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?

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.

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

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

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.

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TopNewestThe 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.

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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

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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

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

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

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Don't know how many are correct

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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.

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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 . . .

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I solved it using pigeon hole principle

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