Our school year started last Monday. My teaching muscles atrophied a bit over the summer, so last week's classes were the pedagogical equivalent of stretching exercises. But starting tomorrow we're really going to hit the ground running. Do you know what that means?

That's right! It means that Problem of the Week makes it's triumphant return. This term's theme: A Tribute to Sam Loyd.

Problem One has just been posted. It's a fairly straightforward algebra problem, but good if you're in the mood for some mental calisthenics. New problems will appear every Monday. Enjoy!

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I solved the problem by brute force algebra, but I can't help thinking I should have found a more insightful approach. Anyway the algebra took much longer than it should have done, because I kept making mistakes! My mathematical muscles have atrophied with age, so some more calisthenics would be useful. Thanks.

X = # eggs bought

Y = normal price eggs per dozen

then:

12 cents = X/12 (Y-1)

and

12 cents + (X/12) = XY/12

Abracadabra : X = 36 eggs, Y = 5 cent/dozen

I had to read it three times before realizing I was assuming that the eggs, being eggs, were sold by the dozen at the start. Then I got stuck trying to determine the approppriate algebraic equations and just used trial and error (what if it was intialle one egg? Two eggs?)

I found that it works with 4 eggs, if I'm interpreting it right. The usual price is 3 cents an egg. He buys 4 eggs for 12 cents, then has them throw in another 2 eggs. This brings the price down to 12 cents for 6 eggs or 2 cents an egg, a savings of one penny compared to the original price of 3 cents.

I assume I'm misinterpreting something.

Glarg, I see it now. "One cent per dozen less", not one cent per egg. That'll teach me to try solving these by half-remembering them while walking my dogs.

Thanks, that was a fun little problem.

n is number of dozen eggs purchased

c is cost per dozen in cents

so

nc = 12

n(c-1) = nc 12n/(12n+2)

...

12n^2 + 2n - 24 = 0

n = 4/3 (or -3/2)

c = 9

so 18 eggs were bought for 12 cents at 8 cents per dozen.

Since I am sitting at my computer:

function eggs()

paid = 12 -- price paid in cents

dozen = 12 -- dozen

for n = 3,144 do -- at least 3, guess less than a gross

pp = dozen*(paid/n) -- price paid per dozen

ap = dozen*(paid/(n-2)) -- asking price dozen

if (ap-pp) == 1 then -- 1 cent difference

print("number of eggs bought: ",n)

print("asking price per dozen: ", ap, "cents")

print("price paid per dozen: ", pp, "cents")

break

end

end

end

I got the same result as rhr using 3 simultaneous equations:

n = original # of eggs purchased (before adding 2 more)

p = original price per egg

q = new price per egg

12p = 12q + 1

pn = q(n+2)

pn = 12

n = 16 p = 3/4 q = 2/3 so that

12*(3/4) = 12*(2/3) + 1 (relating the old price per dozen and the new price per dozen).

The original purchase was to be 16 eggs but with the additional 2, the actual total is 18 eggs so that

16*(3/4) = 18*(2/3) = 12 cents, tying everything together.

A cute problem! I look forward to solving others in the future.

Including the 2 free eggs, 18 eggs were purchased. Generally they would be sold for 9 cents per dozen, so the original 16 eggs cost 12 cents. However, 12 cents for 18 eggs would come out at 8 cents per dozen - 1 cent less per dozen.

I solved this by forming 3 simultaneous equations, where c0 denotes the cost per egg under normal circumstances and c1 denotes the cost per egg in this case, and x denotes the number of eggs purchased (excluding the 2 free eggs):

12c1 + 1 = 12c0

(x+2) * c1 = 12

x * c0 = 12

From there, I solved the latter 2 equations in terms of c0 and c1:

c1 = 12 / (x + 2)

c0 = 12 / x

And inserted them into equation #1:

12 * (12 / (x + 2)) + 1 = 12 * (12 / x)

This was then rearranged until it came to the following quadratic equation:

x^2 + 2 * x - 288 = 0

Then, using the quadratic equation I could solve this to find the following 2 solutions:

x = -18

x = 16

As the number of eggs bought must be a positive number, only the second solution can be correct.

I could then use x to calculate c0 and c1, and arrive at the usual cost of 9 cents per dozen eggs from c0 and the unusual cost of 8 cents per dozen eggs from c1.

I set it up this way:

x = original cost per dozen (in cents)

y = number of eggs originally purchased (not including the two free eggs)

The problem then becomes

x * y/12 = 12 ---> xy = 144

and, since after getting the free eggs, the chef purchased y+2/12 dozen eggs for 12 cents, the cost per dozen is

12/[(y+2)/12] or equivalently 144/(y+2). From the problem this is equal to x-1, so 144/(y+2) = x-1 or (x-1)(y+2) = 144.

We can therefore set xy = (x-1)(y+2) since both of these equal 144. This reduces to 2x-y-2=0. Since 144/x = y, this becomes 2x - 144/x - 2 = 0. Thus 2x^2 - 2x -144 = 0 or

x^2 - x - 72 = 0. This is easily factored into (x-9)(x+8) = 0. Thus x=9 (or x=-8, but this answer doesn't make sense given the problem.) Since y = 144/x, y=16 and therefore 16 eggs were originally purchased (18 were obtained, including the freebies).