Practice (Supposedly) Easy SAT Math Problem?

Question

When x is divided by 3, the remainder is z. In terms of z, which of the following could be equal to x?

the supposedly correct answer is 3+z, and these are the other answers:
z - 3

3 - z

3z

9 + 2z

The explanation says just plug in 5 for x and you'll get 2 for z and it will work, however, this just works with 3, 4, and 5 since they only go in once. Numbers like 7,8, and 100 don't work. Help, this is supposed to be an easy question and I'm (usually) very good at math, I've aced practice math sections before.

Answer 1

The key here is that the question doesn't ask you which of the following IS equal to x, just which of them _could_ be. That said, I really don't like your practice book's explanation, because it suggests that you try to solve by plugging in various values of x, and that won't give you a solution, since you don't know the value of x. What you do on this problem is evaluate each of the expressions with respect to all possible values of z and see which ones give a remainder of z when divided by 3. The best way to go about this is to find out whether the expression is congruent to z mod 3. Thus we have:

3 + z ≡ z mod 3
z - 3 ≡ z mod 3
3 - z ≡ -z mod 3
3z ≡ 0 mod 3
9 + 2z ≡ -z mod 3

This immediately gives you a choice between only two answers. However, since z<3, z-3 can never give you a positive number, and therefore cannot possibly be the right answer. Therefore, by process of elimination, the correct answer is 3+z. Q.E.D.

Answer 2

this question is not looking for one specific value but a range of values.

x can be any number that gives you a value of 3 + the remainder (z), which can only be 1 or 2, because if it were 3 it would be divisible by 3 and therefore leave NO remainder.

So, if you have a remainder of 1, x could be 3 plus 1, or 4

proof: 4/3 = 1 plus 1; 13/3 = 4 plus 1(remainder); 25/3= 8 plus 1(remainder)

7 and 8 should work:

7/3= 2 plus 1(remainder); 8/3= 2 plus 2(remainder)

in each of these, multiplying the number and adding the remainder will give you the original number.

Answer 3

x = nz + 3

then, if Z = 6, then 9 + 2z = 21
divide 21 by 6 and remainder is 3

so 9 + 2z is your answer

Why 6?
look at it as 9 + 2z = 3 + 2(z+3)
This should give you a hint...

i.e. if x = nz+3 and 2z+9 = 2z+6 + 3

nz = 2z + 6
(n-2)z = 6
factorise 6, you have 3&2 or 6&1

now n must be >2, so you can not use 3&2
put z = 6 and n = 3 so that n-2 = 1

Answer 4

(X/3) = (some number) + Z/3

The 3+Z amswer for X is the only one that gives just Z for a remainder (of Z/3) as the fractional part of the answer.

Answer 5

3z = x

z = x/3

Answer 6

None of the other answers can possibly be right, its not which is right all the time, but which can be right.

Answer 7

dfg