Oxford German Olympiads Mathematics 2023 | A beautiful Olympiads Mathematics

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[juergenilse]

We can assue a<=b<=c (otherwise, we can rename the varabes, so that this is given) Now we can tae out 2^a out of the sum:

Since a<0b e hae the possibiities a=b and a<b. et us first oo at thhe case a=b:
2^a*(1+2^0+2^(c-a))=2^2*37
2^a*2*(1+2^(c-a-1))=2^1*2*37
Comparison of the exponents of 2 lead us to a=1 and therefor 1+2^(c-a-1)=37
2^(c-a--1)=36
But 36 is not a power of 2, so a=1 is possible (for whhole nummbers a, b and c).
So we now a<b<=c and
2^ a*(1+2^(b-a)+2^(c--a)=2^2*37
(1+2^ (b-a)+2^(ca)) is an odd number, so compaisoof exponents of 2 on both sides gives us
a=2 and
Now we can use the same trick:..

Again we oo at thhe 2 cases b=c and b<c. The first gives us no solution, so we get b<c and
2^(b-2)*(1+2^(c-b))=2^2*9
Since 1+2^(c-b)is odd, we get
2^b-2)=2^2 and 1+2^(c-b)=9
b-2=2 and 2^(c-b)=8=2^3
b=4 and c-b=3
b=2 and c=7
So the complete solution is a=2, b=4 and c=7.
Chec results in 2^a+2^b+2^ c=2^2+2^4+2^7=4+16+128=148

Why do you rewrite 2^(b-a) to (2^b)/(2^a)? it is unneccesarry. Just tae out 2^(b-a) outt of the su, and ou get

The first factor is a power of 2, the second factor is odd, so we get
2^(b-a)*(1+2^(c-b)=2^2*9
and than 2^(b-a)=2^2 and 1+2^(c-b)=9
b-a=2 and 2^(c-b)=9-1=8=2^3
b=a+2 and c=b+3
Togetthher withh a=2, we have a=2, b=2+2=4 and c=b+3=4+3=7