*The number*n

*expressed in base*b

*has the form 364. In base*(b+3)

*,*n

*is written as 202. Find the decimal values of*n

*and*b

*, which are both positive integers.*

*Solution:*

Without knowing the base, we cannot know the number which is represented by the symbols 364. The place-value system allows us to perceive it as a number having a one's place, a

*b*'s place, and a

*b*

^{2}'s place, just as a three digit decimal number would have a ones' place (10

^{0}), a tens' place (10

^{1}), and a hundreds' place (10

^{2}).

Therefore n=3*b

^{2}+6*b+4

and n also equals 2*(b+3)

^{2}+0*(b+3)+2

so we have

3*b

^{2}+6*b+4=2*(b+3)

^{2}+0*(b+3)+2

3*b

^{2}+6*b+4=2*b

^{2}+12*b+18+2

b

^{2}-6*b-16=0

(b-8)*(b+2)=0

so

*b*=8 or

*b*=-2. The problem says that the base is positive, so we take

*b*=8. Once we have found

*b*, we can put the value into either equation for

*n*.

**We get**

*b*=8,*n*=244.In this problem, I offered the students an easy base that they could find by trial and error. Several students noted that since 364 contains a "6", the base had to be at least 7, a very astute observation. Also, when you solve the quadratic, you get only on positive option.

If the quadratic equation had two positive roots, we could find a number

*n*which had the given representations (for bases

*b*and

*b+3*) for two different values of

*b*. If the base

*b*representation was four digits, there could be as many as three distinct values of

*b*that could solve such a problem. If the base

*b*representation was six digits, we would get a fifth degree equation, which Abel says cannot be solved for the five different

*b*'s that could exist.

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