Olympiad Math | Find the values of a and b | Math olympiad preparation | higherMaths

Find the values of a and b if 2^(a) - 2^(b) = 2016.
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[faithlesshound]

The question involves powers of 2.
Try integer powers first.
2^10 = 1024, so 2^11 = 2048
RHS: 2016 = 2048 - 32 = 2^11 - 2^5
LHS: 2^a - 2^b = 2^11 - 2^5
Therefore a = 11, b = 5.

[devondevon]

a = 11 b=5 Answer
2^a - 2^b =2016
2^3 (2^a-3 - 2^b-3) =2016
factored out 8 (2^3) since 2016 is divisible by 8 since the last three digits, '016' is divisible by 8
2^a -3 - 2^b-3 = 2016/8 = 252
2^2 (2^a-5 - 2^b-5) =252
factor out 4 or 2^4 since 252 is divisible by 4 since the last two digits '52' is divisible by 4
2^a-5 - 2^b-5= 252/4 = 63
2^a -5 - 2^b -5 = 64 - 1
=2^6 - 2^0 since 2^6= 64 and 2^0 =1; hence
a-5 = 6 and b-5 = 0 since both have similar bases
a = 5+6 = 11
b =5+0=5
answer a =11 and b=5; hence
2^11 - 2^5 = 2016
2046 - 32 =2016

[sb]

Since you didn’t say that a and b are integers, a and b have infinity possibility

[joramarentved]

'A, ' 11, 'b:' 5.