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# RMO 2014 Delhi Region Q.4

Is it possible to write the numbers $$17,18,19,\ldots ,32$$ in a $$4\times 4$$ grid on unit squares, with one number in each square, such that the product of the numbers in each $$2\times 2$$ sub-grids $$AMRG, GRND, MBHR$$ and $$RHCN$$ is divisible by 16?

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Note by Aneesh Kundu
2 years, 5 months ago

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The answer is - impossible. Let us multiply all the numbers from 17 to 32 (the product will be equal to the product of four results for the sub-grids). It can easily be shown that the highest power of 2 in the product be 16. But as the number 32 (=2^5) should get in one of the sub-grids, the resultant 11th power of 2 should be distributed among three other grids. But it means that at least one of the products of the last three grids cann't be divisible by the fourth power of 2 equal to 16. · 2 years, 5 months ago

I solved it using pigeon hole principle · 2 years, 5 months ago

it would have been possible if we had one number having 2^1 . by the way .. i got problems 1 ,4 right . it was my first experience in rmo . . . · 2 years, 5 months ago

We have $$18=2\times9$$, $$20=2^{2}\times5$$, $$22=2\times11$$, $$24=2^{3}\times3$$, $$26=2\times13$$,$$28=2^{2}\times7$$, $$30=2\times15$$, $$32=2^5$$.

I will consider the set of powers of 2,$$A$$= $${1,2,1,3,1,2,1,5,0,0,0,0,0,0,0,0}$$ ($$0$$ corrsponds to the power of $$2$$ in the odd numbers).

Suppose we can partition the set $$A$$ into $$4$$ subsets $$X_{1},X_{2},X_{3},X_{4}$$ where $$|X_{i}|=4$$ where $$i=1,..,4$$ such that $$k_{i}≥4$$ where $$k_{i}$$ denotes the sum of the elements of $$X_{i}$$.

Note that the sum of the elements of $$A$$ is $$16$$.

Clearly $${5}$$ belongs to one of the $$4$$ subsets,say $$X_{1}$$.Then $$k_{1}≥5$$

Hence,$$k_{2}+k_{3}+k_{4}=16-k_{1}≤11$$ contradiction!(since sum of them is $$≥12$$)

So, the arrangement given in the problem is impossible. · 2 years, 5 months ago

yes , $$2^{4} =16$$

First let us write the prime factorization of all the even numbers ,

$$18 = 2 * 9$$

$$20 = 2^{2} * 5$$

$$22 = 2 * 11$$

$$24 = 2^{3} * 11$$

$$26 = 2 * 13$$

$$28 = 2^{2} * 7$$

$$30 = 2 * 15$$

$$32 = 2^{5}$$

The product of the numbers in 2x2 square to be divisible by 16 , each square should contain at least 2 raise to the power 4 . thus if in one 2x2 square if only 32 and other odd numbers is there - its satisfys . similarly - (18 , 20 ,22) , (20 ,28) ,(18 , 22 , 26 ,30) etc · 2 years, 5 months ago

Nope If u place 32 in one of the sub grids u still need 12 factors of 2 (to be placed in the 3 other sub grids)but we are left with only 11 factors so such an arrangement is not possible · 2 years, 5 months ago

Oh yes sorry i used 18 2 times Thank you . How much you are expecting to get right?@Aneesh Kundu · 2 years, 5 months ago

I solved 4

Don't know how many are correct · 2 years, 5 months ago