GATE1991-01,xiii

Question

The number of integer-triples $(i,j,k)$ with $1 \leq i,j,k \leq 300$ such that $i+j+k$ is divisible by 3 is________

Answer 1

For $i$ and $j$ we have $300$ combinations each as they can repeat and $(x,y,z)$ and $(y,x,z)$ are different as order inside triplets are to be considered. Now, once we have chosen $x,y,$ $z$ must be chosen so that $x+y+z$ is divisible by 3. From $1-300$, we have 100 multiples of 3. So, we get $300/3 = 100$ choice for $z$ making total no. of triplets $= 300 \times 300 \times 100 = 9 \times 10^6.$

Comments

sir,as per your method value of i and j can be from 1-300  value of k can be from set

{3,6,9,....,300}

let i=5,j=5,k=3

then (i+j+k)=(5+5+3)=13 which is NOT divisible by 3

plz help me understand

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case -1. when i%3=j%3=k%3=0.

Total combinations= 100*100*100

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case 2- when i%3=j%3=k%3=2.

Total combinations= 100*100*100.

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case 3- when i%3=j%3=k%3=1.

Total combinations= 100*100*100.

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case 4- when i%3=0,j%3=1,k%3=2.// total 6 combination

Total combinations= 6*100*100*100.

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Total = 9 * 10^6.

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@Aksh if i=5 and j=5 then we will not choose k=3 but k=2,5,8,....,299 (which are 100 terms)

Hence for every combination of i and j we have 100 'k' terms for which i+j+k is divided by 3.

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@Aksh  what Arjun Sir has said is that u choose any x and y , depending on this u choose z, so if say x+y =5+5 =10 then u choose z such that (10 +z ) mod 3 =0 ie. z mod 3=-10%3=(-10+3*4)%3=2 ie. z mod 3 =2, so u will get ultimately 100 numbers for z between 1 and 300 .

Answer 2

@Arjun sir: why is this wrong?

number of numbers (1 to 300) giving remainder 0 when divided by 3 =300/3=100
number of numbers (1 to 300) giving remainder 1 when divided by 3 =300/3=100
number of numbers (1 to 300) giving remainder 2 when divided by 3 =300/3=100


sum of 3 numbers would be divisible by 3 as:
(consider the remainders that each number(i/j/k) produces when divided by 3)
 

1) 1 1 1 ==>100 * 99 * 98

2) 0 1 2 / 0 2 1 / 1 0 2 / 1 2 0 / 2 0 1 / 2 1 0 ==> 100 * 100 * 100

3) 3 3 3( I mean...0 0 0) ==> 100 * 99 * 98

so totally:
2*100*99*98 + 100*100*100

=2940400

 

Comments

you missed the case for (2,2,2). Also, there is no need for the numbers to be distinct as per question. So, sum will be

$3 * 100 * 100 * 100 + 6 * 100 * 100 * 100  = 9 * 10^6$

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perfect! thanks.