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Problem 422

Algebra Difficulty 2.7 Multiple choice CEMC Gauss (Grade 7) · Canada · 2019

Each of a, b, c,a,~b,~c, and dd is a positive integer and is greater than 3. If 1a2=1b+2=1c+1=1d3\dfrac{1}{a-2}=\dfrac{1}{b+2}=\dfrac{1}{c+1}=\dfrac{1}{d-3} then which ordering of these four numbers is correct?

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Official solution

Solution 1

Each of the four fractions is equal to the other three fractions, and each fraction has numerator 1, so then each denominator must be equal to the other three denominators.

That is, a2=b+2=c+1=d3a-2=b+2=c+1=d-3.

If we let b=4b=4, then a2=4+2=c+1=d3a-2=4+2=c+1=d-3 or a2=6=c+1=d3a-2=6=c+1=d-3, and so a=8,c=5,a=8, c=5, and d=9d=9.

Thus, the correct ordering is b<c<a<db<c<a<d.

Solution 2

Each of the four fractions is equal to the other three fractions, and each fraction has numerator 1, so then each denominator must be equal to the other three denominators.

That is, a2=b+2=c+1=d3a-2=b+2=c+1=d-3, and by adding 3 to each, we get a+1=b+5=c+4=da+1=b+5=c+4=d.

Since d=a+1d=a+1, then dd is one more than aa and so a<da<d.

Since a+1=c+4a+1=c+4, then aa is 3 more than cc and so c<ac<a.

Since c+4=b+5c+4=b+5, then cc is 1 more than bb and so b<cb<c.

Thus, the correct ordering is b<c<a<db<c<a<d.

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