Ron made up remainder problems

[wind]

On his study hall November 2011 he gave these 2 problems which I could not solve :

a) If a and b are positive integers and a/b = 87.75 b=?
25
26
27
28
30

b) the set S= 6 consecutive positive integers ,positive integer n =?

1-when each integer in S is divided by n the sum of the remainders = 11
2-when each integer in S is divided by n the remainders include 5 different values

Any clues ??? thanks in advance

[jlucero]

1. a/b = 87.75
a/b = 87 + 3/4
a = 87b + (3/4)b

From this equation we see that b must be a number that when multiplied by 3/4 will give us a whole integer. Therefore, b must be divisible by 4 and the answer must be D.

2. Not sure if there's an algebraic solution to this, but what's most important about a question like this is when you divide consecutive integers by a number (n), you will get remainders that increase from zero to n-1... divide 0, 1, 2, ....10 by 5 and you'll get a pattern of remainder = 0, 1, 2, 3, 4, 0, 1, 2, 3, 4. So if you were dividing these consecutive integers by:

1: remainders = 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
2: remainders = 0, 1, 0, 1, 0, 1, 0, 1, 0, 1
3: remainders = 0, 1, 2, 0, 1, 2, 0, 1, 2, 0
4: remainders = 0, 1, 2, 3, 0, 1, 2, 3, 0, 1
5: remainders = 0, 1, 2, 3, 4, 0, 1, 2, 3, 4
6: remainders = 0, 1, 2, 3, 4, 5, 0, 1, 2, 3
etc.

With the first statement, you need to find six numbers in a row that will add up to 11. Note that these can be any six numbers, not just ones that start with zero. You should be able to quickly eliminate the numbers 1, 2 and 3 (sum is too small), and any number that's 6 or larger (sum of 1-5 is too large). So focusing on dividing by 4 and 5 we find:

4: remainders = 0, 1, 2, 3, 0, 1, 2, 3, 0, 1
5: remainders = 0, 1, 2, 3, 4, 0, 1, 2, 3, 4

This eliminates AD from our grid.

Looking at statement 2, in order to have 5 unique remainders, you would need to have one number repeat. This can only happen when the first and the last number in the sequence are the same and the ones in the middle are all different. And that will only happen when dividing by 5. So statement 2 is sufficient and the answer is B.

[wind]

Looking at statement 2, in order to have 5 unique remainders, you would need to have one number repeat. This can only happen when the first and the last number in the sequence are the same and the ones in the middle are all different. And that will only happen when dividing by 5. So statement 2 is sufficient and the answer is B.

thanks a million for the great explanation .. but one last question : why should the 1st and last should be the same and the in between are different??

[RonPurewal]

why should the 1st and last should be the same and the in between are different??

try writing out some lists of the remainders you get when you divide consecutive integers by some fixed number (or, take a look at the lists that joe has already made in his post above). you'll notice that the remainders always go 0, 1, 2, ..., until they get to one less than the number you're dividing by, and then they "roll back" to 0 again.

if all of the remainders need to be different except for one matching pair, then the matching pair has to be the first and last remainders.

try it in the lists above -- if any other remainders match, then others will match, too. for instance, if the first and next-to-last remainders are the same, then the second and last remainders will also be the same. etc.

[wind]

why should the 1st and last should be the same and the in between are different??

try writing out some lists of the remainders you get when you divide consecutive integers by some fixed number (or, take a look at the lists that joe has already made in his post above). you'll notice that the remainders always go 0, 1, 2, ..., until they get to one less than the number you're dividing by, and then they "roll back" to 0 again.

if all of the remainders need to be different except for one matching pair, then the matching pair has to be the first and last remainders.

try it in the lists above -- if any other remainders match, then others will match, too. for instance, if the first and next-to-last remainders are the same, then the second and last remainders will also be the same. etc.

thanks a lot , I got it now :)

[jnelson0612]

Great to hear! :-)