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How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation?

 

A) 5

B) 6

C) 7

D) 8

E) 9

 

The answer is apparently A) 5, but I am not sure how to get to it.

 

Any help is appreciated. 

 

Thanks!

- PM

 Jan 26, 2020
edited by PartialMathematician  Jan 26, 2020
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How many two digit positive integers in the base-four representation are
twice the two digit positive integers in the base-three representation?

 

I assume:

 

\(\begin{array}{|rl|c|c|c|c|c|c|l|} \hline & & 10 & 11 & 12 & 20 & 21 & 22 & \text{base}~3 \\ & & =3 & =4 & =5 & =6 & =7 & =8 & \text{base}~10 \\ \text{base}~4 & \text{base}~10 & 6 & 8 & 10 & 12 & 14 & 16 & 2x \\ \hline 10 & = 4 & & & & & & & \\ 11 & = 5 & & & & & & & \\ 12 & = 6 &6=6\checkmark& & & & & & \\ 13 & = 7 & & & & & & & \\ 20 & = 8 & &8=8\checkmark& & & & & \\ 21 & = 9 & & & & & & & \\ 22 & = 10 & & &10=10\checkmark& & & & \\ 23 & = 11 & & & & & & & \\ 30 & = 12 & & & &12=12\checkmark& & & \\ 31 & = 13 & & & & & & & \\ 32 & = 14 & & & & &14=14\checkmark& & \\ 33 & = 15 & & & & & & & \\ \hline \end{array}\)

 

The answer is 5.

 

\(\begin{array}{|ll|} \hline 1. &12_4 = 2\times 10_3 \\ 2. &20_4 = 2\times 11_3 \\ 3. &22_4 = 2\times 12_3 \\ 4. &30_4 = 2\times 20_3 \\ 5. &32_4 = 2\times 21_3 \\ \hline \end{array}\)

 

laugh

 Jan 26, 2020

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