1. Suppose a and b are positive integers
bigger than 1 but with sum less than 100.
There are two people : Mr. Sum
and Mr. Product. Now Mr. Sum knows only
a+b and Mr. Product knows only
a*b. The following conversation takes place.
Mr. Sum : "You don't know a
and b."
Mr. Product : "Now I do."
Mr. Sum : "Now I do, too."
What are a and b?
Solution : We make several observations:
1. For Mr. Product to look at the
product a*b and know what a and b are
requires that
a*b satisfy one of
(a) a*b is the product of two primes
(b) a*b is the cube of a prime(e.g. if a*b = 27
then Mr. P knows that a and b
are 3 and 9)
(c) a*b has a prime factors greater than 50(e.g.
if a*b = 236, then Mr. P
knows that a and b are 4 and 59, since this is
the only factor reaction of
236 into two factors whose aum is less than 100.
2. Since Mr. Sum claimed that Mr.
P didn't know a and b we deduce that Mr. S
couldn't be looking at a sum a+b for which the corresponding
product a*b
might satisfy one of (a),(b),(c).
3. This rules out for the sum a+b
all but 11 possibilities:
11,17,23,27,29,35,37,41,47,51,53.
E.g.: if Mr.S were looking at the sum 46, it would
be possible that Mr.P is
looking at 205 = 41*5 in which case Mr. S could
not say (as he did) that
Mr. P doesn't know a and
b. Hence a+b cannot be 46.
E.g: if Mr. S were looking at the sum 57, it would
be possible that MR.P is
looking at 212 = 53*4 in which case Mr.P would
know a and b (see 1.(c)
above ). Hence a+b cannot he 57.
4. Now Mr. P hears that Mr. S says.
So Mr. P (being sharp) knows (as we do)
that a+b must he one of the 11 possibilities enumerated
in 3.
5. To illustrate the technique
we will employ in determining a and b, let's show
now for example ,that it is not possible that a
and b are 4 and 19.
We first observe that a*b = 76 does not meet any
of the three criteria laid out
in 1, so indeed Mr. S (looking
at the sum 23)could make the first statement he
did, now Mr. P is looking at
the number 76 which has two factorizations: 76 =
38*2 and 76 = 19*4. Since 38+2
= 40 is not one of the possibilities for the sum
a+b, Mr. P knows that a and
b are not 38 and 2.Therefore, he would know
that a and b are 19 and 4 and
would therefor have made the statement he did.
So far, so good. But now, Mr.S
looking at his sum 23 would not be able to
make his final statement. For
as for as he can determine there are two distinct
possililities for a and b at this stage.
(I)a and b are 19 and 4
(II)a and b are 16 and 7
The reader can check that in both these cases,
Mr.P would know a and b, and
hence say what he did. Moreover, in all other possibilities
(E.g.: a = 18 b = 5,
a = 13 b = 10 etc). Mr. P would be unable to determine
a and b from the
product a*b and hence would not make the statement
that we know he did
make. Thus Mr. S could not make the final statement
that he did make and we
conclude that a = 19, b = 4 (or vice versal is not
a possibilitity .)
6. Using the technique we can now
systematically examine the 11 possibilities.
Case
1: Can a+b = 11?
possibility
a*b
Would Mr. P know a and b after hearing Mr. S's first
statement
9+2
18
Yes
8+3
24
Yes
7+4
28
Yes
6+5
30
No
Since there is more than
one "yes" in the last column, we conclude that Mr.S
could not know a and b and therefore not make the
final statement that he did.
So a+b cannot be 11.
Case 2: Can a+b = 17?
Possibility
a*b
Would Mr. P know a and b after hearing Mr. S's first
statement
15+2
30
No
14+3
42
No
13+4
52
Yes
12+5
60
No
11+6
66
No
10+7
70
No
9+8
72
No
Since there is only one "yes" in the last column, we
conclude that if a and b are
13 and 4 then Mr. P would know a and b, and that
Mr. S would subsquently
himself know a and b. Thus 4 and 13 is a solution.
Continuing in this way throuhg each of the remaining
9 possibilites for the sum
a+b we are able to conclude that this is the only
solution.
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