Prove divisibility

[Geronimas]

1. It is given that 3a is divisible by n and (12a+5b) is divisible by n. Prove that 10b is divisible by n.

 

[Deleted member 4993]

1. It is given that 3a is divisible by n and (12a+5b) is divisible by n. Prove that 10b is divisible by n.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this assignment.

 

[HallsofIvy]

How can you combine (add, subtract, multiply by constants, ...) 3a and 12a+ 5b to get 10b?

 

[Geronimas]

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this assignment.

I have the solution, but I do not understand it: it says if 3a is divisible by n, then 12a is divisible by n, and (12a+5b-12a) is divisible by n and equal to 5b. From this comes 5b is divisible by n and 10b is divisible by n. Care to explain?

 

[Deleted member 4993]

1. It is given that 3a is divisible by n and (12a+5b) is divisible by n. Prove that 10b is divisible by n.

3a is divisible by n \(\displaystyle \ \ \to \ \ \) 3*a/n = p [where 'm' is an integer] \(\displaystyle \ \ \to \ \ \) 3*a = p * n \(\displaystyle \ \ \to \ \ \) 12 * a = q*n [where 'q' = 4*p & is an integer]

(12a + 5b) is divisible by n \(\displaystyle \ \ \to \ \ \) (12a + 5b)/n = r [where 'r' is an integer] \(\displaystyle \ \ \to \ \ \)

(12*a + 5*b)/n = r \(\displaystyle \ \ \to \ \ \) (q*n + 5*b) = n*r \(\displaystyle \ \ \to \ \ \) 10*b = 2*n*(r - q) .......... continue

 

[Dr.Peterson]

I have the solution, but I do not understand it: it says if 3a is divisible by n, then 12a is divisible by n, and (12a+5b-12a) is divisible by n and equal to 5b. From this comes 5b is divisible by n and 10b is divisible by n. Care to explain?

In words, rather than symbols, here is what they are saying:

if 3a is divisible by n, then 12a is divisible by n:

since 12a is 4 times 3a (that is, 12a is a multiple of 3a), it is divisible by any divisor of 3a.​

(12a+5b-12a) is divisible by n.

if we subtract two multiples of n, namely 12a+5b and 12a, the result is also a multiple of n.​

5b is divisible by n.

12a+5b-12a = 5b, so the previous statement means that 5b is divisible by n.​

10b is divisible by n.

since 10b is 2 times 5b, it too is divisible by n.​

Of course, if you are taking a course in this, you also need to be able to work with the symbols easily.

What facts have you learned about divisibility?

 

[Steven G]

You need to think about this a bit. It really is not that difficult.

7 goes into 56. In fact 7 goes into 56 8 times. So does 7 go into a multiple of 56? That is does 7 go into 56*2 =112, 56*6=336, 56*10=560, etc. Yes of course! 7 goes into 56*2, 882=16 times. 7 goes into 56*10=560, 8*10=80 times. Do you see this now??

So if a goes into b, then a goes into kb.

 

[JeffM]

In your other question, I tried to explain this to you in a way that would let you think about this from the ground up, but you clearly did not make any effort to think at all.

We are dealing with integers. When we say that integer n divides 3a evenly, where a is an integer, what do we mean? We mean that there exists an integer p such that

[MATH]p =\dfrac{3a}{n} \implies np = 3a.[/MATH]
Can you follow that? To give an example, if 10 divides evenly into 60, then there is some integer that when multiplied by 10 results in 60, namely 6. It is not a terribly difficult concept, but if you need a few more examples, we shall be happy to supply them.

Similarly, if n divides evenly 12a + 5b, that means that there exists an integer q such that

[MATH]q = \dfrac{12a + 5b}{n} \implies nq = 12a + 5b.[/MATH]
Are you still following?

Ok so

[MATH]np = 3a \implies 4 * np = 4 * 3a \implies 12a = 4np.[/MATH]
Is everything still clear? Let us know if you have got lost; we want to help.

[MATH]\therefore (12a + 5b) - 12a = nq - 4np \implies 5b = n(q - 4p).[/MATH]
Not too advanced algebra. But 4 is an integer as is p so 4p is also an integer. And q is an integer.
So r = q - 4p means r is an integer.

[MATH]\therefore 5b = n(q - 4p) = nr \implies \dfrac{5b}{n} = r, \text { AN INTEGER.}[/MATH]
But that means n divides evenly into 5b by definition.

Now if there is any place where this seems the least bit obscure to you, tell us where it is.