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### Prices in Store 88

```
Date: 07/05/99 at 17:56:03
From: Darwin Fu
Subject: Prices in Store 88

I really like this question my dad give me. Do you know the answer?

Store 88 is on 8th street. Inside they sell exactly ten items. Some
items may have the same price as others, but the only the digit on the
price tags of each item is 8. If you buy one of each item, it will
cost exactly \$1000.

Darwin Fu
```

```
Date: 07/06/99 at 12:06:59
From: Doctor Peterson
Subject: Re: Prices in Store 88

Hi, Darwin. You're right, this is a good question! I hadn't seen it
before.

I'll show you my reasoning as I work through this, and see how I do.

With ten items totaling \$1000, they must average \$100 each, and at
least one must cost more than that. Since there can be only 8's in any
price, our first item must be \$888, which leaves 112 as the total of
the other 9 items. They must average over \$12 each, so there must be
at least one \$88 item, which will leave \$24 for the other 8 items. But
that means they must all cost \$3, which isn't allowed. Well, we could
make \$24 using three \$8 items, with the other five items labeled
"free"; maybe that's legal, since there are no digits other than 8 in
it, but I'll assume I can't do that.

Ah, I've forgotten about cents. Presumably \$888 would not have to be
written as \$888.00, so what we've done so far is legal, but we may be
able to add .88 to some or all prices. A tag could say \$x, or
\$x.88, or 88 cents, or 8 cents. This will make a big difference.

Suppose we have two \$8 items; then we've used four items and have \$8
left to divide among 6 items, which average over \$1. How can I do
that?

Let's back up a moment. We're going to end up with some extra cents if
we keep going this way, and we have to end up with an even number of
dollars. If we have to use only 8 cent and 88 cent items to make a
whole number of dollars, how can we do that? We'll have only 8's in
the cents column, so we must have either five 8's (40) or ten 8's (80)
to give us a zero in that column. Either all ten prices end in .88 (or
are 8 cents alone), or five of them do.

Let's suppose they all end in .88:

888.88
88.88
?.88
?.88
?.88
?.88
?.88
?.88
?.88
?.88
------
984.80 + ?

That's .80 too high to be an even number of dollars, so we'll have to
change either one or six .88's to .08, saving either .80 or 4.80.
First, let's change only one. That will leave us with seven unknown
dollar amounts that have to add up to \$16; some can be 0 and some can
be 8. That's easy: use two 8's and five 0's:

888.88
88.88
8.88
8.88
0.88
0.88
0.88
0.88
0.88
.08
------
1000.00

That works!

Now, let's change six .88's to .08 and see if that can work too. That
will leave us with two unknown dollar amounts that have to add to \$20:

888.88
88.88
?.88
?.88
.08
.08
.08
.08
.08
.08
------
980.00 + ?

I can't do that with just two 8's, so this won't work.

This means we have one solution: the prices are 888.88, 88.88, two
8.88, five 88 cents, and one 8 cents. This doesn't depend on being
able to write a price as \$88 with no cents; if I'd known for sure that
I didn't have to consider that case, I could have solved the problem
with a little less work.

But I haven't really convinced myself that this is the only solution,
since I only supposed that there are ten, and not five, 8's in the
cents column. Suppose there were only five. Each of them can be either
0.08, or 0.88, or .88 added to a dollar amount. Let's write it this
way, though the cents could be added on to the 888 and 88 rather than
all being separate:

888.00
88.00
?.00
?.00
?.00
?.?8
?.?8
?.?8
?.?8
?.?8
------
976.40 + ?

The five unknown numbers in the ten-cent column, which are all either
0 or 8, have to add up to a number ending in 6 in order to make this
an even number of dollars; they must be two 8's and three 0's, adding
up to 16:

888.00
88.00
?.00
?.00
?.00
?.88
?.88
.08
.08
.08
------
978.00 + ?

The five unknown dollar amounts add up to \$22; this is not a multiple
of 8, so I can't do it. Looks like the first answer was the only one,
and we can really say we know what the prices in the store are.

There are two hard parts in this sort of problem. One is to interpret
what the words mean: in what ways can a price tag contain only 8's? I
had to figure that out step by step as I went through the problem,
because I didn't take enough time at the beginning. The other hard
part is to be sure we've found all possible answers, because if we've
found AN answer, we can't say for sure that it's the ONLY answer, so
we can't say "that price tags HAVE to say..."

I hope you enjoyed the puzzle as much as I did.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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