Remainders?

useless · Long-term Member

Remainders?

webmaster · Administrator

Go here!

I don't think you will ever need the rule of dividing by 7, I just put it in the website because I thought it was cool. Anyway, they are very easy and quick if you know how to do them.

fht · Long-term Member

Thanks, Mr. Jones! for giving me the chance to find the shortcut for #77.

I think the most frequently encountered are 9 and 11; for the former, you add all the digits up and the remainder is the same as that of the sum, if you see some numbers add up to 9, cross them out in your head, e.g. 3769382, cross the 9, 3and6, 2and7 out, leaves you with 8 and 3, yielding the remainder of 2. For 11, you subtract the summation of the odd ordinal digits(the 1st, 3rd, 5th, 7th... the direction in which you start counting doesn't matter) from that of the even ordinal digits(2nd, 4th, 6th...), the difference should be the remainder, e.g. 142857: 4+8+7-(1+2+5)=11=0mod11, so 142857 is divisible; 124857: 2+8+7-(1+4+5)=7, thus the remainder is 7.

Look at the last digits for the remainder of 5, and the last 3 digits for 8

Hope it helps and Best Wishes

P.S: Someone please tell me when the next test is scheduled.

webmaster · Administrator

Well, I guess you liked #77. Whether Mr. White uses it or not, I will not know. He had asked me to look for other methods to possibly use. I gave it to him, but I don't know if it will ever show up. But it is still pretty cool!

vvr1590 · Senior Member

how 'bout for 6?

fht · Long-term Member

public int remainderBySix(int N){

int summation = the sum of all digits in N;

boolean even = (N%2 == 0);

if(even){

switch(summation%3){

case 1: return 4; case 2: return 2; default: return 0;}}

else{

switch(summation%3){

case 0: return 3; case 1: return 1; case 2: return 5; }}}

}

useless · Long-term Member

Your program would be slightly more accurate had you included break statements...

Thanks for the explanation, though. That trick was one of the few things that wasnt outlined in ELEMENTS OF NUMBER SENSE. So thank you.

-REGARDED!!!-

zack

fht · Long-term Member

Ah-Oh, I forgot I was writing a program to an expert.

Thanks for the break; statement, though.

See, I really have never seen a problem like "123457 / 6 has a remainder of ___." Most of them time it's "(12 + 34 * 5) / 6 has a remainder of ___." In that case, just figure out the remainder of every term and then do the operations and it's quite simple.

And, As I have asked a few times before, WHEN IS THE NEXT TEST!

webmaster · Administrator

I will do my best to make a test a week, and Zack should have them by Thurs night. If you find someone else to write tests, you can do more than 1 a week, but like Megan said, they take a while to do. Therefore, with my very busy work schedule, I can only commit to one a week. I believe Zack will post them every Friday to my knowledge.

fht · Long-term Member

As I have said, I would never mean to frustrate you, Mr. Jones. It's just that the tests are so good that I wanted to have it a bit more often. It was just a thought.

So the test would be posted every Friday morning, it seems.

I deeply appreciate what you do for us, Mr. Jones, I won't ask for more anymore.

Best Wishes

vvr1590 · Senior Member

or you could do this

public int remainderbysix (int n)
{
return n%6;
}