Elementary Mathematics 2020 (II)

Step-by-step solution

1. 1
   $x^3+x^2+16=x\left(x^2+x\right)+16$

   Why: Pull out every term that already carries a factor of $x$. What survives outside the bracket is the only thing that can obstruct the division.

   Dividend $=$ divisor $\times$ quotient $+$ remainder
2. 2
   $x\mid x^3+x^2+16⟺x\mid 16$

   Why: Since $x$ divides $x\left(x^2+x\right)$ exactly, the whole expression leaves the same remainder on division by $x$ as $16$ does. So the condition is simply that $x$ is a divisor of $16$.

3. 3
   $16=2^4$, so its positive divisors are $1,2,4,8,16$

   Why: A prime power $p^4$ has exactly the divisors $p^0,\dots ,p^4$ - five of them.

   Prime factorisation
4. 4
   Number of possible values of $x$ is $5$

   Why: Every positive divisor of $16$ works and nothing else does, so the count is exactly the number of divisors.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $30=2\times 3\times 5$

   Why: Three distinct primes, each to the first power. This is the fact that makes the count easy.

   Prime factorisation
2. 2
   Each of $2$, $3$, $5$ must go whole to exactly one of $a$, $b$, $c$

   Why: A prime appearing to the first power cannot be split between two factors, so for each prime there are exactly $3$ choices of where it lands.

   Count ordered factorisations by handing out the primes
3. 3
   $3\times 3\times 3=27$

   Why: The three choices are made independently, and two different assignments give two different triples $(a,b,c)$, so no triple is counted twice and none is missed.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $x^2-4x-{\log }_{10}N=0$ has $a=1,b=-4,c=-{\log }_{10}N$

   Why: Identify the coefficients before touching the discriminant; the constant term carries a minus sign that is easy to lose.

   Quadratic formula
2. 2
   $b^2-4ac\ge 0\Rightarrow 16+4{\log }_{10}N\ge 0$

   Why: Real roots means the discriminant is not negative. Here $-4ac=-4(1)\left(-{\log }_{10}N\right)=+4{\log }_{10}N$.

3. 3
   ${\log }_{10}N\ge -4$

   Why: Divide by $4$ and transpose. The question has become: how small can $N$ be if its logarithm is at least $-4$?

4. 4
   $N\ge 10^{-4}=\frac{1}{10000}$

   Why: ${\log }_{10}$ is increasing, so the inequality carries straight over to $N$ once each side is written as a power of $10$. The least value is attained at equality.

   Logarithm to exponential form

Common mistakes

Skills used

Step-by-step solution

1. 1
   $x$ can be $1,2,\dots ,10$

   Why: All three are at least $1$, so $x\le 12-1-1=10$.

   Count integer solutions by fixing one variable
2. 2
   For a fixed $x$, $y+z=12-x$ with $y\ge 1$ has $y=1,2,\dots ,11-x$

   Why: Once $y$ is chosen, $z=12-x-y$ is forced, so the number of solutions is the number of admissible $y$. The largest $y$ is $12-x-1=11-x$, which keeps $z\ge 1$.

3. 3
   Total $=10+9+8+\cdots +1$

   Why: Substituting $x=1,2,\dots ,10$ gives the counts $10,9,\dots ,1$.

4. 4
   $1+2+\cdots +10=\frac{10\times 11}{2}=55$

   Why: Sum of the first ten natural numbers.

   Sums of the first $n$ natural numbers, their squares and their cubes

Common mistakes

Skills used

Step-by-step solution

1. 1
   $a=n,b=n+1,c=ab=n^2+n$

   Why: 'Consecutive integers' fixes $b$ in terms of $a$; naming $a=n$ puts everything in one variable.

2. 2
   Put $m=n^2+n$; then $a^2+b^2=n^2+(n+1{)}^2=2n^2+2n+1=2m+1$ and $c^2=m^2$

   Why: Both pieces are built from the same quantity $n^2+n$, so naming it $m$ turns a quartic in $n$ into a quadratic in $m$.

3. 3
   $I=m^2+2m+1=(m+1{)}^2={\left(n^2+n+1\right)}^2$

   Why: $m^2+2m+1$ is the square of a sum. So $I$ is always a perfect square - that settles two of the four options at once.

   Square of a sum and of a difference
4. 4
   $n^2+n=n(n+1)$ is even, so $n^2+n+1$ is odd

   Why: Of two consecutive integers one is even, so their product is even and one more than it is odd. Hence $I$ is the square of an odd integer.

   Write an odd number as $2m+1$

Common mistakes

Skills used

Step-by-step solution

1. 1
   Odd numbers in order: $1$ (neither prime nor composite), $3,5,7$ (prime), $9=3^2$ (composite)

   Why: The smallest odd composite number is $9$, not $3$ - $3$ is prime, and $1$ is neither prime nor composite.

   Prime, composite, and the status of $1$
2. 2
   $2+3+6+2+9+7+1+3+3+5=41$

   Why: Add every digit except $P$. A number is divisible by $9$ exactly when its digit sum is, so the digit sum is all that matters.

   A number is congruent to its digit sum modulo $9$
3. 3
   $41+P$ must be a multiple of $9$

   Why: The full digit sum of $23P62971335$ is $41+P$.

4. 4
   $45$ is the only multiple of $9$ in range, so $41+P=45$ and $P=4$

   Why: $P$ is a single digit, so $41+P$ lies between $41$ and $50$; the only multiple of $9$ there is $45$. That also shows the answer is unique.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $1^5\equiv 1$; $2^5=32\equiv 0$; $4^5\equiv 0$

   Why: Any even number raised to a power at least $2$ is a multiple of $4$, so the two even terms contribute nothing.

   Reduce a large power by the cycle of its base modulo $n$
2. 2
   $3^2=9\equiv 1\Rightarrow 3^5={\left(3^2\right)}^2\cdot 3\equiv 1\cdot 3=3$

   Why: Powers of $3$ modulo $4$ cycle with period $2$: $3,1,3,1,\dots$, so an odd exponent leaves $3$.

3. 3
   $5\equiv 1\Rightarrow 5^5\equiv 1$

   Why: A base congruent to $1$ stays congruent to $1$ at every power.

4. 4
   Sum $\equiv 1+0+3+0+1=5\equiv 1(\mathrm{mod}4)$

   Why: Remainders add, and the total $5$ is then reduced again to bring it below the divisor.

   Dividend $=$ divisor $\times$ quotient $+$ remainder

Common mistakes

Skills used

Step-by-step solution

1. 1
   $3^1=3,3^2=9,3^3=27,3^4=81$: unit digits $3,9,7,1$

   Why: The unit digit of a power depends only on the previous unit digit, so once a digit repeats the whole pattern repeats. Here the cycle length is $4$.

   Unit digit of a power from its cycle
2. 2
   $99=4\times 24+3$

   Why: Divide the exponent by the cycle length. The remainder says how far into a fresh cycle the exponent lands.

3. 3
   Remainder $3\Rightarrow$ the same unit digit as $3^3$, which is $7$

   Why: The remainder is counted from $3^1$, so remainder $3$ means the third entry of the cycle, not the fourth. A remainder of $0$ would mean the last entry, $3^4$.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let $H$ be the HCF; then the LCM is $28H$

   Why: '$28$ times their HCF' is the only relation between them, so one letter suffices.

   Turn a worded condition into a linear equation in one unknown
2. 2
   $H+28H=1740\Rightarrow 29H=1740\Rightarrow H=60$

   Why: The sum condition now reads as a single linear equation. $1740=29\times 60$.

3. 3
   LCM $=28\times 60=1680$

   Why: Both quantities are now known.

4. 4
   $H\times \text{LCM}=60\times 1680=100800=240\times (\text{other number})$

   Why: For two numbers, HCF $\times$ LCM equals their product. The product is known and one factor is given.

   HCF $\times$ LCM = product of the two numbers
5. 5
   Other number $=\frac{100800}{240}=420$

   Why: Divide. Cancelling a zero first keeps this in small numbers: $\frac{10080}{24}=420$.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $x^n-a^n=(x-a)\left(x^{n-1}+x^{n-2}a+\cdots +a^{n-1}\right)$

   Why: The second bracket has $n$ terms, all with a plus sign, and is a polynomial for every positive integer $n$. So $x-a$ is a factor whatever $n$ is.

   $a-b$ divides $a^n-b^n$ for every $n$
2. 2
   By the factor theorem: put $x=a$ in $f(x)=x^n-a^n$, giving $f(a)=a^n-a^n=0$

   Why: The same conclusion with no factorisation to write out: $x-a$ divides $f(x)$ exactly when $f(a)=0$, and that holds for every natural $n$.

   Factor and remainder theorem
3. 3
   Hence $(x-a)\mid \left(x^n-a^n\right)$ for every natural number $n$

   Why: No parity or primality condition appears anywhere in the argument.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $17=18-1$, so $17\equiv -1(\mathrm{mod}18)$

   Why: Working with $-1$ instead of $17$ is the whole trick: a base of $-1$ makes any power trivial.

   Reduce a large power by the cycle of its base modulo $n$
2. 2
   $17^{2020}\equiv (-1{)}^{2020}=1(\mathrm{mod}18)$

   Why: $2020$ is even, so the sign comes out positive.

3. 3
   Remainder $=1$

   Why: $1$ lies between $0$ and $17$, so it is a legitimate remainder and no further reduction is needed.

   Dividend $=$ divisor $\times$ quotient $+$ remainder

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}$

   Why: Multiplying above and below by the conjugate turns the denominator into a difference of squares, which clears the surds.

   Rationalise a denominator using the conjugate
2. 2
   $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$

   Why: The denominator is exactly $1$, so each term is simply the difference of the two roots in it.

3. 3
   Sum $=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\cdots +\left(\sqrt{100}-\sqrt{99}\right)$

   Why: Writing the terms out shows every interior root appearing once with each sign.

   Telescoping a sum by writing each term as a difference
4. 4
   $=\sqrt{100}-\sqrt{1}=10-1=9$

   Why: Only the last root and the first survive the cancellation. The $-\sqrt{1}$ is the piece most often dropped.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $\sqrt{x}=x^{1/2}$, so $x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$

   Why: Start at the innermost radical. Adding indices is the only law needed when the bases match.

   Power of a power, and the move to a common base
2. 2
   $\sqrt{x\sqrt{x}}={\left(x^{3/2}\right)}^{1/2}=x^{3/4}$

   Why: A square root is the index $\frac{1}{2}$, and a power of a power multiplies the indices.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
3. 3
   $x\sqrt{x\sqrt{x}}=x\cdot x^{3/4}=x^{7/4}$

   Why: The outer $x$ inside the fourteenth root contributes an index of $1$.

4. 4
   $\sqrt[14]{x^{7/4}}={\left(x^{7/4}\right)}^{1/14}=x^{7/56}=x^{1/8}$

   Why: The fourteenth root is the index $\frac{1}{14}$; $\frac{7}{4}\times \frac{1}{14}=\frac{7}{56}=\frac{1}{8}$.

5. 5
   $x^m=x^{1/8}\Rightarrow m=\frac{1}{8}$

   Why: Equal powers of the same base force equal indices.

Common mistakes

Skills used

Step-by-step solution

1. 1
   The ten numbers are $1,-1,2,-2,3,-3,4,-4,5,-5$, and their sum is $0$

   Why: Each number pairs with its negative, so the total is zero. This is the fact the identity will exploit.

2. 2
   ${\left(\sum a_i\right)}^2=\sum a_i^2+2{\sum }_{i<j}{a}_i{a}_j$

   Why: Squaring a sum produces every square once and every product two at a time twice - which is exactly the quantity asked for.

   Square of a sum of several terms
3. 3
   $\sum a_i^2=2\left(1^2+2^2+3^2+4^2+5^2\right)=2\times 55=110$

   Why: A number and its negative have the same square, so each square appears twice; $1+4+9+16+25=55$.

   Sums of the first $n$ natural numbers, their squares and their cubes
4. 4
   $0=110+2{\sum }_{i<j}{a}_i{a}_j\Rightarrow {\sum }_{i<j}{a}_i{a}_j=-55$

   Why: Substitute the sum $0$ and the sum of squares $110$, then solve for the pairwise sum.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Distance $=110+165=275$ m

   Why: Crossing a bridge is complete when the rear of the train leaves it, so the train travels its own length in addition to the bridge's.

   What a train covers when it crosses something
2. 2
   $132$ km/hr $=132\times \frac{5}{18}=\frac{110}{3}$ m/s

   Why: The lengths are in metres, so the speed must be in metres per second; multiplying by $\frac{5}{18}$ does that.

   km/hour to m/second
3. 3
   Time $=\frac{275}{110/3}=\frac{275\times 3}{110}=7.5$ s

   Why: Time is distance over speed. Cancelling $55$ from $275$ and $110$ keeps the arithmetic small: $\frac{5\times 3}{2}$.

   Speed, distance and time

Common mistakes

Skills used

Step-by-step solution

1. 1
   Let the number of years be $n$; then the rate is also $n\%$

   Why: 'Numerically equal' is the single condition that makes one unknown do for both.

2. 2
   $\text{SI}=\frac{P\times n\times n}{100}=\frac{P{n}^2}{100}$

   Why: Simple interest is $\frac{PRT}{100}$, with $R=T=n$.

   Simple interest $=\frac{PRT}{100}$
3. 3
   $\frac{P{n}^2}{100}=\frac{P}{4}$

   Why: The interest is given as one-fourth of the sum. The principal appears on both sides and cancels, so its value is never needed.

4. 4
   $n^2=25\Rightarrow n=5$

   Why: Multiply by $\frac{100}{P}$ and take the positive root; a number of years cannot be negative.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Original space $=60n$ lines

   Why: $60$ pages at $n$ lines each.

   Turn a worded condition into a linear equation in one unknown
2. 2
   New space $=(60+10)(n-3)=70(n-3)$

   Why: Three fewer lines per page, ten more pages. 'To give the same writing space' is what makes these two equal.

3. 3
   $70(n-3)=60n$

   Why: Set the two totals equal - this is the one stated relation.

4. 4
   $70n-210=60n\Rightarrow 10n=210\Rightarrow n=21$

   Why: Expand, collect the $n$ terms on one side, divide.

Common mistakes

Skills used

Step-by-step solution

1. 1
   Work $=c\times (\text{men})\times (\text{hours per day})\times (\text{days})$

   Why: The three factors enter multiplicatively, which is the content of the work-equivalence rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$ written with a constant.

   The work-equivalence chain rule $M_1{D}_1{H}_1=M_2{D}_2{H}_2$
2. 2
   $x=c\cdot x\cdot x\cdot x=c{x}^3$

   Why: The first statement supplies all four quantities, so it determines $c$.

3. 3
   $c=\frac{1}{x^2}$

   Why: Divide both sides by $x^3$: $c=\frac{x}{x^3}$.

4. 4
   $k=c\cdot y\cdot y\cdot y=\frac{y^3}{x^2}=y^3{x}^{-2}$

   Why: The same constant governs the second workforce. A negative index is just a reciprocal, so $\frac{1}{x^2}=x^{-2}$.

   Negative exponent is a reciprocal

Common mistakes

Skills used

Step-by-step solution

1. 1
   $5$ and $11$ are prime, so $d(5)=d(11)=2$

   Why: A prime has exactly two divisors, $1$ and itself. Statement $1$ is therefore correct.

   Prime factorisation
2. 2
   $55=5\times 11\Rightarrow d(55)=(1+1)(1+1)=4$

   Why: For $n=p_1^{e_1}{p}_2^{e_2}\cdots$ the number of divisors is $(e_1+1)(e_2+1)\cdots$. Here the divisors are $1,5,11,55$.

3. 3
   $d(5)\cdot d(11)=2\times 2=4=d(55)$

   Why: Statement $2$ is correct. This is the multiplicative property of $d$ for coprime arguments, and $5$ and $11$ share no prime.

4. 4
   $16=2^4\Rightarrow d(16)=4+1=5$

   Why: The divisors are $1,2,4,8,16$ - five of them. The exponent is $4$, so the count is $5$, not $4$.

5. 5
   $d(5)+d(11)=4\ne 5=d(16)$

   Why: Statement $3$ fails. So statements $1$ and $2$ hold and $3$ does not.

Common mistakes

Skills used

Step-by-step solution

1. 1
   $P_n=2\times 3\times 5\times \cdots$ always includes the prime $2$, so $P_n$ is even

   Why: $2$ is the first prime, so it is a factor of $P_n$ for every $n\ge 1$. This one observation drives the whole question.

   Prime, composite, and the status of $1$
2. 2
   $A_n=P_n+1$ is therefore odd

   Why: One more than an even number is odd.

   Write an odd number as $2m+1$
3. 3
   $A_n+2$ is odd $+$ even $=$ odd: statement $2$ is correct

   Why: Adding $2$ does not change parity.

4. 4
   $A_n+1$ is odd $+$ odd $=$ even: statement $3$ is correct

   Why: One more than an odd number is even.

5. 5
   $n=1$: $P_1=2$ and $A_1=3$, which is prime - statement $1$ fails

   Why: A single counterexample is enough to reject an 'always' claim. ($n=2$ gives $7$ and $n=3$ gives $31$, both prime as well; the first composite case is $n=6$, where $30031=59\times 509$.)

Common mistakes

Skills used