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

[vivekma]

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

Consider the set (4,3,1,8 ) and (5,2,6,7). In both cases, a=(b+c+d)/3. All numbers upto 8 are present.

Similarly, for (12,11,9,16) and (13,10,14,15) the conditions are satisfied. So for each consecutive group of 8, the equation is satisfied.

So we can generalize as (4+8n,3+8n,1+8n,8+8n) and (5+8n,2+8n,6+8n,7+8n) for all n>=0.

Hence it satisfies for all sets of multiples of 8.

correct option is (3).

[javed_nitb]

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 answer is n=12
.
.
since a= (b+c+d)/3 So b+c+d will be a multiple of 3
i.e b+c+d = 3,6,9,12... and correspondin a=1,2,3,4...

Now
a=1, not possible since b+c+d can't be = 3 being disjoint
a=2, not possible since no disjoint set possible
a=3, the only disjoint set possible is (3,1,2,6)
a=4, disjoint sets possible are(4,1,2,9), (4,1,3,8)(4,1,5,6)(4,3,2,7)
a=5, disjoint sets possible are(5,1,2,12), (5,1,3,11),(5,1,4,10)(5,1,6,8) (5,3,2,10), (5,4,2,9),(5,6,2,7)

we have used upto n=12

total no. of disjoint sets=12
So option (2)12

[shantanugangal]

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

a + b + c + d = a + 3a = 4a
Therefore, 1 + 2 + .. + n should be divisible by 4.
For n = 6, 12, 36 : Sum is not divisible by 4.
For n = 16, sum = 126.
Ans: (3) 16

[killerkart19]

Originally Posted by shantanugangal

a + b + c + d = a + 3a = 4a
Therefore, 1 + 2 + .. + n should be divisible by 4.
For n = 6, 12, 36 : Sum is not divisible by 4.
For n = 16, sum = 126.
Ans: (3) 16

The disjoint subset's sum should be divisible by 4 as u say, and not the sum of the whole set. We talking about subsets here and sum of whole set may be irrelevant..

[implex]

Originally Posted by javed_nitb

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 answer is n=12
.
.
since a= (b+c+d)/3 So b+c+d will be a multiple of 3
i.e b+c+d = 3,6,9,12... and correspondin a=1,2,3,4...

Now
a=1, not possible since b+c+d can't be = 3 being disjoint
a=2, not possible since no disjoint set possible
a=3, the only disjoint set possible is (3,1,2,6)
a=4, disjoint sets possible are(4,1,2,9), (4,1,3,(4,1,5,6)(4,3,2,7)
a=5, disjoint sets possible are(5,1,2,12), (5,1,3,11),(5,1,4,10)(5,1,6, (5,3,2,10), (5,4,2,9),(5,6,2,7)

we have used upto n=12

total no. of disjoint sets=12
So option (2)12

what about a=6 for example (3,7,

[implex]

a=(b+c+d)/3
(b+c+d)=3a
(a+b+c+d)=4a
since we get n disjoint subsets
sum of the disjoint subsets
4(a1+a2+..an)=n(n+1)/2
n= is of form 8k

so n=16

[Aarav]

Monday morning blues or after mock-affects? This problem is simple to invite more right responses.

[javed_nitb]

Originally Posted by implex

what about a=6 for example (3,7,8)

Thanks implex!
...i identified my mistake...
...its in the def. of disjoint set itself...
...Ur approach is damn cool...
...each disjoint set vl ve divisible by 4...so their summation vl be of the form...4k
which implies... 4k = (summation of the elements in the set (1,2,3,4,...n))

4k = Sum of n natural no. = n(n+1)/2
n(n+1) = 8k implies n vl be multiple of 8
Hence answer is (c) 16
....thanks once again...

[javed_nitb]

Originally Posted by killerkart19

The disjoint subset's sum should be divisible by 4 as u say, and not the sum of the whole set. We talking about subsets here and sum of whole set may be irrelevant..:bigear:

...the problems says that the set (1,2,3,4...n) can be split into n disjoint elements subset (a,b,c,d)...
..since summation of elements of each subset is divisible by 4
...and they all together form the original set...disjointly
..hence...the summation of elements of original set must be divisilble by 4...

[Varun Khullar]

Originally Posted by implex

a=(b+c+d)/3
(b+c+d)=3a
(a+b+c+d)=4a
since we get n disjoint subsets
sum of the disjoint subsets
4(a1+a2+..an)=n(n+1)/2
n= is of form 8k

so n=16

nice .. solution..