RMO 2014 Delhi Region Q.4

Is it possible to write the numbers 17,18,19,,3217,18,19,\ldots ,32 in a 4×44\times 4 grid on unit squares, with one number in each square, such that the product of the numbers in each 2×22\times 2 sub-grids AMRG,GRND,MBHRAMRG, GRND, MBHR and RHCNRHCN is divisible by 16?


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Note by Aneesh Kundu
4 years, 8 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.

Сергей Кротов - 4 years, 8 months ago

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yes , 24=16 2^{4} =16

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

18=2918 = 2 * 9

20=225 20 = 2^{2} * 5

22=21122 = 2 * 11

24=231124 = 2^{3} * 11

26=21326 = 2 * 13

28=22728 = 2^{2} * 7

30=215 30 = 2 * 15

32=2532 = 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

sandeep Rathod - 4 years, 8 months ago

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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

Aneesh Kundu - 4 years, 8 months ago

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Yes, I tried combinations, but all fell in vain.

Ritu Roy - 4 years, 8 months ago

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Oh yes sorry i used 18 2 times Thank you . How much you are expecting to get right?@Aneesh Kundu

sandeep Rathod - 4 years, 8 months ago

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@Sandeep Rathod I solved 4

Don't know how many are correct

Aneesh Kundu - 4 years, 8 months ago

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We have 18=2×918=2\times9, 20=22×520=2^{2}\times5, 22=2×1122=2\times11, 24=23×324=2^{3}\times3, 26=2×1326=2\times13,28=22×728=2^{2}\times7, 30=2×1530=2\times15, 32=2532=2^5.

I will consider the set of powers of 2,AA= 1,2,1,3,1,2,1,5,0,0,0,0,0,0,0,0{1,2,1,3,1,2,1,5,0,0,0,0,0,0,0,0} (00 corrsponds to the power of 22 in the odd numbers).

Suppose we can partition the set AA into 44 subsets X1,X2,X3,X4X_{1},X_{2},X_{3},X_{4} where Xi=4|X_{i}|=4 where i=1,..,4i=1,..,4 such that ki4k_{i}≥4 where kik_{i} denotes the sum of the elements of XiX_{i}.

Note that the sum of the elements of AA is 1616.

Clearly 5{5} belongs to one of the 44 subsets,say X1X_{1}.Then k15k_{1}≥5

Hence,k2+k3+k4=16k111k_{2}+k_{3}+k_{4}=16-k_{1}≤11 contradiction!(since sum of them is 12≥12)

So, the arrangement given in the problem is impossible.

Souryajit Roy - 4 years, 8 months ago

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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 . . .

Nihar Mahajan - 4 years, 8 months ago

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I solved it using pigeon hole principle

Sauditya YO YO - 4 years, 7 months ago

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