MGMAT Advanced Quant. Ch. 8 - Problem set #9 (p.222)

[jonvindjohnsen]

Eight consecutive integers are selected from the integers 1 to 50, inclusive. What is the sum of the remainders when each of the integers is divided by x?


1. The remainder when the largest of the consecutive integers is divided by x is 0.
2. The remainder when the second largest of the consecutive integers is divided by x is 1.



I need a little bit more help to understand this problem.
Thanks!

[tim]

Before we help with this question, we need you to show some effort of your own. What did you try on this question? Where did you get stuck?

[jonvindjohnsen]

What I did, I picked numbers. Say the largest term is 8. For st. 1) We then have 1, 2, 4, and 8 all possible Xs. Giving different remainders, hence insuff.

2) x can then be 3, 2, 14 etc. same conlusion. --> insuff.

1+2) We see that only 2 fits both statements, hence we can find the rest of the remainders and then calc. the sum.

BUT, then I started to think: 'Is this true for every number?'.. Is this enough to conclude that the ans. is c) ?

[tim]

you haven't explained how you get different remainders. can you explain further? when you respond it will also be helpful if you let us know what part of the solution didn't make sense to you so we can help you understand it better..

[jonvindjohnsen]

8/8 = 1q + 0/8
8/4= 2q + 0/4
8/2 = 4q + 0/2
8/1 = 8q + 0/1

All with a 0 remainder.

For st. 2) to be true for this sequence starting from 8. Then x could be 2, 3 or 6 etc.
7/2 = 3q + 1/2 and so on..

All with a remainder of 1.

Together: only 2 fits both statements for this sequence. But what I'm not quite figuring out without testing is, what if the sequence start with 9 or 10 or 11 etc. Or is the sequence always a multiple of 8? If it is, how do we know that it is?


Thanks.

[tim]

i'm totally lost now. in one post you claim that statement 1 gives all different remainders, then you say all the remainders are the same. you still have not justified why your choice of numbers (regardless of whether you process them correctly) is appropriate. can you start again with your approach, and be careful to explain your process thoroughly and proofread your post before you send it over? this looks like an interesting problem, and i'm interested in helping you with it once i have a better idea what you've tried..

[jonvindjohnsen]

What I meant in the previous post is that if we pick x=8 we get different remainders when we go down the set. 7/8, 6/8, 5/8 etc. Hence, we will get different sums depending on what x is. So if x can be 8, 4, and 2 etc. Then remainder when divided by 8 is 0, hence statement is satisfied. But it will lead to different sums. Hence insufficient.

Why I picked 8 as a starting number for the set? Well, that was only because it was 8 consecutive integers to be selected from stem. But, it is here I'm a little confused. Because x will change depending on the starting term of the set. Right? So, if starting term is 9, 10, 11, 12 etc up to 15. Then, for these sets, they will violate the condition in statement 2 when 1+2) together. And if we look at all possible starting numbers in the set, 1 to 50. We see that 1+2) only fits starting multiples of 8. BUT, on the test I don't have time to check all numbers. So that is why I'm wondering how I can quickly come to the same conclusion without having to check all numbers.

My process from the start: First I picked 8 as starting term for the set. Only because it said 8 consecutive integers in stem, and I thought it would be easy to work with.
st. 1) Picked numbers strategy: say x=8, then rem= 8/8 = 0, 7/8= 7.. --> sum= (8+7+6+...+1)
Say x=2, then rem = 8/2= 0, 7/2= 1, 6/2=0 ---> sum= (1+0+...+1)

Insufficient.

st. 2) Say x=3, then rem= 7/6= 1 --> satisfy the condition from statement. 8/6=2, 6/6=0 ---> sum= (2+1+0+...+ 0)
Say x= 2. Still satisfies statement. But different sum.

Insufficient.

1+2) We see that only the number 2 satisfies both statements. Hence, sufficient.

I'm sorry for my bad explanations.

[tim]

although your explanation is riddled with typos (i say that as a caution to other readers who are trying to follow your work), your logic in eliminating 1 and 2 individually is sound. you haven't provided a convincing explanation though for why they are sufficient together. care to try that part?

[jonvindjohnsen]

Well, that was what I hoped you could help me with.
I compared the different Xs from st. 1) and st. 2). If x=3 it meets the condition in st. 2, but not in st. 1. (Since 8/3= rem. of 2. and not 0). If x=4, it is the reverse. Only x=2 fits both statements conditions. If x=2 then 8/2= rem. of 0 and 7/2= rem. of 1. And we can figure out the sum of the remainders.

[RonPurewal]

ok, the deal is basically this.

if you are counting up consecutive integers, then ...
... the remainders when you divide those integers by 2 are 1, 0, 1, 0, 1, 0, etc.
... the remainders when you divide those integers by 3 are 1, 2, 0, 1, 2, 0, 1, 2, 0, etc.
... the remainders when you divide those integers by 3 are 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, etc.

now, take a look at the 2 statements. together, they say that one of your remainders is 1, and that the NEXT remainder is 0.
if you see the patterns here, it should be clear pretty quickly that only x = 2 is going to do that.
(if you are dividing by any bigger number, then, after getting a remainder of 1, the next remainder you get will be 2, not 0.)

so, with both statements together, you know that x has to be 2.
once you've got that, you actually have all eight remainders (in order, they would be 1, 0, 1, 0, 1, 0, 1, 0).