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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#Combinatorics #NumberTheory #RMO #Divisibility

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Comments

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 - 6 years, 6 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

sandeep Rathod - 6 years, 6 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

Aneesh Kundu - 6 years, 6 months ago

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

Ritu Roy - 6 years, 6 months ago

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

sandeep Rathod - 6 years, 6 months ago

@Sandeep Rathod – I solved 4

Don't know how many are correct

Aneesh Kundu - 6 years, 6 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.

Souryajit Roy - 6 years, 6 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 . . .

Nihar Mahajan - 6 years, 6 months ago

I solved it using pigeon hole principle

Sauditya YO YO - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$