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# Modulus Operator

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Hi there, I have read through the search result of the Modulus operator and could not find the answer to my question, so, I post my question here.
In mathematic term: a = dq+r, where "a" is an integer and "d" is a dividend and q is a quotient, "r" is the remainder. In java, we can use "%" to get a reminder of a division. e.g
m=101, n=11, 101%11=2, if we just know "m=100" and the reminder (2), how can we get "n" in java?
In reality, everyone can calculte in their head and get the solution, how can I write the code in java to get the solution? e.g 101%n=2, how to solve n?
Waiting for help!
Thanks!
Mindy

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Use your example, 101%n =2 -> 101-2 = 99 will be divisble by n.
In other words n will be factor of 99 (n>1).
so 99%n == 0 for all n >1 should be the answer set you are looking for.

HTH.

Originally posted by Mindy Wu:
Hi there, I have read through the search result of the Modulus operator and could not find the answer to my question, so, I post my question here.
In mathematic term: a = dq+r, where "a" is an integer and "d" is a dividend and q is a quotient, "r" is the remainder. In java, we can use "%" to get a reminder of a division. e.g
m=101, n=11, 101%11=2, if we just know "m=100" and the reminder (2), how can we get "n" in java?
In reality, everyone can calculte in their head and get the solution, how can I write the code in java to get the solution? e.g 101%n=2, how to solve n?
Waiting for help!
Thanks!
Mindy

Greenhorn
Posts: 28
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101%n=2
so 101=q*n+2 we have q*n=99
so n is the factor of 99
implementation :
for(int i=1;i<=99;i++)
if((99/i)==0){
n=i;
System.out.println("n is : "+n);
}
}
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Mindy Wu
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actually, my problem a little bit more complicate than this, how to solve if x%10=8, where 8 is a reminder, 10 is the dividen, x must be less than or equal to 16.
Thanks!
[ January 19, 2003: Message edited by: Mindy Wu ]

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I am totally confused.
The first time you ask 101%n = 2, how to get n?
Now you ask n%10 = 8, how to get n? (n<=16)
The previous posts have show how to solve question #1.
For #2, n= 10 * m + 8 (m is int.)
and,
0<m<=2 (based on requirement and definition)
So, n=8 is the only solution.

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n = 18 is also a solution. In fact, there are infinite solutions such that n = 10*m + 8, where m is an integer.

John Lee
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But the requirement says n <=16, so n=18 is not good.

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