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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 12510 = ?2**

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**We know 125 = 1(102) + 2(101) + 5(100) = 100 + 20 + 5, but what is it equal to in base 2?**

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12510 = ?(2n) + ?(2n-1) + … + ?(20)

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**The largest power of 2 in 125 is 64 = 26, so we now know our base 2 number will be:**

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**?(26) + ?(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20) and it will have 7 digits**

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**Since there is one 64, we have: 1(26) + ?(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20)**

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**We now have 125 – 64 = 61 left over, which is one 32 = 25 and 29 left over,**

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**so we have: 1(26) + 1(25) + ?(24) + ?(23) + ?(22) + ?(21)+ ?(20)**

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**In the left-over 29, there is one 16 = 24, with 13 left over,**

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**so we have: 1(26) + 1(25) + 1 (24) + ?(23) + ?(22) + ?(21)+ ?(20)**

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**In the left-over 13, there is one 8 = 23, with 5 left over,**

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**so we have: 1(26) + 1(25) + 1(24) + 1(23) + ?(22) + ?(21)+ ?(20)**

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**In the left-over 5, there is one 4 = 22, with 1 left over,**

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**so we have: 1(26) + 1(25) + 1(24) + 1(23) + 1(22) + ?(21)+ ?(20)**

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**In the left-over 1, there is no 2 = 21, so we still have 1 left over,**

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**and our expression is: 1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+**

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**The left-over 1 is one 20, so we finally have:**

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**1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ 1(20) =**

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**1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ 1(20) = 11111012**

1(26) + 1(25) + 1(24) + 1(23) + 1(22) + 0(21)+ 1(20) =

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12510 =

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Fini!

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