Maths interview questions: 12 practice problems with solutions

Answer: Two solutions exactly when \(-\tfrac13\lt k\lt 1\).

Replacing \(x\) by \(-x\) changes nothing, so the graph is symmetric about the \(y\)-axis. It crosses the \(x\)-axis at \(x=\pm1\) and the \(y\)-axis at \(y=-\tfrac13\).

Write \(y=1-\dfrac{4}{x^2+3}\). As \(x^2\) grows from \(0\), the fraction \(\dfrac{4}{x^2+3}\) falls from \(\tfrac43\) towards \(0\), so \(y\) rises from \(-\tfrac13\) towards \(1\) and never reaches it. The minimum is \((0,-\tfrac13)\), there is no other turning point, and \(y=1\) is a horizontal asymptote approached from below on both sides. The range is \(-\tfrac13\le y\lt 1\).

A horizontal line \(y=k\) meets this graph twice exactly when \(-\tfrac13\lt k\lt 1\); once when \(k=-\tfrac13\) (at \(x=0\)); and not at all otherwise.

What the interviewer listens for: Say what you are checking before you draw: symmetry, intercepts, behaviour for large x. Rewriting the fraction is the step that turns guesswork into an argument; say why the range stops short of 1.

Follow-up question: How does the picture change for \(y=\dfrac{x^2-1}{x^2-3}\)?

Answer: \(n\equiv3\) or \(4 \pmod 5\); always odd; never a multiple of \(3\).

Work modulo \(5\): try \(n\equiv0,1,2,3,4\). The values of \(n^2+3n+7\) are \(7,11,17,25,35\), with remainders \(2,1,2,0,0\). So it is divisible by \(5\) exactly when \(n\) leaves remainder \(3\) or \(4\) on division by \(5\).

\(n^2+3n=n(n+3)\), and \(n\) and \(n+3\) have opposite parity, so one of them is even and \(n(n+3)\) is even. Adding \(7\) makes the total odd.

Modulo \(3\), \(3n\equiv0\) and \(7\equiv1\), so \(n^2+3n+7\equiv n^2+1\). A square leaves remainder \(0\) or \(1\) on division by \(3\), so \(n^2+1\) leaves \(1\) or \(2\): never \(0\).

What the interviewer listens for: Reducing an infinite question to a few cases is the key idea; say it out loud. Checking small n first is fine, as long as you then explain why the pattern must continue.

Follow-up question: Is there an \(n\) for which \(n^2+3n+7\) is divisible by \(25\)? By \(125\)?

Answer: Least value \(8\) at \(x=3\); for \(x\lt 0\) the greatest value is \(-4\) at \(x=-3\).

For \(x\gt 0\), multiplying by \(x\) keeps the inequality the same way round: \(x+\tfrac9x\ge6\iff x^2-6x+9\ge0\iff(x-3)^2\ge0\), which is true, with equality only at \(x=3\).

So \(\dfrac{x^2+2x+9}{x}=x+2+\dfrac9x\ge8\) for \(x\gt 0\), with equality at \(x=3\): the least value is \(8\).

For \(x\lt 0\), put \(x=-t\) with \(t\gt 0\): \(x+2+\tfrac9x=2-\left(t+\tfrac9t\right)\le2-6=-4\). So on the negative side the expression is at most \(-4\), reached at \(x=-3\), and there is no least value (it tends to \(-\infty\) as \(x\to0^-\)).

What the interviewer listens for: Notice the condition x > 0 and say where you use it. Completing the square, AM-GM and calculus all work; naming more than one route is a plus.

Follow-up question: Find the least value of \(x+\dfrac{9}{x^2}\) for \(x\gt 0\) without calculus.

Answer: \(u_n=1+4\cdot3^{n-1}\); a multiple of \(7\) exactly when \(6\mid n\).

The terms are \(5,13,37,109,\dots\); subtracting \(1\) gives \(4,12,36,108\), which multiply by \(3\) each time. So guess \(u_n=1+4\cdot3^{n-1}\).

Proof by induction: true for \(n=1\) since \(1+4=5\). If \(u_k=1+4\cdot3^{k-1}\), then \(u_{k+1}=3(1+4\cdot3^{k-1})-2=1+4\cdot3^{k}\), which is the formula for \(n=k+1\).

(Another route: \(u_{n+1}-1=3(u_n-1)\), so \(u_n-1\) is geometric with ratio \(3\).)

\(7\mid u_n\) means \(4\cdot3^{n-1}\equiv-1\equiv6\pmod7\). Multiply by \(2\) (since \(2\cdot4=8\equiv1\)): \(3^{n-1}\equiv12\equiv5\pmod7\). The powers \(3^0,3^1,\dots\) leave remainders \(1,3,2,6,4,5\) and then repeat with period \(6\), so \(3^{n-1}\equiv5\) exactly when \(n-1\equiv5\pmod6\), that is, when \(n\) is a multiple of \(6\). Check: \(u_6=1+4\cdot243=973=7\times139\).

What the interviewer listens for: Spotting the pattern is not enough; the interviewer wants to hear why it holds (induction, or the substitution v = u - 1). For the remainder part, organising the powers of 3 in a table is a good habit.

Follow-up question: Which recurrences \(u_{n+1}=au_n+b\) have a closed form like this? What goes wrong when \(a=1\)?

Answer: \(1\)

First check the integrand is defined on \([0,2]\): \(x^3+(2-x)^3=6x^2-12x+8=6(x-1)^2+2\gt 0\).

Let \(f(x)=\dfrac{x^3}{x^3+(2-x)^3}\). Then \(f(2-x)=\dfrac{(2-x)^3}{(2-x)^3+x^3}\), so \(f(x)+f(2-x)=1\).

Let \(I=\int_0^2 f(x)\,dx\). The substitution \(u=2-x\) gives \(I=\int_0^2 f(2-u)\,du\). Adding the two forms, \(2I=\int_0^2\big(f(x)+f(2-x)\big)\,dx=\int_0^2 1\,dx=2\), so \(I=1\).

A picture says the same: the graph of \(f\) is symmetric under a half-turn about \((1,\tfrac12)\), so the area under it is half the area of the \(2\times1\) rectangle.

What the interviewer listens for: Many candidates start expanding. It is fine to try that briefly and say it looks messy, then look for structure. Checking the denominator is never zero shows care.

Follow-up question: Show that \(\displaystyle\int_0^a\frac{g(x)}{g(x)+g(a-x)}\,dx=\frac a2\) whenever the denominator is never zero. Can you invent another integral this does?

Answer: \(\tfrac{37}{216}\); the most likely largest value is \(6\).

All three dice show at most \(k\) with probability \(\left(\tfrac k6\right)^3=\tfrac{k^3}{216}\). The largest is exactly \(k\) when all are at most \(k\) but not all at most \(k-1\), so

For \(k=4\): \(\dfrac{64-27}{216}=\dfrac{37}{216}\).

\(k^3-(k-1)^3=3k^2-3k+1\) increases with \(k\), so the most likely largest value is \(6\), with probability \(\dfrac{91}{216}\).

What the interviewer listens for: Counting the outcomes with largest exactly 4 directly is possible but error-prone; the complement-style trick is what they hope you find, perhaps after a hint. Sanity-check: the six probabilities should add to 1.

Follow-up question: What is the probability that the smallest number showing is \(4\)? What is the expected value of the largest?

Answer: \(126\) and \(1287\).

A \(0\) could only be the first digit, which is not allowed, so the digits come from \(1\) to \(9\).

Strictly increasing: choose any \(5\) different digits from the \(9\); there is exactly one increasing order. That is \(\binom95=126\).

Never decreasing: the number is fixed by how many times each digit \(1,\dots,9\) appears, with the counts adding to \(5\). Placing \(5\) identical counters into \(9\) boxes is choosing positions for \(5\) counters among \(5+8=13\) symbols (counters and the \(8\) dividers between boxes): \(\binom{13}5=1287\).

What the interviewer listens for: Reducing the order question to a choice question is the insight; say it before computing. If you do not know the counters-and-dividers argument, try a smaller case (two-digit numbers) and look for the pattern.

Follow-up question: How many five-digit numbers have digits that strictly increase and add up to \(25\)?

Answer: The circle with centre \(\left(\tfrac{16}3,0\right)\) and radius \(\tfrac83\); it does not meet the \(y\)-axis.

Squaring, \(x^2+y^2=4\big((x-4)^2+y^2\big)\), so \(3x^2+3y^2-32x+64=0\), that is \(x^2+y^2-\tfrac{32}3x+\tfrac{64}3=0\). Completing the square:

So \(P\) lies on the circle with centre \(\left(\tfrac{16}3,0\right)\) and radius \(\tfrac83\). It crosses the \(x\)-axis at \(x=\tfrac83\) and \(x=8\): check \(PA=\tfrac83=2\cdot\tfrac43\) and \(PA=8=2\cdot4\).

It does not meet the \(y\)-axis: the circle lies in \(\tfrac83\le x\le8\). Geometrically, a point on the \(y\)-axis is never further from \(A\) than from \(B\), so it cannot be twice as far.

What the interviewer listens for: Squaring is safe here because both sides are non-negative; saying so earns credit. The geometric reason for the last part is a nice touch after the algebra.

Follow-up question: What is the set of points with \(PA=\lambda\,PB\) for other \(\lambda\gt 0\)? What happens at \(\lambda=1\)?

Answer: \(2026\): no. \(2027=1014^2-1013^2\). \(2028\): three ways. Exactly the integers not \(\equiv2\pmod4\).

\(a^2-b^2=(a-b)(a+b)\), and \(a-b\) and \(a+b\) differ by \(2b\), so they are both odd or both even. If both are odd the product is odd; if both are even it is a multiple of \(4\). So a number that leaves remainder \(2\) on division by \(4\) is never a difference of two squares. \(2026=4\cdot506+2\): impossible.

Every odd number works: \(2m+1=(m+1)^2-m^2\). So \(2027=1014^2-1013^2\). Since \(2027\) is prime (it has no prime factor up to \(43\), and \(45^2\gt 2027\)), this is the only way with \(a\gt b\ge0\).

Every multiple of \(4\) works: \(4m=(m+1)^2-(m-1)^2\). For \(2028\), write \(a-b=2d\), \(a+b=2e\) with \(de=507=3\cdot13^2\) and \(d\lt e\): \((d,e)=(1,507),(3,169),(13,39)\), giving \(508^2-506^2\), \(172^2-166^2\) and \(52^2-26^2\).

So a positive integer is a difference of two squares exactly when it does not leave remainder \(2\) on division by \(4\).

What the interviewer listens for: Factorising a^2 - b^2 and thinking about parity is the expected first move. A full answer has both directions: why remainder 2 fails, and a construction for everything else.

Follow-up question: In how many ways can \(2^{10}\) be written as \(a^2-b^2\) with \(a\gt b\ge0\)?

Answer: \(3^{19}\) is larger.

\(2^{30}=8^{10}=8\cdot8^9\) and \(3^{19}=3\cdot9^9\). So

By the binomial theorem, keeping only the first three (positive) terms,

So \(\dfrac{3^{19}}{2^{30}}\ge\dfrac38\cdot\dfrac{43}{16}=\dfrac{129}{128}\gt 1\). Hence \(3^{19}\) is larger, though only by a few per cent.

What the interviewer listens for: Logarithms you happen to remember are not a proof unless you can justify the precision. A clean exact comparison like this, and saying how close the two numbers are, is what impresses.

Follow-up question: Which is larger, \(2^{100}\) or \(3^{63}\)? How confident can you be without a calculator?

Answer: \(0\) for \(k\lt 0\); \(1\) for \(k=0\); \(3\) for \(0\lt k\lt 4e^{-2}\); \(2\) for \(k=4e^{-2}\); \(1\) for \(k\gt 4e^{-2}\).

Let \(f(x)=x^2e^{-x}\). Then \(f(x)\ge0\), with \(f(0)=0\), and \(f'(x)=x(2-x)e^{-x}\). So \(f\) decreases on \((-\infty,0]\), increases on \([0,2]\) and decreases on \([2,\infty)\): a minimum \(f(0)=0\) and a maximum \(f(2)=4e^{-2}\).

As \(x\to-\infty\), \(f(x)\to\infty\); as \(x\to\infty\), \(f(x)\to0\) from above.

Counting where \(y=k\) meets the graph:

• \(k\lt 0\): no solutions;
• \(k=0\): one solution (\(x=0\));
• \(0\lt k\lt 4e^{-2}\): three solutions (one negative, one in \((0,2)\), one greater than \(2\));
• \(k=4e^{-2}\): two solutions (\(x=2\) and one negative);
• \(k\gt 4e^{-2}\): one solution (negative).

What the interviewer listens for: This is really a curve-sketching question. The interviewer listens for the limits at both ends and for the careful treatment of the boundary cases k = 0 and k = 4/e^2.

Follow-up question: For which \(k\) does \(x^3e^{-x}=k\) have exactly two solutions?

Answer: \(f(f(x))=x\): the graph is symmetric in \(y=x\). \(g\) is its own inverse for every \(a\ne-1\).

\(f(f(x))=\dfrac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=\dfrac{(x+1)+(x-1)}{(x+1)-(x-1)}=\dfrac{2x}{2}=x\) for \(x\ne1\).

So \(f\) is its own inverse, and its graph is symmetric in the line \(y=x\) (reflecting it in \(y=x\) gives the graph of the inverse, which is \(f\) again). It meets \(y=x\) where \(x^2-2x-1=0\), at \(x=1\pm\sqrt2\).

For \(g\), put \(g(x)\) into \(g\) and multiply top and bottom by \(x-1\): the top becomes \((x+a)+a(x-1)=(1+a)x\) and the bottom becomes \((x+a)-(x-1)=a+1\). So \(g(g(x))=x\) whenever \(a\ne-1\). When \(a=-1\), \(g(x)=1\) for every \(x\ne1\), a constant, which is not its own inverse.

What the interviewer listens for: Careful algebra out loud, then the geometric meaning. The special case a = -1 is exactly the kind of detail interviewers probe.

Follow-up question: For which \(p,q,r,s\) is \(h(x)=\dfrac{px+q}{rx+s}\) its own inverse?