CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

[nv_kkd]

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Quantitative Question # 062
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For which positive integer values of n the set {1, 2, 3, ..., n} can be split into n disjoint elements subsets {a, b, c, d} such that a = (b+c+d)/3?

(1) 6 (2) 12 (3) 16 (4) 36 (5) at least two of the foregoing

[nv_kkd]

sorry wrong post

[killerkart19]

Originally Posted by nv_kkd

when N-6 ,no.of Subsets Satisfying The Cond Is 2
N-12,no.of Subsets>12
N-16,no.of Subsets>16
-> Option 5 Is Ruled Out

Hence I Go With Option 4 (36)

Can U Please Elaborate More On Ur Approach Mate....

[nbangalorekar]

this question can be solved by simple elimination:
since we have 4 distinct elements in each subset, 'n' shud be divisible by 4.
directly option (1) is eliminated.
now let a1, a2, a3.... be the different values 'a' can take.
so we get 4*(a1+a2+a3......)=n(n+1)/2 {since all subsets r disjoint}
so n(n+1) shud be divisible by 8
answer can be only option (3)
however will try to come out with a more refined method...

[Montubhardwaj]

For which positive integer values of n the set {1, 2, 3, ..., n} can be split into n disjoint elements subsets {a, b, c, d} such that a = (b+c+d)/3?

(1) 6 (2) 12 (3) 16 (4) 36 (5) at least two of the foregoing

I think the ans fr this is n=6.
coz only fr n=6,the set satisfies the condition.
the set ehich satisfies d condition is (4,5,6,1)
where 4= (5+6+1)/3

Pl correct me if am wrong

[shivam_01]

Originally Posted by nv_kkd

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Quantitative Question # 062
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For which positive integer values of n the set {1, 2, 3, ..., n} can be split into n disjoint elements subsets {a, b, c, d} such that a = (b+c+d)/3?

(1) 6 (2) 12 (3) 16 (4) 36 (5) at least two of the foregoing

My take :-

since the (b+c+d)/3 =a hence (b+c+d) should be a multiple of 3
say 3,6,9,12,15 etc
but for 3 no disjoint set can be found
hence we should go with remainining values
now
n=6
a=(b+c+d)/3 which gives solution as a=4 and b=1,c=5,d=6
similarly for n=9 we can write it as (6,1,8,9) or (6,2,7,9) but it is not a disjoint set hence neglected

now in case of 12 solution set is (8,2,12,10) which is a disjoint set

in case of 16 we cannot find a disjoint set hence not possible

for n=36 set can be (24,7,36,29) which is a disjoint set hence

my take is option e (5) at least two of the foregoing

correct me if i am wrong

[Fabio capello]

The only answer could be 6

[madadi]

Originally Posted by Fabio capello

The only answer could be 6

Hey Fabio, how can you say that? wat wud you say about the disjoint set formed by taking n=12, then?? the set comes out to be-->(7,8,9,4)....

[madadi]

The answer according to me should be--> option 5, i.e, at least two of the foregoing...
this is because, if you take n=6, the subset which complies by the given condition is (3,6,2,1)...
for n=12, the disjoint set formed is--> (7,8,9,4)...
for n=16, no such subset can be formed....
for n=36, the subset formed is (14,11,15,16)...

[mba.yodha]

Originally Posted by nbangalorekar

this question can be solved by simple elimination:
since we have 4 distinct elements in each subset, 'n' shud be divisible by 4.
directly option (1) is eliminated.
now let a1, a2, a3.... be the different values 'a' can take.
so we get 4*(a1+a2+a3......)=n(n+1)/2 {since all subsets r disjoint}
so n(n+1) shud be divisible by 8
answer can be only option (3)
however will try to come out with a more refined method...

Can you plz explain the bold part