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Published byPiper Goodspeed Modified over 2 years ago

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MATHCOUNTS TOOLBOX Facts, Formulas and Tricks

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Lesson 5: Different Bases base 10 to base 2

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Base 10: decimal – only has digits 0 - 9

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Changing from Base 10 to Base 2: binary – only has digits 0, 1

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What is the base 2 representation of 125 base 10 or = ? 2

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We know 125 = 1(10 2 ) + 2(10 1 ) + 5(10 0 ) = , but what is it equal to in base 2?

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= ?(2 n ) + ?(2 n-1 ) + … + ?(2 0 )

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The largest power of 2 in 125 is 64 = 2 6, so we now know our base 2 number will be:

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?(2 6 ) + ?(2 5 ) + ?(2 4 ) + ?(2 3 ) + ?(2 2 ) + ?(2 1 )+ ?(2 0 ) and it will have 7 digits

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Since there is one 64, we have: 1(2 6 ) + ?(2 5 ) + ?(2 4 ) + ?(2 3 ) + ?(2 2 ) + ?(2 1 )+ ?(2 0 )

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We now have 125 – 64 = 61 left over, which is one 32 = 2 5 and 29 left over,

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so we have: 1(2 6 ) + 1(2 5 ) + ?(2 4 ) + ?(2 3 ) + ?(2 2 ) + ?(2 1 )+ ?(2 0 )

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In the left-over 29, there is one 16 = 2 4, with 13 left over,

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so we have: 1(2 6 ) + 1(2 5 ) + 1 (2 4 ) + ?(2 3 ) + ?(2 2 ) + ?(2 1 )+ ?(2 0 )

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In the left-over 13, there is one 8 = 2 3, with 5 left over,

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so we have: 1(2 6 ) + 1(2 5 ) + 1(2 4 ) + 1(2 3 ) + ?(2 2 ) + ?(2 1 )+ ?(2 0 )

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In the left-over 5, there is one 4 = 2 2, with 1 left over,

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so we have: 1(2 6 ) + 1(2 5 ) + 1(2 4 ) + 1(2 3 ) + 1(2 2 ) + ?(2 1 )+ ?(2 0 )

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In the left-over 1, there is no 2 = 2 1, so we still have 1 left over,

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and our expression is: 1(2 6 ) + 1(2 5 ) + 1(2 4 ) + 1(2 3 ) + 1(2 2 ) + 0(2 1 )+ ?(2 0 )

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The left-over 1 is one 2 0, so we finally have:

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1(2 6 ) + 1(2 5 ) + 1(2 4 ) + 1(2 3 ) + 1(2 2 ) + 0(2 1 )+ 1(2 0 ) =

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1(2 6 ) + 1(2 5 ) + 1(2 4 ) + 1(2 3 ) + 1(2 2 ) + 0(2 1 )+ 1(2 0 ) =

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=

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Fini!

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