Maths Olympiad Prep

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, 2023

Algebra Difficulty 1.0 Junior Prove it Canada

The average of nn, 2n2n, 3n3n, 4n4n, 5n5n is 18. What is the value of nn?
Suppose that 2x+y=52x+y=5 and x+2y=7x+2y=7. What is the average of xx and yy?
The average of t2t^2, 2t2t and 33 is 9. If t<0t<0, determine the value of tt.

Solution

Since the average of the 5 numbers nn, 2n2n, 3n3n, 4n4n, and 5n5n is 18, we obtain the equation n+2n+3n+4n+5n5=18\dfrac{n+2n+3n+4n+5n}{5} = 18.

Therefore, 15n5=18\dfrac{15n}{5} = 18 and
so 3n=183n = 18 or n=6n = 6.
Solution 1

Adding the equations 2x+y=52x + y = 5
and x+2y=7x + 2y = 7, we obtain (2x+y)+(x+2y)=5+7(2x+y)+(x+2y)=5+7 and so 3x+3y=123x + 3y=12.

Therefore, the average of xx and
yy is $x+y2=3x+3y6=126\$\dfrac{x+y}{2} = \dfrac{3x+3y}{6} = \dfrac{12}{6}
= 2$.

Solution 2

Since 2x+y=52x + y = 5, then 4x+2y=104x + 2y = 10.

Subtracting the second equation, we obtain (4x+2y)(x+2y)=107(4x+2y)-(x+2y) = 10-7 which gives 3x=33x = 3 and so x=1x = 1.

Thus, y=52x=3y = 5-2x =3.

The average of xx and yy is thus 1+32=2\dfrac{1+3}{2} = 2.
Since the average of the three numbers t2t^2, 2t2t and 33 is 9, then t2+2t+33=9\dfrac{t^2 + 2t + 3}{3} = 9.

Therefore, t2+2t+3=27t^2 + 2t + 3 = 27 and so
t2+2t24=0t^2 + 2t - 24 = 0 which gives (t+6)(t4)=0(t + 6)(t - 4) = 0.

Since t<0t<0, then t=6t = -6.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.