Thursday, March 26, 2009

solution: base b & b+3

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 b2's place, just as a three digit decimal number would have a ones' place (100), a tens' place (101), and a hundreds' place (102).

Therefore n=3*b2+6*b+4
and n also equals 2*(b+3)2+0*(b+3)+2

so we have

3*b2+6*b+4=2*(b+3)2+0*(b+3)+2
3*b2+6*b+4=2*b2+12*b+18+2
b2-6*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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