Question bank SEAMO 2024

Simplify $\dfrac{11}{2\sqrt5-3}+\dfrac5{\sqrt5-2}-7\sqrt5$.

Evaluate $\sqrt{14+6\sqrt5}-\sqrt{14-6\sqrt5}$.

Find $\cos^2\theta+\cos^2(\theta+2\pi/3)+\cos^2(\theta-2\pi/3)$.

Factorise $x^3+4x^2+5x+2$.

All three sides of a right triangle are integers. One of the two legs is $35$. Find the minimum possible perimeter.

Rectangle $ABCD$ has diagonals meeting at $O$. Point $E$ lies on $AB$ and $CE$ bisects $\angle BCD$. Given $\angle ACE=15^\circ$, find $\angle BOE$.

Suppose $xy=144$ and $\log_yx+\log_xy=10/3$, with $x,y\gt 0$. Find $(x+y)/2$ in simplest form.

Given $x+1/x=3$, evaluate $x^{10}+x^5+1/x^5+1/x^{10}$.

Find the remainder when $(257^{33}+46)^{26}$ is divided by $50$.

A positive integer is friendly if divisible by the sum of its digits; for example $111$ is friendly but $123$ is not. How many two-digit friendly integers are there?

A candidate answers $7$ of $15$ questions split into groups $A,B,C$ with $4,5,6$ questions respectively. At least two must be chosen from each group. How many choices are possible?

In square $ABCD$, $M\in BC$, $N\in CD$ and $\angle MAN=45^\circ$. Given $BM=3.5$, $DN=2.5$, find $MN$.

A paper has $6$ questions. Exactly $500$ participants answered each question correctly. For any two participants, at least one question was answered incorrectly by both. Find the least number of participants.

$ABCD$ is a trapezium with $AD\parallel BC$, $AB=BC$, $AD=AE$, $\angle ABC=90^\circ$. Point $E$ lies on $AB$ with $\angle BEC=75^\circ$. Find $\angle DCE$.

Two intersecting circles have radii $8$ and $6$, and their centres are distance $12$ apart. A line through intersection $P$ meets the circles again at $Q,R$ with $QP=PR$. Find $(QP)^2$.

Find the remainder when $2024^{2024}$ is divided by seven.

Triangle ABC has $AB=3,BC=4,AC=5$. If $\sin2A=x/y$ in lowest positive terms, find x+y.

Find the remainder when $x^3+3x^2+2x+1$ is divided by $x-2$.

Evaluate $\sqrt[256]{(2+1)(2^2+1)(2^4+1)(2^8+1)\cdots(2^{256}+1)+1}$.

For $a\gt 1$, find the least value of $2\log_{100}a-\log_a(0.0001)$.