Grinch's Toy Store

[bonanova]

After disappointing Black Friday sales, management at Grinch's Toy Store found a way to boot profits.

Following an approach used in some amusement parks, GTS now sells coupons, which customers then use for payment at checkout.

The store's inventory comprises 22 toy models, each clearly marked with a unique, whole-dollar price.

Coupons are available in $6, $9 and $20 denominations.

Grinch's strategy is threefold.

[1] At checkout, extra coupon value is forfeit; no change is given.

Example: your total at checkout is $5. You pay with a $6 coupon. Grinch's keeps the $1 change.

[2] Prices have been set to ensure there is extra coupon value when toys are bought singly.

[3] Each toy model is limited one to a customer.

Your task is to visit Grinch's, browse the shelves, and find the minimum number of toys which, when bought together, using coupons, will ensure a fair purchase at checkout.

That is, there will be no change for Grinch's to keep.

This just in - Grinch's heard you were coming.

Fearing your intellectual prowess, they have replaced all price tags with UPC stickers, which only the checkout register can read.

[Guest]

As for 9 items --- Without the costs of the items we cannot solve this problem.

On the contrary, I believe I did just that. If you see a flaw in my analysis, please point it out. (I'll be the first to thank/congratulate you!)

[Guest]

spoxjox has it. 43 is the highest number that can't be made with a combination of 6s, 9s, and 20s. Every number higher than that can. 43 also happens to be the lowest total that can be had using 8 coupons where there will be extra coupon value. Since you can't get a total of 43 or less with 9 of the lowest possible coupon amounts, 9 must be the answer.

[Guest]

Ok, you already established I was wrong to set a minimum price on any item. If 9 is true, then I should not be able to come up with 9 prices which are not a multiple of 6, 9, or 20, or any sum thereof, which in total would not be a multiple of 6, 9, 20 or a sum thereof.

Try $301, $323, $401, $452, $501, $502, $601, $777, $998. Those are all valid prices.

That is 9 items. You will still give Grinch change. Total = $4856. 242 x $20. 1 x $9. 1 x $6. Grinch keeps $1.

The answer must allow 0 possibility of giving grinch any money from change. 9 cannot be the answer.

[Guest]

Ok, you already established I was wrong to set a minimum price on any item. If 9 is true, then I should not be able to come up with 9 prices which are not a multiple of 6, 9, or 20, or any sum thereof, which in total would not be a multiple of 6, 9, 20 or a sum thereof.

Try $301, $323, $401, $452, $501, $502, $601, $777, $998. Those are all valid prices.

That is 9 items. You will still give Grinch change. Total = $4856. 242 x $20. 1 x $9. 1 x $6. Grinch keeps $1.

The answer must allow 0 possibility of giving grinch any money from change. 9 cannot be the answer.

241($20) + 4($9) = $4856.

Read through my solution. I demonstrate that any integer greater than 43 can be expressed as a sum of multiples of 6, 9, and 20. Since it requires a maximum of nine items to exceed a value of $43, nine is the magic number.

[Guest]

Try $301, $323, $401, $452, $501, $502, $601, $777, $998. Those are all valid prices.

None of them are valid prices. There is no whole number over 43 that can't be made with a combination of 6s, 9s, and 20s.

$301 can be paid exactly with 29 $9 coupons and 2 $20 coupons.

$323 can be paid exactly with 27 $9 coupons and 4 $20 coupons.

$401 can be paid exactly with 29 $9 coupons and 7 $20 coupons.

$452 can be paid exactly with 34 $9 coupons and 7 $20 and 1 $6 coupon.

$501 can be paid exactly with 29 $9 coupons and 12 $20 coupons.

and so on.

That is 9 items. You will still give Grinch change. Total = $4856. 242 x $20. 1 x $9. 1 x $6. Grinch keeps $1.

But if you give him 241 x $20 and 6 x $6, the Grinch keeps nothing.

[Guest]

Ok, took me a while but I smell what you are cooking now. Math's my weak area. Sorry I was so dense on that one. LOL

[bonanova]

43 can't be matched with coupons; any number greater than 43 can.

You can verify that for 44, 45, 46, 47, 48 and 49 - then just add $6 coupons for higher numbers.

There are 22 numbers less than 44 that can't be matched with coupons, so you know the 22 toy model prices. And WB is correct - they do start with $1.

Starting with the lowest numbers, you can stay below 44 with eight of them, but not with nine.

[Guest]

Following an approach used in some amusement parks, GTS now sells coupons, which customers then use for payment at checkout.

You know what? I think this implies no cash. I didn't read it as meaning that initially, but on rereading, I think I agree with Martini that this precludes cash usage.

[unreality]

0

[bonanova]

0

Your task is to visit Grinch's, browse the shelves, and find the minimum number of toys which, when bought together, using coupons, will ensure a fair purchase at checkout.

LTNS amigo....!

whichever one you use Grinch will keep the entire amount.

Not a fair purchase.

[unreality]

Sorry- I guess I just assumed there was a nil set item or whatever it's called... like a $0 coupon by the absence of a coupon

[bonanova]

Instant replay -- on further review ...

I might not have closed the loophole on that point - should have said using at least one coupon.

Zero may be a valid number of coupons to use.

[Guest]

Zero may be a valid number of coupons to use.

No, it's not. The riddle states "find the minimum number of toys which, when bought together, using coupons...". Toys must be bought and coupons must be used.

[bonanova]

That was my original stance.

I decided, in the spirit of the "gift-giving season" perhaps, that UR had earned some slack.

The OP might have been stated more specifically about that point.

[1] One way would state that you must select at least one toy to purchase.

[2] Another way might state: You must pay for all your toys using coupons.

Alternative [1] does not permit UR's answer, while [2] does. Universal quantifiers do not possess existential import.

The statement "All idiots with IQ greater than 200 and three lefts hands are male" does not imply that at least one of them exists.

If All is replaced by Some, the statement does imply at least one exists.

I cut UR the slack because the OP does not express the idea of "some" toys or "some" coupons.

Absent that language, I didn't explicitly rule out buying zero toys using zero coupons.

That being the case, it seems more a matter of semantics than of logic.

[Guest]

That was my original stance.

I decided, in the spirit of the "gift-giving season" perhaps, that UR had earned some slack.

That's what I figured, but I've always had a bit of Grinch in me.

I cut UR the slack because the OP does not express the idea of "some" toys or "some" coupons.

Sure it does. "When bought together" refers only to situations where toys are actually bought and we know that coupons must...there I go again.

[unreality]

lol my answer seems to have stoked some discussion

Merry Christmas everyone! (or whatever you celebrate!)

[bonanova]

For sure ... it was relatively quiet while you were "away".

Merry Christmas* to all ... !

* or equivalent.

[unreality]

Yeah I didnt come to this forum for a while, real life stuff

[Guest]

excuse my rekindling of a dead topic, but i believe i have found useful/less information

since no item can be above 43,

the prices are (in dollars)

1,2,3,4,5,7,8,10,11,13,14,16,17,19,22,23,25,31,34, 37, 41, 43,

So if you randomly picked the lowest

1. 1 (1)

2. 2 (3)

3. 3 (6)

4. 4 (10)

5. 5 (15)

6. 7 (22)

7. 8 (30)

8. 10 (40)

9. 11 (51)

And we stop when we hit above 43

So yes, the answer is 9. *whew*

but if you say: couldn't the answer be 7 if you take away the 4,3, and 1 to make 43?

But then what if you just picked the lowest 7. Then you can't.

So I'm not sure if I am restating the answer but that is how I see it.

[Guest]

If five of the 22 item were marked at 7 dollars then five would be the answer

[bonanova]

If five of the 22 item were marked at 7 dollars then five would be the answer

The OP doesn't allow for more than one item marked the same price:

The store's inventory comprises 22 toy models, each clearly marked with a unique, whole-dollar price.

[bonanova]

excuse my rekindling of a dead topic, but i believe i have found useful/less information

since no item can be above 43,

the prices are (in dollars)

1,2,3,4,5,7,8,10,11,13,14,16,17,19,22,23,25,31,34, 37, 41, 43,

So if you randomly picked the lowest

1. 1 (1)

2. 2 (3)

3. 3 (6)

4. 4 (10)

5. 5 (15)

6. 7 (22)

7. 8 (30)

8. 10 (40)

9. 11 (51)

And we stop when we hit above 43

So yes, the answer is 9. *whew*

but if you say: couldn't the answer be 7 if you take away the 4,3, and 1 to make 43?

But then what if you just picked the lowest 7. Then you can't.

So I'm not sure if I am restating the answer but that is how I see it.

You have it correctly.

Seven won't work because you can find 7 items that would lose you some money and the OP asks that you ensure that does not happen.

find the minimum number of toys which, when bought together, using coupons, will ensure a fair purchase at checkout.

[Guest]

I'm totally confused, what's the prices of the toys?

If they went from 1-22 then surely you buy 22 toys and which would make 253 then you pay with 11 20 vouchers, 3 9 vouchers and a 6 voucher if I'm making any sense at all which is unlikely.

If the prices were 1,2,3,4,5,7,8,10,11,13,14,16,17,19,22,23,25,31,34, 37, 41, 43, as suggested then you buy all 22 making the price 386 so you use 19 20 vouchers and a 6 voucher.

I don't know if I'm understanding this but thought I'd chip in.

(sorry typo, all better now)

Edited March 14, 2008 by chicken-licken

[bonanova]

I'm totally confused, what's the prices of the toys?

If they went from 1-22 then surely you buy 22 toys and which would make 253 then you pay with 11 20 vouchers, 3 9 vouchers and a 6 voucher if I'm making any sense at all which is unlikely.

If the prices were 1,2,3,4,5,7,8,10,11,13,14,16,17,19,22,23,25,31,34, 37, 41, 43, as suggested then you buy all 22 making the price 386 so you use 19 20 vouchers and a 6 voucher.

I don't know if I'm understanding this but thought I'd chip in.

(sorry typo, all better now)

The prices of the toys can be deduced from the information given.

[1] the prices are unique

[2] none of the items can be bought singly without losing money - none of them were the exact value of any combination of coupons.

The puzzle asks for the minimum number of toys. 22 works, but there are smaller numbers that work.

Does that help?

[Guest]

Ok thank you very much. I'll have a think.