When 2021 is written in base 21

If 202 1 21 2021_{21} is written in base 20 20 , what is the digit-sum of the converted number?

202 1 21 \red{ 2021_{21} } means that the number written in base 21 \red {21}

* The digit-sum of 432 = 4 + 3 + 2 = 9 432=4+3+2=9 . *


The answer is 21.

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2 solutions

Braden Dean
Dec 29, 2020

202 1 21 = 2 2 1 3 + 2 21 + 1 2021_{21} = 2⋅21^3 + 2⋅21 + 1

202 1 21 = 2 ( 20 + 1 ) 3 + 2 ( 20 + 1 ) + 1 2021_{21} = 2(20+1)^3 + 2(20+1) + 1

202 1 21 = 2 ( 2 0 3 + 3 2 0 2 + 3 20 + 1 ) + 2 20 + 2 + 1 2021_{21} = 2(20^3 + 3⋅20^2 + 3⋅20 + 1) + 2⋅20+2+1

202 1 21 = 2 2 0 3 + 6 2 0 2 + 6 20 + 2 + 2 20 + 2 + 1 2021_{21} = 2⋅20^3 + 6⋅20^2 + 6⋅20 + 2 + 2⋅20+2+1

202 1 21 = 2 2 0 3 + 6 2 0 2 + 8 20 + 5 2021_{21} = 2⋅20^3 + 6⋅20^2 + 8⋅20 + 5

202 1 21 = 268 5 20 2021_{21} = 2685_{20}

2 + 6 + 8 + 5 = 21 2+6+8+5 = \fbox{21}

David Vreken
Dec 27, 2020

202 1 21 = 2 2 1 3 + 0 2 1 2 + 2 2 1 1 + 1 2 1 0 = 18522 2021_{21} = 2 \cdot 21^3 + 0 \cdot 21^2 + 2 \cdot 21^1 + 1 \cdot 21^0 = 18522

18522 = 2 2 0 3 + 6 2 0 2 + 8 2 0 1 + 5 2 0 0 = 268 5 20 18522 = 2 \cdot 20^3 + 6 \cdot 20^2 + 8 \cdot 20^1 + 5 \cdot 20^0 = 2685_{20}

2 + 6 + 8 + 5 = 21 2 + 6 + 8 + 5 = \boxed{21} .

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