ketkoag wrote:A and b are divided by 8, respectively. Are the two remainders different from each other?
1) When 3a and 3b are divide by 8, the remainders are different.
2) When 5a and 5b are divide by 8, the remainders are different.
Answer: D
my question is: Will it always be true that if the numbers 3a and 3b or 5a and 5b or any respective multiple is divided by 8 and the remainders come to be different, then when a and b are divided by 8, then the respective remainders will be different. Please elaborate... also lemme know whether this is true always for any number in the denominator..
coz i took some values and checked, i think it is true, but m not sure about all the values..
The generalization you seek can be established along these routes.
First, a,b CANNOT be a multiple of 8 or else the remainder in both cases will be equal, ie the remainder will be 0.
Suppose a and b are a multiples of 8. Then the remainder of 3a/8 and a/8 and 3b/8 and b/8 will be equal to zero hence will NOT be different. This must mean that in I a and b are not multiples of 8. But if they are not multiples of 8 are there no conditions under which their remainders can be equal to zero or some positive number?
Let’s consider some restrictions.
Let a,b>8. We use a from now on but result follows for b.
a/8 can be written as a mixed fraction of the form p(r/8) where p is the whole number and r is the remainder. Now let’s examine 3a/8
3 x p(r/8). Remembering how to multiply mixed fractions you know this is
3p + 3r/8 . So lets compare
3p +3r/8 with p(r/8)
Since r is remainder we know r must be less than 8 because our divisor is 8. So r can be
1, 2, 3, 4, 5, 6, 7
We can see that for no value of r will 3r/8 be an integer. And we know r/8 is a fraction. So we can conclude that for a>8 3a/8 and a/8 will have remainders Different from 0. But note that for r of 4 5 6, 7. 3r>8 and you will have a quotient that must be added to 3p. Whatever is left will be the remainder for 3a/8. Take a=31>8 as an example. 3 x 31/8= 3 x 3(7/8)
=9+21/8=9+2+5/8=11(5/8). In this example r =7 and we end up with a remainder < than 7. Continuing as in the above we know that for all r= 4, 5, 6, 7, it is true that we get some remainder q<r, so an integer is never possible neither can q=r. Since q and r will never equal zero and will always be different , this establishes our proof in this case. 7 was the remainder for a/8 and 5 is the remainder for 3a/8 in the example.
For a<8, a/8 CANNOT be a mixed fraction but just a proper fraction,
3a/8 will have the form 3a/8 and in this case the remainder will be a. And of course it will be the same values as above (1, 2, 3, 4, 5, 6, 7). From this we know 3a/8 is never an integer and similar arguments can show that the remainder 0 is not possible a/8 and 3a/8 will have different remainders. Take a=7<8 as another example.
a/8 =7/8
3a/8= 21/8= 2(5/8)
The same argument above holds for a<8 and the Desired Generalization Can Be Made.
