Problem 27: Australian Mathematics Competition 2023 – Junior Year

Let's discuss a problem from the AMC 2023 Junior Category: Problem 27 which revolves around basic algebra.

Question

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Amelia noticed that the names of three friends, Mei, Emma and Liam, were all made from letters of 'Amelia'. She chose different values from 0 to 9 for each of (A, E, I, L) and (M) so that

\(M+E+I=E+M+M+A=L+I+A+M\)

What is the largest possible value of \(\mathrm{A}+\mathrm{M}+\mathrm{E}+\mathrm{L}+\mathrm{I}+\mathrm{A}\) ?

Solution

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From the given expression : \(M + E + I = L + I + A + M\)

If we cancel out \(M\) and \(I\) then we are left with: \(E = L + A\)...(1)

Also, \(M + E + I = E + M + M + A\)

Again, if we cancel out \(M\) and \(E\) then, \(I = M + A\).............(2)

So in, \(\mathrm{A}+\mathrm{M}+\mathrm{E}+\mathrm{L}+\mathrm{I}+\mathrm{A} = I + E + I + E\).

\(\mathrm{A}+\mathrm{M}+\mathrm{E}+\mathrm{L}+\mathrm{I}+\mathrm{A} = 2(E + I)\)

Thus, to get the maximum value, \(E\) must be equal to 9 or 8 and \(I\) must be equal to 8 or 9.
Thus the maximum value for \(\mathrm{A}+\mathrm{M}+\mathrm{E}+\mathrm{L}+\mathrm{I}+\mathrm{A} = 2 \times 17 = 34\).

Thus the maximum value is \(34\).

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