United States🇺🇸Maths Olympiad | A Tricky Algebraic Maths Olympiad Question.

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[GilbertCholet]

you can express 148 in base 2 which gives 10010100 which equals 1*2^7 +0*2^6 + 0*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 0* 2^1 + 0*2^0 = 2^7 +2^4 + 2^2 . So triplet (a, b, c) = ( 7.4.2) and permutation of these numbers are one and unique solution.

[drummerboy]

2^a+2^b+2^c=148
Convert 148 to base 10:
148=10010100 (in base 2)

a=7, b=4, c=2
So therefore, (a, b, c)=(7, 4, 2)

[marceliusmartirosianas]

2^(a+b+c)=2^74 abc= 37=36-1=36+1=36= [ abc=4^9]=[ abc=9 ] =3^2=5]=[ a+b+c=2^3]=[ tada a+vb+c=3 ] =a+b=2 ]=[2+c=3 =[ c=1] =[ =c=1 a+b=2]=[ b^a=2 a=2 b=2-2=1 b=1==== AcademiC Marcelius Martirosianas Mathematics econmoics.

[lubomirzak]

Technická poznámka : musíte psát na papíře který pruží, není pevně podložen?

[wes]

Largest power of 2 in 148 is 7
Largest power of 2 in 148–128 is 4
Power of 2 in 148–128–16 is 2

[marceliusmartirosianas]

=rezultat 2(a+b+c)=148= 2=148=146 abc=146 ab^c=146 c=146=c-146=c/146=1 146=c ab=c a^b=b a=1 tada b=c 2ca=ca=1 c146 a=1 b=146

[chance]

You can solve this by inspection, once you realize that 148 is the sum of 128 + 16 + 4, all powers of two.