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How can I find the missing base?
28= 34 ___
28= 26 ___
23 in base twelve = 43 ____
28= 34 ___
28= 26 ___
23 in base twelve = 43 ____
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- Feb 4, 2004
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Experimentation would be one method.bware said:
I will assume that the original values are written in base-10.
Suppose we guess the new base for the first exercise to be, I dunno, 5. Then "34" represents 3×5<sup>1</sup> + 4×5<sup>0</sup> (just like "28" represents, in base-10, 2×10<sup>1</sup> + 8×10<sup>0</sup>), which simplifies as 15 + 5 = 21, in base-10.
So the new base of the first exercise can't be 5.
Now you try something else.
Eliz.
Hello, bware!
For example: .34<sub>5</sub> means: (3 x 5) + 4 .= .19
. . and: .213<sub>6</sub> means: (2 x 6<sup>2</sup>) + (1 x 6) + 3 .= .81
The first one says: .28 .= .34<sub>b</sub> . . . but 34<sub>b</sub> = 3b + 4
. . So we have: .28 .= .3b + 4 . ---> . b = 8
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check: .Does 28 = 34<sub>8</sub> ?
. . . . 34<sub>8</sub> .= .(3 x 8) + 4 .= .28 . . . Yes!
.
You're expected to know how numbers are constructed in other bases.
For example: .34<sub>5</sub> means: (3 x 5) + 4 .= .19
. . and: .213<sub>6</sub> means: (2 x 6<sup>2</sup>) + (1 x 6) + 3 .= .81
The first one says: .28 .= .34<sub>b</sub> . . . but 34<sub>b</sub> = 3b + 4
. . So we have: .28 .= .3b + 4 . ---> . b = 8
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check: .Does 28 = 34<sub>8</sub> ?
. . . . 34<sub>8</sub> .= .(3 x 8) + 4 .= .28 . . . Yes!
.