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When the base-$b$ number $11011_b$ is multiplied by $b-1$, then $1001_b$ is added, what is the result (written in base $b$)?

 May 27, 2019
 #1
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Since you don't specify base b, that means the answer will be the same in any base from 2 to 10:

 

11011_2 * (2 - 1) + 1001 = 100100 = 36_10

11011_3 * (3 - 1) + 1001 = 100100 = 252_10

11011_4 * (4 - 1) + 1001 = 100100 = 1040_10

11011_5 * (5 - 1) + 1001 = 100100 = 3150_10

11011_6 * (6 - 1) + 1001 = 100100 = 7812_10

11011_7 * (7 - 1) + 1001 = 100100 = 16856_10

11011_8 * (8 - 1) + 1001 = 100100 = 32832_10

11011_9 * (9 - 1) + 1001 = 100100 = 59130_10

11011_10*(10-1) + 1001 = 100100 = 100100_10 

 May 27, 2019
 #2
avatar+23181 
+2

When the base-$b$ number $11011_b$ is multiplied by $b-1$, then $1001_b$ is added, what is the result (written in base $b$)?

 

\(\begin{array}{|rcll|} \hline \mathbf{11011_b }&=& \mathbf{1}\cdot b^4 +\mathbf{1}\cdot b^3 +\mathbf{0}\cdot b^2 +\mathbf{1}\cdot b^1 +\mathbf{1}\cdot b^0 \\ \hline (11011_b)(b-1) &=&( \mathbf{1}\cdot b^4 +\mathbf{1}\cdot b^3 +\mathbf{0}\cdot b^2 +\mathbf{1}\cdot b^1 +\mathbf{1}\cdot b^0)(b-1) \\\\ &=& \mathbf{1}\cdot b^5+\mathbf{1}\cdot b^4 +\mathbf{0}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{1}\cdot b^1 \\ && \qquad\quad -\mathbf{1}\cdot b^4 -\mathbf{1}\cdot b^3 -\mathbf{0}\cdot b^2 -\mathbf{1}\cdot b^1-\mathbf{1}\cdot b^0 \\\\ &=& \mathbf{1}\cdot b^5 +\mathbf{0}\cdot b^4 -\mathbf{1}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{0}\cdot b^1-\mathbf{1}\cdot b^0 \\ \hline (11011_b)(b-1)+1001_b &=& \mathbf{1}\cdot b^5 +\mathbf{0}\cdot b^4 -\mathbf{1}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{0}\cdot b^1-\mathbf{1}\cdot b^0 \\ && \qquad \qquad\qquad + \mathbf{1}\cdot b^3 +\mathbf{0}\cdot b^2 +\mathbf{0}\cdot b^1+\mathbf{1}\cdot b^0 \\\\ &=& \mathbf{1}\cdot b^5 +\mathbf{0}\cdot b^4 +\mathbf{0}\cdot b^3 +\mathbf{1}\cdot b^2 +\mathbf{0}\cdot b^1+\mathbf{0}\cdot b^0 \\\\ &=& 100100_b \\ \hline \end{array} \)

 

for every base.

 

laugh

 May 28, 2019

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