Question 1134705: In what bases, b, does (b+6) divide into (5b+6) without any remainder?
Note: I’m guessing that there are many bases that this could function because the question says in what BASES!
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52852)   (Show Source):
You can put this solution on YOUR website! .
The problem in the post is formulated by the very curved and twisted way, using outdated and antiquated language.
In my opinion, the correct (adequate; straightforward and mathematically correct) formulation is AS FOLLOWS:
Find integer non-negative numbers "b" such that
(5b + 6) = 0 mod (b+6).
So, the problem is about solving congruences.
Solution
5b + 6 = 5(b+6) - 30 + 6,
5b + 6 = 5*(b+6) - 24.
Therefore,
5b + 6 = 0 mod (b+6) (1)
if and only if
-24 = 0 mod (b+6),
which is equivalent to
24 = 0 mod (b+6). (2)
From (2), you easily find b = 0, 2, 6, 18.
Check.
a) b = 0. Then 5b + 6 = 5*0 + 6 = 0 + 6 = 6. From the other side, b+6 = 6, and 6 = 0 mod 6. ! correct !
b) b = 2. Then 5b + 6 = 5*2 + 6 = 10 + 6 = 16. From the other side, b+6 = 8, and 16 = 0 mod 8. ! correct !
c) b = 6. Then 5b + 6 = 5*6 + 6 = 30 + 6 = 36. From the other side, b+6 = 12, and 36 = 0 mod 12. ! Correct !
d) b = 18. Then 5b + 6 = 5*18 + 6 = 90 + 6 = 96. From the other side, b+6 = 24, and 96 = 0 mod 24. ! Correct !
So, the problems has 4 solutions b = 0, 2, 6 and 18. ANSWER
Solved.
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The formulations and wordings as in the original post, are used now in sections and circles of amateurs of the OLD ENGLISH only.
In professional Math it is not in use just at least 200 years,
and in the School Math it is not in use more than 100 years.
It is some "before-Newton's era" language . . .
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Tutor @ikleyn didn't like the way the problem was stated, so she stated it in a different way and solved it. But her restated problem has four solutions (0, 2, 6, and 18); the original problem only has three solutions, because b=0 is not a valid base.
So here is a solution of the problem as it was posted.
For base b, we want the number 5b+6 to be divisible by b+6; that is, we want
where k is an integer.
It is easy (with experience!) to find the values of b that make k an integer in that equation.
The general technique is to perform the indicated division to get a result that is an integer plus a remainder:
For this expression to be an integer, (b+6) has to be a divisor of 24. And since b in this problem is a number base, b has to be 2 or greater.
If b is 2 or greater and b+6 is a divisor of 24, then b+6 is at least 8; the divisors of 24 that are 8 or greater are 8, 12, and 24, giving us three solutions to the problem:
(1) b+6 = 8 --> b = 2 --> k = 5-(24/8) = 5-3 = 2
Check: (5b+6)/(b+6) = 16/8 = 2
(2) b+6 = 12 --> b = 6 --> k = 5-(24/12) = 5-2 = 3
Check: (5b+6)/(b+6) = 36/12 = 3
(3) (b+6) = 24 --> b = 18 --> k = 5-(24/24) = 5-1 = 4
Check: (5b+6)/(b+6) = 96/24 = 4
So....
ANSWER: There are three bases in which (b+6) divides into (5b+6) without any remainder: 2, 6, and 18
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