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# An oldie a professor once told me on Area and Perimeter

## Question

Two smart mathematicians (Ben and Jen) are told that a rectangle with integer sides L and W has been drawn, having perimeter less than 200, where L > W > 1 . Ben is told the area, Jen is told the perimeter. They now say:
Ben: "I can't determine the dimensions." Jen: "I knew that."
Ben responds, "Now I can determine them." Jen: "So can I."
Given these are true statements, what are the length and the width?

## Recommended Posts

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This is actually a duplicate puzzle as the one I posted a long while ago on these forums:

The only difference is the perimeter would obviously be twice the "SUM" in the problem I posted...but just divide by 2, and it's the same logic to figure out...I love this puzzle...

4 x 13

Edited by Pickett
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Let me do a bonanova -

For starters

L ans W both can't be prime numbers, otherwise Ben could easily find out the values of L & W.

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Further

This is just a guess, since Jen's 1st statement was enough for Ben to determine the numbers, i think, the number of ways 2 integers can be multiplied to get the area, should be only 2, if we apply the constraint that L+W<100.

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^ On secons thought, post 3 may not be right.

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I got 3 solutions: (4,13), (4,61), (16,73).

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A plumb bob 100 inches long i.e. fixed on its end is hanging.

then if we move the string to the right 1in ..the plumb bob

becomes 99 in hanging..(2 in..98in hanging).. the tip goes 45deg

..Let right move = W & hang length =L 100=half perimeter P

Area A= (100-W)W , if given a perimeter P.. A=(P/2-W)W

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We eliminate sets of L, W values according to the following reasoning.

Ben is told LW. He says "I can't determine L and W."

Suppose he is told LW = 21

Because 21 has only prime factors, he would know L=7 and W=3.

So we know, and now Jen knows, that LW is not 21, or any area with (only) prime factors.

Suppose Ben is told LW = 24.

LW could be (12x2), (8x3), or (6x4) now Ben can make his statement.

But from these possibilities, we cannot work back to get a unique area:

(12x2) => L+W=14 => 12x2, 11x3, 10x4, 9x5, 8x6; LW = 24, 33, 40, 45 or 48

( 8x3) => L+W=11 => 9x2, 8x3, 7x4, 6x5; LW = 18, 24, 28 or 30

( 6x4) => L+W=10 => 8x2, 7x3, 6x4; LW = 16, 21 or 24

How can Jen reason at this point?

Taking a different situation, suppose Jen is told L+W=10.

Then L=7 W=3 is possible.

But after Ben makes his statement, it becomes impossible.

So when Jen says "I already knew that" she is telling Ben that

there are no prime (L,W) compositions for his LW value.

So now Ben knows that L+W is NOT 10, or ANY sum of primes.

So if Ben is told LW=24, he knows after Jen's statement that

the rectangle is NEITHER 6x4 (7,3) NOR 12x2 (11,3).

He knows that it is 8x3.

So he can say "Now I can determine them."

But is this the solution?

Only if it solves L W for Jen as well.

What can she infer from Ben's ability to solve?

Can she infer the product from her sum?

Is there one and only one product that gives her sum?

Jen has been told L+W=11.

She knows LW is 18, 24, 28 or 30.

We know that if LW=24 Ben can find L and W.

Is there another LW that has only one (L,W) possibility?

If not, then Jen can make her last statement.

Let's look:

Suppose LW=18 the dimensions could be 9x2 or 6x3 and Ben knows L+W is 11 or 9.

If 11, then Jen could suspect 9x2, 8x3, 7x4, 6,5 and know Ben does not know

If 9, then Jen could suspect 7x2, 6x3, 5x4; because of 7x2, Jen would not know Ben does not know.

So her statement would tell Ben that L+W=11.

So Ben can solve if LW=24 (as we have seen) but he can also solve if LW=18.

If LW=24 he can deduce 8x3 but if LW= 18 he can deduce 6x3.

So Jen is at least uncertain whether LW=24 or LW=18

It could also be 28 (but not 30.)

28: 14x2, 7x4 => L+W is 16 or 11

16: 14x2, 13x3, 12x4, 11x5, 10x6, 9x7 ELIMINATE DUE TO 13x3

11: 9x2, 8x3, 7x4, 6x5 OK

30: 15x2, 6x5 => L+W is 17 or 11

17: 15x2, 14x3, 13x4, 12x5, 11x6, 10x7, 9x8 OK

11: 9x2, 8x3, 7x4, 6x5 OK

So if LW=28 Ben can decide

Summary:

If L=8 W=3, Ben would be told 24 and Jen would be told 22.

Ben could deduce L+W=11, but Jen could only deduce LW was 18, 24 or 30.

So L=8 W=3 is not a solution.

• L and W cannot both be prime
• Ben can solve only if his LW product allows only one set of LxW dimensions that does not include two primes.
• Jen can solve only if her L+W sum allows only one LW product that allows only one set of LxW dimensions that does not include two primes.

Gives L=13, W=4

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I am not surprised. As I stated in the forum title this is an old riddle. Sadly my forum seeking ability for old posts is not as strong as it should be.

This is actually a duplicate puzzle as the one I posted a long while ago on these forums:

The only difference is the perimeter would obviously be twice the "SUM" in the problem I posted...but just divide by 2, and it's the same logic to figure out...I love this puzzle...

4 x 13

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Though Pickett got the right answer bonus points go to Bonanova for an awesome explanation.

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Anyone cares to show why the other two solutions I've found (in post #5) are wrong?

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