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1) We have positive integers a, b and c such that a>b>c. When a, b and c are divided by 19, the remainders are 4, 2 and 18 respectively. When the number 2a+b-c is divided by 19, what is the remainder?

 

2) Given that m and n are positive integers such that m=6 (mod 9) and n=0 (mod 9), what is the largest integer that mn is necessarily divisible by?

MathCuber  Nov 30, 2018
 #1
avatar+784 
+5

a/19 has a remainder of 4. b/19 has a remainder of 2. c/19 has a remainder of 18. 

 

2a/19 then must have a remainder of 8. (4 + 4 = 8) 

 

The remainder of \(2a + b - c\) when divided by 19 is just \(8 + 2 - 18 = -8\). We can get the real remainder by adding -8 to 19, which equals 11. the remainder is \(\boxed{11}\).  

 

Hope this helps,

- PM

PartialMathematician  Nov 30, 2018
 #2
avatar+784 
+5

Since \(m = 6 (mod 9)\)  \(n = 0 (mod 9)\) and , we know that \(m\) is a multiple of 3 and \(n\) is a multiple of 9. \(m\) can also sometimes be a multiple of 6. (For example, m = 24, 24 (mod 9) = 6). To make sure that the largest integer that \(mn\) is necessarily divisible by is always true, we must assume that m is a multiple of 6. The largest integer that \(mn\) is necessarily divisible by is \(3\cdot3\cdot3 = \boxed{27}\).

 

Hope this helps,

 

- PM

PartialMathematician  Nov 30, 2018
 #3
avatar+784 
+5

I think my answer above might be a little confusing, so I'll re-state it. 

 

Odd number * odd number = Odd. 

 

Since n is a multiple of 9, the prime factorization of n (other than 0) has at least two 3s. 

 

Since m is a multiple of 3, the prime factorization of m (other than 0) has at least one 3. 

 

We cannot consider the possibility that m and n are even because they are not always even. 

 

The largest integer that mn is always divisible by is \(3\cdot3\cdot3 = \boxed{27}\), which is proven by prime factorization. 

 

Hope this helps more, 

 - PM

PartialMathematician  Nov 30, 2018
 #4
avatar+784 
+5

Go try it out yourself. It really works for all m = 6 (mod 9) and n = 0 (mod 9).

 

- PM

PartialMathematician  Nov 30, 2018
 #5
avatar+27237 
+2

Here's another way of looking at 1).

 

Alan  Nov 30, 2018
 #6
avatar+20683 
+8

1) 

We have positive integers a, b and c such that a>b>c.

When a, b and c are divided by 19, the remainders are 4, 2 and 18 respectively.

When the number 2a+b-c is divided by 19, what is the remainder?

 

\(\begin{array}{|rcll|} \hline a & \equiv & 4\pmod{19} \quad & | \quad \cdot 2 \\ \mathbf{2a} & \mathbf{\equiv} & \mathbf{8\pmod{19}} \\\\ \mathbf{b} & \mathbf{\equiv} & \mathbf{2\pmod{19}} \\\\ c & \equiv & 18\pmod{19} \\ c & \equiv & 18-19\pmod{19} \\ \mathbf{c} & \mathbf{\equiv} & \mathbf{-1\pmod{19} } \\ \\ \hline \\ 2a+b-c & \equiv & 8+2-(-1) \pmod{19} \\ \mathbf{2a+b-c} & \mathbf{\equiv} & \mathbf{{\color{red}11} \pmod{19}} \\ \hline \end{array}\)

 

The remainder is 11

 

laugh

heureka  Nov 30, 2018
edited by heureka  Nov 30, 2018
 #7
avatar+20683 
+8

2)
Given that m and n are positive integers such that m=6 (mod 9) and n=0 (mod 9),
what is the largest integer that mn is necessarily divisible by?

 

\(\begin{array}{|rcll|} \hline m & \equiv & 6\pmod{9} \\ \text{or} \\ m &=& 6 + z\cdot 9 ,\quad z \in \mathbb{Z} \\\\ n & \equiv & 0\pmod{9} \\ \text{or} \\ n &=& 0 + z_2 \cdot 9, \quad z_2 \in \mathbb{Z} \\\\ \hline \\ mn &=& (6 + z\cdot 9)\cdot z_2 \cdot 9 \\ mn &=& 3\cdot(2 + z\cdot 3)\cdot z_2 \cdot 9 \\ \mathbf{mn} &\mathbf{=}& \mathbf{{\color{red}27}\cdot(2 + 3z)\cdot z_2} \\ \hline \end{array}\)

 

The largest integer is 27

 

laugh

heureka  Nov 30, 2018
 #8
avatar+27237 
+1

I guess 27 is the answer being sought here. However, it seems to me that if m and n are positive integers, then the largest integer that mn is necessarily divisible by is mn.  !!!

Alan  Nov 30, 2018
 #9
avatar
-1

I agree with Alan.

 

m mod 9 =6    the smallest m that will satisfy the congruence =9 + 6 =15, and:

n mod 9  =0    the smallest n that will satisfy the congruence  =9. Then:

15 x 9 = 135 - and the divisors of 135 =1, 3, 5, 9, 15, 27, 45, 135 (8 divisors).

So, the largest divisor of 135 is 135 !!!.

Guest Nov 30, 2018
 #10
avatar+195 
0

It says necessarily divisible by, so every number possible has to be divisible by it, I was confused there too. Thank Y'all for helping me!

MathCuber  Dec 2, 2018

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