# ZIO 2019 Discussion

UPDATE2:questions: https://www.iarcs.org.in/inoi/2019/zio2019/zio2019-question-paper.pdf

Hi how was Zio if possible post it answers

Post ur zio scores here

Mostly accepted answers(from discuss and ICO whatsapp group)

1a.69 1b.105 1c.133

2a.8 2b.7 2c.11

3a.198 3b.520 3c.1363

4a.144 4b. 8192 4c. 1638400

Question 4: N×N array with -1 and 1

It is magical if exactly 1 column, 1 row has product =-1 all other column and row product =1.

For a given N find no magical grids.

4a.N=3 4b. N=4 4c. N=5

Solution 4: Fix the ROW and COLUMN that have the Product of elements =-1;

Now RANDOMLY fill 1,-1 in remaining cells.

The no of such possibility =2^{(n-1)^2};

Now leave the intersection cell(of selected row and column) other than that for each cell in the selected row and column the value of that cell is fixed. ( Since the product of elements in REMAINING row or column needs to be =1)

Now sign of intersection cell is also fixed.

Proof:Product of all elements in grid EXPECT than selected ROW and COLUMN and Product of elements in ROW except the INTERSECTION is SAME.

Because the product of all other columns except selected COLUMN is 1

Similarly
Product of all elements in grid EXPECT than select ROW and COLUMN and Product of elements in COLUMN except the INTERSECTION is same.

Which implies
Product of elements in COLUMN except for the INTERSECTION
And
Product of elements in ROW except the INTERSECTION is same

So if that product of elements in Selected ROW expect INTERSECTION is -1 value of INTERSECTION cell is 1
Else it is -1

So if we fix ROW and COLUMN whose product is -1 and the fix the REMAINING cells then selected Row and column can be filled in only one way.

so Total no such GRIDS possible =no of ways of choosing row and column × no ways of filling remaining cells.

Total no such GRIDS possible=n^22^{(n-1)^2}

question 1: Given integer N.
Take first N natural numbers

Find no subsequence such that in that subsequence a[i-1] divides a[i] for all i in that subsequence?

1a.N=15 1b.N=19 1c.N=22

Solution 1: Have dp[i]=no of such subsequence ending with the last element as i;
If i is prime dp[i]=2.

else dp[i]=2+dp[all the factors of i];

ex:dp[6]=2+dp3+dp2;

ans=summation all dp[] till N;

Thanks, @kayak for the solution for question 4.

3 Likes

@lokesh2002 I also got the same answers for Q1, 2, 3 (3b was 520). For fourth, I think the answer for a is 72. I don’t know about 4b and 4c.

Can anyone give full question 4 statement

@ssp547 hope now question 4 it is clear

In your opinion what will be the cutoff this year?

The answer for 4a is 144.

is there some specific reason for my answers in ques 1 to be less by 6 than the most accepted answers in each case.
1=>63
2=>99
3=>127

@gutloo I have updated question and soln for question 1.it might help u.

Any guesses on what the cut off is going to be for 11th? It was 40 last year IIRC and IMO, this year the test was way harder so 35-40?

1] 70,
2] 102,
3] 117,
4] 7 ,
5] 7,
6] 11,
unfortunately, I got further answers are wrong, the answers were too small to compare with!

What would the cutoffs be for 9th grade?

could somebody please send me the link to the ICO whatsapp group?