SOLUTION: In what bases, b, does (b + 6) divide into (5b + 6) without any remainder?

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Question 1200551: In what bases, b, does (b + 6) divide into (5b + 6) without any remainder?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13215)   (Show Source): You can put this solution on YOUR website!


The requirement is to find integers b for which (5b+6) divided by (b+6) is some integer k:








"Perform the division" like this:



So





In that equation, -6 is an integer, and b must be an integer. That means must be an integer; and since b must be positive, must be greater than 6. Trying values of k that make an integer greater than 6...

k=1; ; b = -6+6 = 0 (not possible)

k=2; ; b = -6+8 = 2

CHECK: (5b+6)/(b+6) = 16/8 = 2

k=3; ; b = -6+12 = 6

CHECK: (5b+6)/(b+6) = 36/12 = 3

k=4; ; b = -6+24 = 18

CHECK: (5b+6)/(b+6) = 96/24 = 4

ANSWERS: bases 2, 6, and 18
Answer by ikleyn(52905)   (Show Source): You can put this solution on YOUR website!
.

The problem's wording makes me shudder - so unprofessionally it sounds.


The normal mathematical formulation should sound this way 

    Find integer numbers b such that  (b+6)  divides  (5b+6)  with no remainder.


The "bases" are irrelevant.

All these tricks with "divide into" were used at the times of Shakespeare and Dickens.

It is just 100 years nobody uses such outdated form in mathematical texts,
excluding the lovers of old English, who want to show in this way how smart they are.



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