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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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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## Comments

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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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