^3\right)$

   Why: Cubing $-2$ gives $-8$, a common factor of all three terms.

3. 3
   Since $x+y+z=0$, $x^3+y^3+z^3=3xyz$

   Why: The identity $a^3+b^3+c^3-3abc=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$ has a vanishing first factor when the three add to zero.

4. 4
   $-8\times 3xyz=-24xyz$

   Why: Combine the two previous lines.

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Step-by-step solution

1. 1
   Let $f(x)=x^3+3x^2+4x+k$. Then $(x+3)$ is a factor $\iff f(-3)=0$.

   Why: The factor theorem converts a divisibility statement into a single evaluation, so no long division is needed.

   Factor and remainder theorem
2. 2
   $f(-3)=(-3{)}^3+3(-3{)}^2+4(-3)+k=-27+27-12+k$

   Why: Evaluate term by term, watching that $(-3{)}^3=-27$ is negative while $3(-3{)}^2=+27$ is positive.

3. 3
   $-12+k=0\Rightarrow k=12$

   Why: The first two terms cancel, leaving a one-line equation for $k$.

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Step-by-step solution

1. 1
   A four-digit integer is one with $1000\le N\le 9999$, so the target is the least perfect square $N\ge 1000$.

   Why: Fixes exactly what has to be minimised; 'smallest four-digit' is a lower bound, not the number $1000$ itself.

   Least perfect square not below a given bound
2. 2
   $31^2=961<1000$ and $32^2=1024\ge 1000$

   Why: Squares increase with the base, so the first base whose square clears $1000$ gives the first four-digit square; $31$ is short and $32$ clears it.

3. 3
   $1000=2^3\times 5^3$ has odd indices, so it is not a perfect square; $1024=2^{10}={\left(2^5\right)}^2$ is.

   Why: Prime factorisation settles both candidates outright: a perfect square has every prime index even.

   Prime factorisation
4. 4
   Every integer strictly between $961$ and $1024$ lies between two consecutive squares, so none is a square. The answer is $1024$.

   Why: This is what makes $1024$ the smallest rather than merely a four-digit square.

Common mistakes

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Step-by-step solution

1. 1
   Let $p(x)=3x^3+4x^2-7$. A zero is any $r$ with $p(r)=0$.

   Why: This is the definition the factor theorem rests on, and it makes the search a matter of evaluation.

   Factor and remainder theorem
2. 2
   The constant term is $-7$ and the coefficients sum to $3+4-7=0$, so $p(1)=0$.

   Why: For any polynomial, $p(1)$ is just the sum of the coefficients, so a coefficient sum of zero exhibits $x=1$ as a root immediately.

3. 3
   $x=1$ is a zero, and correspondingly $3x^3+4x^2-7=(x-1)\left(3x^2+7x+7\right)$

   Why: Dividing out confirms the root and shows the remaining quadratic has discriminant $49-84<0$, so $1$ is the only real zero.

Common mistakes

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Step-by-step solution

1. 1
   $\text{HCF}=21$, so the numbers are $21m$ and $21n$ with $\text{HCF}(m,n)=1$.

   Why: Pulling the HCF out of both numbers leaves a coprime pair, since any shared factor of $m$ and $n$ would enlarge the HCF.

   Coprime means no shared prime factor
2. 2
   $\text{HCF}\times \text{LCM}=21\times 3003=21m\times 21n$, so $mn=\frac{3003}{21}=143$

   Why: The product identity turns the two given quantities into a single equation for $mn$.

   HCF $\times$ LCM = product of the two numbers
3. 3
   $143=11\times 13$, so the coprime factorisations of $143$ are $1\times 143$ and $11\times 13$.

   Why: Both $11$ and $13$ are prime, so these are the only two ways to split $143$, and both splits happen to be coprime.

   Prime factorisation
4. 4
   $m=1$ would give the number $21$, which is not greater than $21$; so $\{m,n\}=\{11,13\}$ and the numbers are $231$ and $273$.

   Why: The condition 'both greater than $21$' is what eliminates the trivial split and makes the answer unique.

5. 5
   $231+273=504$

   Why: The sum asked for.

Common mistakes

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Step-by-step solution

1. 1
   $(\alpha +1)(\beta +1)=\alpha \beta +\alpha +\beta +1$

   Why: Expanding first means only symmetric functions of the roots are needed, so the roots never have to be found.

   Sum and product of roots (Vieta)
2. 2
   For $a{x}^2+bx+c=0$: $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

   Why: Vieta's relations, with the minus sign sitting on the sum, not on the product.

3. 3
   $(\alpha +1)(\beta +1)=\frac{c}{a}-\frac{b}{a}+1=\frac{c-b+a}{a}=\frac{a-b+c}{a}$

   Why: Putting everything over $a$ collects the three terms into one fraction.

Common mistakes

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Step-by-step solution

1. 1
   Let $f(x)=3x^3+k{x}^2+5x-6$. Dividing by $(x+1)$ leaves the remainder $f(-1)$.

   Why: The remainder theorem replaces a long division by a single evaluation.

   Factor and remainder theorem
2. 2
   $f(-1)=3(-1{)}^3+k(-1{)}^2+5(-1)-6=-3+k-5-6=k-14$

   Why: Only the $k{x}^2$ term keeps its sign, since $(-1{)}^2=1$ while $(-1{)}^3=-1$.

3. 3
   $k-14=-7\Rightarrow k=7$

   Why: Set the evaluated remainder equal to the given remainder.

Common mistakes

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Step-by-step solution

1. 1
   Every term of $f$ has degree at most $p$ and every term of $g$ at most $q$, so every term of $f\pm g$ has degree at most $\max (p,q)$.

   Why: Addition never creates a power that was not already present in one of the summands, so the degree cannot rise.

   Degree of a product of polynomials
2. 2
   When the leading terms do not cancel, $\mathrm{deg}(f\pm g)=\max (p,q)$; e.g. $f=x^3$, $g=x^2$ gives degree $3$.

   Why: Equality is attained, so the bound cannot be tightened to a strict inequality.

3. 3
   When they do cancel the degree drops: $f=x^3+x$, $g=x^3$ gives $f-g=x$, of degree $1<\max (3,3)$.

   Why: This single example kills 'greater than $\min (p,q)$' and 'equal to $\min (p,q)$' alike, since here $\min (p,q)=3$ while the degree is $1$.

4. 4
   Hence $\mathrm{deg}\{f(x)\pm g(x)\}\le \max (p,q)$, and nothing stronger holds in general.

   Why: The bound is exactly the one that is always true and sometimes tight.

Common mistakes

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Step-by-step solution

1. 1
   Write $A=\sqrt{5}-\sqrt{3}$ and $B=\sqrt{5}+\sqrt{3}$; the expression is $\frac{A}{B}-\frac{B}{A}=\frac{A^2-B^2}{AB}$.

   Why: Naming the two conjugates turns a messy surd expression into one difference of squares over one product.

   Rationalise a denominator using the conjugate
2. 2
   $AB={\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2=5-3=2$

   Why: A conjugate pair multiplies to a difference of squares, which clears both surds at once.

   Difference of two squares
3. 3
   $A^2-B^2=(A-B)(A+B)=\left(-2\sqrt{3}\right)\left(2\sqrt{5}\right)=-4\sqrt{15}$

   Why: Factorising as a difference of squares avoids expanding $A^2=8-2\sqrt{15}$ and $B^2=8+2\sqrt{15}$ separately.

4. 4
   $\frac{-4\sqrt{15}}{2}=-2\sqrt{15}$

   Why: Divide; the result is negative because $B>A>0$, so $\frac{A}{B}<1<\frac{B}{A}$.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$

Common mistakes

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Step-by-step solution

1. 1
   First term: multiply numerator and denominator by $x^a$, giving $\frac{x^a}{x^a+x^a\cdot x^{b-a}+x^a\cdot x^{c-a}}=\frac{x^a}{x^a+x^b+x^c}$

   Why: The indices in the denominator are all differences measured from $a$, so multiplying by $x^a$ restores them to $x^a$, $x^b$ and $x^c$.

   Power of a power, and the move to a common base
2. 2
   Second term, multiplied above and below by $x^b$: $\frac{x^b}{x^b+x^a+x^c}$

   Why: Same move with $b$ as the reference index, and the denominator is the same three-term sum.

3. 3
   Third term, multiplied above and below by $x^c$: $\frac{x^c}{x^c+x^a+x^b}$

   Why: And again with $c$, which is why the three fractions can now simply be added.

   Negative exponent is a reciprocal
4. 4
   Sum $=\frac{x^a+x^b+x^c}{x^a+x^b+x^c}=1$

   Why: The numerators add to the common denominator, which is non-zero for $x>0$.