[abhi2400]

For 2day's question :

a^-1+b^-1+c^-1=(a+b+c)^-1

This term can be simplified to give the equation:
(a+b)(a+c)(b+c)=0

that means either (a+b), or(a+c) or(b+c)=0 at a time

so for (a+b)=0, a can be chosen in 10C1 ways, b can be chosen in 1 way and c can be chosen in 8C1 ways(as a,b,c, are distinct)= total 10.1.8= 80

the same number of ways for (b+c) and (a+c)=0

Therefore the total umber of ways=3.(80)=240

Hence the answer is (c)

[sastry751]

a,b,c in these 3 numbers 2 has to get cancel out for the conditiion to be true if we select (-5,5) we can select one from remaning 8 and also these 3 can be arranged in 3 fact ways.so we can select -4,4 until -1,1 we get 48*5 so 240

[v-factor]

Can somebody please post today's question on the thread??

[Jaskiran]

Let a, b, c be distinct non-zero integers such that -5 <= a, b, c <= 5. How many solutions (a, b, c) does the equation 1/a + 1/b + 1/c = 1/(a+b+c) have?

a) 72 (b) 120 (c) 192 (d) 240 (e) none of the foregoing

[debashis_dan]

1/a + 1/b + 1/c = 1/(a+b+c)
this holds true in 3 cases : (i) when (b+c)=0 (ii) when (c+a)=0 (iii) when (a+b)=0
[ this is because when b+c=0, RHS becomes 1/a and LHS becomes 1/a + (b+c)/bc = 1/a ]
CASE 1: WHEN b+c=0

here b can be chosen in 10C1 ways [we've 10 options intially from (-5,5) ]
for every value of b there remains only 1 value of c that makes b+c=0
a can be chosen in 8C1 ways
so, no. of ways = 10C1 * 1 * 8C1 = 80

CASE 2: when c+a=0

no. of ways = 10C1 * 1 * 8C1 = 80 [similarly]

CASE 3: when a+b=0

no. of ways = 10C1 * 1 * 8C1 = 80 [similarly]

Hence, TOTAL NUMBER OF CASES = 3*80 = 240
correct answer is (d)

[Superstar]

Hi Guys,

I think the answer is d)

I got 240 as the solution.

I solved the equation and got

(a+c) (b+c) (a+b) = 0

which would lead to 80*3 non distinct non-zero solutions

Open to discussion again...

[navigator09]

In a class there are 100 students. A division of the students in n sections is good if:
1) the sections have different numbers of students
2) for any partition of one of the sections in 2 smaller sections, among the (n+1) sections you get 2 with the same number of students (any section has at least 1 student).
The positive difference between the maximal and minimal possible value of n such that the division is good is

(a) 2 (b) 3 (c) 7 (d) 8 (e) none of the foregoing
Solution :
Well, the language of the question wasn't perfect and people interpreted this question in different manners.
Some assumed "one of the sections" as - some one, while others assumed it as each one. Then, "you get 2 with the same number", some assumed atleast 2, while some assumed exactly 2. We give the answer of all possible 4 cases.
Case 1: when the question meant each one + atleast 2 (debashis_dan was the 1st to get this right)
Max = [1, 2, 3, ..., 12, 22]. Min = [1, 3, 5, ..., 19]. Thus, answer is 13-10 = option (a).
Case 2: when the question meant each one + exactly 2 (sampatha and Superstar were the 1st to get this right)
Max = [1, 3, 5, ..., 19]. Min = [1, 3, 5, ..., 19]. Thus, answer is 10-10 = option (e).
Case 3: when the question meant some one + exactly 2
Max = [1, 3, 4, ..., 12, 24]. Min = [1, 3, 96]. Thus, answer is 12-3 = option (e).
Case 4: when the question meant some one + atleast 2
Max = [1, 2, 3, 4, ..., 12, 22]. Min = [1, 3, 96]. Thus, answer is 13-3 = option (e).


hey in the above solution I could understand the first 2 cases but the last 3 cases has me stumped...

Case 3:when the question meant some one + exactly 2
Max = [1, 3, 4, ..., 12, 24]. Min = [1, 3, 96].

when n=3 (in min) 96 can be split into 47 and 49 hence the split will not make any two sections with the same number of students..
I am confused... Can anyone explain this????:

[abhi2400]

i m sorry....but the choice 240 is in option (d) and NOT (c) as i had written in my previous post

[N-Ever Give up!]

Hi folks, This is just my thought...I feel that 240 cannot be the answer..The question says that each of a, b, c is distinct so we can have only 10C3 --> 120 ways of selecting the distinct combinations of a,b,c from 10 numbers (-5,-4,-3,-2,-1,1,2,3,4,5)...

Hence in my opinion the answer could be 72 or none of the foregoing...as 120 is too optimistic answer

I think the 240 condition includes for example (1,-1,1),(2,-2,2),(3,-3,3),(4,-4,4),(5,-5,5) wherein we dont have a distinct set of a,b,c
Please correct me if I am wrong ...Again its only my opinion...

[vineet.nitd]

Originally Posted by N-Ever Give up!

I think the 240 condition includes for example (1,-1,1),(2,-2,2),(3,-3,3),(4,-4,4),(5,-5,5) wherein we dont have a distinct set of a,b,c

Are you sure it does? .
Had it included those triplets , there would have been 300 solutions.Is not it?

Generally the best way to deal with triplet problems of this sort is to take a+b=p,b+c=q and a+c=r.I had mentioned this in one of the past QQAD problems as well.
And there are more than 1 or 2 ways of doing this problem.