remainder left type questions

[what?]

hey guys,
came across a few questions and am not sure how to solve them.

1. Twice the sum of 3 integers x ,y and z when divided by 7 gives remainder 1
what is the remainder when x+y+z is divided by 7?

2. x when divided by 7 leaves remainder 4.What is the remainder when 2x + 5 is divided by 7?

Please tell me the method to solving these, or similar concept questions.

Thanks.

[DWarrior]

1. I couldn't set it up algebraically, so I had to plug in numbers. let S be x+y+z, so we're looking for:
2S/7=n/7+1

This works when 2S=8, because then n=7 and there's a remainder of 1.

S=4, so 4/7=0+4/7

So the remainder is 4.

2. x/7=n/7+4/7
x=n+4, where n is divisible by 7. Substitute that into the new equation:
[2(n+4)+5]/7=(2n+8+5)/7=2n/7+13/7=2n/7+7/7+6/7. 13 can't be the remainder, since remainders must be less than the divisor.

So the remainder is 6

[what?]

Dwarrior: thanks for the explanation. your answers are right.
I am still a little confused, but maybe thats just because it is very difficult to put a question like this into an equation. There is no clear cut way to handle this.
But I do get the idea from your answer. Hope i can use it to figure out a similar problem.

[DWarrior]

I was asking on IRC how to solve problem #1 methodically, and somebody posted an interesting analysis that I think might be more helpful than my solution:

First, you should realize that we'll always be working with (x+y+z) and never with the individual variables, so you should simplify the problem by substitution (S for x+y+z). That way, there's a greater chance that you'll "see" the solution.

Second, set up the original problem:
2S=7p+1, where p is some multiple of 7 divided by 7. We know 2S must be even, and therefore p must be odd (do you see why?).

So now just plug in any odd number for p and you'll get a valid S. For example, use p=3:
2S=21+1=22
S=11
11/7 leaves remainder of 4.

This setup actually seems like a good method to solve all problems of this type, where you're given a remainder and then have to find remainder of another related number.

Set up the original remainder equality. p will always be a positive integer. Look for any restrictions, like it having to be odd/even, or anything else. Then plug a value for p that meets all restrictions to get a sample x, plug the x back into the new number and find the remainder.

For example, do #2:
x=7p+4
In this case there are no other restrictions on p (aside from it being a positive integer).
x=7*3+4
x=21+4
x=25

Now we need the remainder of 2x+5:
50+5=55
55/7 leaves remainder of 6.

[what?]

Sir that is a Great solution. easy to understand and pretty methodological.

Thank you very much for sharing.

[drosera01]

These questions are just like plug and play electronic devices now. Plug the number and you will get the answer. right?

[learner85]

drosera hitted the bulls eye. I agreed with him . This are simple plug-in number problems. Practice plugging-in and back-solving, they are really helpful when you are stuck in a problem. Actually almost all of us are supposed to be stucked once or more on GRE exam.

Regards,

learner85