Each of the integers $1,$ $2,$ $3,$ $\dots,$ $16$ is written on a separate slip of paper and these slips are placed in a pile. Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of the numbers she has drawn from the pile have a product that is a perfect square. What is the maximum number of slips that Jillian can draw without obtaining a product that is a perfect square?

Question

03-01-2020
Mathematics

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Each of the integers $1,$ $2,$ $3,$ $\dots,$ $16$ is written on a separate slip of paper and these slips are placed in a pile. Jillian will randomly draw slips from the pile without replacement and will continue drawing until two of the numbers she has drawn from the pile have a product that is a perfect square. What is the maximum number of slips that Jillian can draw without obtaining a product that is a perfect square?

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ibtehajhassan ibtehajhassan · Answer 1 · 2020-01-03T09:32:13+01:00

Answer:

11 slips

Step-by-step explanation:

A perfect square is a positive integer that is the square of another integer. For example, 25 is a perfect square of 5. See the calculation below

.

[tex]5^{2} = 5*5\\5^{2} = 25[/tex]

There is a rule that should be kept in mind

1. When two perfect squares are multiplied by each other (e.g. 4 * 9), the result is a perfect square ([tex]36 = 6^{2}[/tex])

Let's identify the combination of numbers that result in perfect square, when multiplied with each other. These combinations are as follows,

• 1*4, 1*9, 1*16

• 2*8

• 3*12

• 4*9, 4*16

• 9*16

From the list of numbers 1, 4, 9 and 16 are already perfect square e.g. [tex]2^{2} = 4, 3^{2} = 9[/tex]. If they are multiplied by each other, the result will also be a perfect square. Let’s assume that our first number is 1. Now we can't have any of the three numbers (except for 1), mentioned above. This rule out these three numbers.

Next, from 2, 8, 3 and 12 we can only draw two numbers. e.g. if we take 2, we can’t take 8 as it will give a perfect square. Same goes with 3 and 12. Hence from these four numbers we can discard two of them.

We discarded three numbers initially and two now. Therefore, out of 16 slipds we can draw a maximum of 11 slips without obtaining a product that is a perfect square.