Problem 27: Australian Mathematics Competition 2023 – Junior Year
Join Trial or Access Free ResourcesJoin Trial or Access Free ResourcesLet's discuss a problem from the AMC 2023 Junior Category: Problem 27 which revolves around basic algebra.
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 \(\mathrm{A}+\mathrm{M}+\mathrm{E}+\mathrm{L}+\mathrm{I}+\mathrm{A}\) ?
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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