Divisibility in the Integers
Algebra
Level
pending
First prove that there exists a number n in Z(integers) such that (n/2),((n-1)/3),(n/4) are in Z. Then find a formula such that if you plug in a value k where k is in Z, it pops out a value for n. If the function of n can be written as n=mk-b, what is m+b?
The answer is 20.
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1 solution
If n/2, n-1/3, and n/4 are in Z, then their sum is in Z. Once summing the fractions we get (13n-4)/12. Splitting up 13n into 9n and 4n, we get (9n/12)+(4(n-1)/12). Simplify it to (3n/4)+((n-1)/3). Thus n has to be divisible by 4 and n-1 has to be divisible by 3. Some values of n that work are 4, 16, or 28. All these numbers can be made through a function: 4(3k-2)=n. Distributing the 4, we get n=12k-8. Thus m=12 and b=8 and m+b=20.