| Answer: 4 and 13
Solution:
I will present a two-part solution. First, I will show that 4 and
13 is a pair of numbers that yields the four answers shown above; Second I
will show that no other pair of numbers works.
4 and 13 is a pair that works
Suppose the 2 whole numbers are 4 and 13.
Statement 1 is from Peter, who knows only the product, 52: "I
don't know what the numbers are". That's because 52 is the
product of either 4*13 or 2*26. So this statement is true.
Statement 2 is from Lois, who knows only the sum, 17: "I knew you
wouldn't know what the numbers were." This statement requires a
detailed explanation. Lois reasoned that the numbers were 2,15 or
3,14 or 4,13 or 5,12 or 6,11 or 7,10 or 8,9. So their products were 30, 42, 52, 60, 66, 70,
or 72. If any one of these products gave away the two numbers (the
way 77 gives away 7,11, for example) then Lois wouldn't know for sure that
Peter would be unable to discern the numbers. But, as it turns out,
each of these seven products Lois imagined can be expressed as products at
least two different ways. To summarize Lois' discovery, the sum, 17,
has the property -- I'll call it the "Lois Property" that if a+b=17
(and a,b are in the interval [2,99]) then the product ab can be expressed
as cd, with c,d also in [2,99] and c,d are different numbers from a,b.
Statement 3 is from Peter, who quickly determines that the only sums
with the Lois Property are 11, 17, 23, 27, 29, 35, 37, 41, 47, and 53.
Since Peter knows the product is 52, he knows the numbers are either 4,13
or else they are 2,26. Of these two pairs, only 4+13=17 has the Lois
Property, so Peter is able to say he now knows what the numbers are.
To summarize Peter's discovery, the product, 52, has the property -- you
guessed it, the "Peter Property" that if ab=52 (a,b in [2,99]),
and a+b has the Lois Property, then no other numbers c,d in [2,99] exist
such that cd=52 and c+d has the Lois Property.
Statement 4, is from Lois, who already knew the product is one of {30, 42, 52, 60, 66, 70,
or 72}. She checks each one of these products to see if it has the
Peter Property.
30=5*6 or 2*15 and both 5+6=11 and 2+15=17 have the Lois Property, so
30 doesn't have the Peter Property.
42=3*14 or 2*21 and both 3+14=17 and 2+21=23 have the Lois Property,
so 42 doesn't have the Peter Property.
52=4*13 and 4+13=17, which has the Lois Property. No other
factorization of 52 gives numbers whose sum has the Lois Property, so 52
has the Peter Property! For completeness, Lois checked the other
possible products...
60=5*12 or 3*20, and both 5+12=17 and 3+20=23 have the Lois Property,
so 60 doesn't have the Peter Property.
66=6*11 or 2*33, and both 6+11=17 and 2+33=35 have the Lois Property,
so 66 doesn't have the Peter Property.
70=7*10 or 2*35, and both 7+10=17 and 2+35=37 have the Lois Property,
so 70 doesn't have the Peter Property.
72=8*9 or 3*24, and both 8+9=17 and 3+24=27 have the Lois Property,
so 72 doesn't have the Peter Property.
4 and 13 is the only pair that works
Statement 1 doesn't really matter. The first statement that
restricts the initial pair of numbers is Lois' statement, "I knew you
wouldn't know what the numbers were." This restricts the pairs
to just those whose sums have the Lois Property -- that is pairs whose sum
is in
The Lois Set: {11, 17, 23, 27, 29, 35, 37, 41, 47, 53}
The sum s is in the Lois Set iff a+b=s implies the product ab can be
expressed as cd with a,b,c,d in [2,99] and c,d are different numbers
from a,b.
Put differently, it means that a number is in the Lois Set if, no
matter how you partition it into a sum a+b, the product ab can be
expressed as the product of two different numbers cd. It is a
remarkable thing, I think, that each of the 25 different choices of a,b
such that a+b=53 are such that the product ab can be factored
differently as cd. That is, each of the 25 products 102, 150, 196, 240, 282, 322, 360, 396, 430, 462, 492, 520, 546, 570, 592, 612, 630, 646, 660, 672, 682, 690, 696, 700,
and 702 can be factored at least two different ways.
I don't intend to prove here that each of the sums from 8 to 198 that
aren't listed among those in the Lois Set are, in fact, not in the Lois
Set, although it's not hard to do with the aid of a computer. All
you have to do to show a number is not in the Lois Set is find a
partition a+b of that number such that ab can be factored only one
way. If s is an even number, then it's not hard to find coprime a,b
such that a+b=s and the smallest factor of a multiplied by b is larger
than 99, so no even numbers are in the Lois Set. If s is 2+p, where
p is prime, then it's not in the Lois set, because there's no other way to
factor 2p. If s is odd, and greater than 53 then you can partition s
into 4+b, where b>49. There's no other way to factor 4b such that
b is in [2,99], so 53 is the largest Lois Number. That just leaves
just one number to test:
If a+b=51 then let a,b=17,34. ab=578, which can be factored
only as 1*578, 2*289, and 17*34, so no other c,d exist in [2,99] such
that cd=578.
Statement 3 is Peter's comment: "Now I know what the numbers
are." He knows because Lois just told him the sum is in the
Lois set, {11, 17, 23, 27, 29, 35, 37, 41, 47, 53}, and he knows the
product. This must mean Peter's product is in
The Peter Set: {18, 24, 28, 50, 52, 54, 76, 92, 96, 100, 110, 112, 114, 124, 130, 138, 140, 148, 152, 154, 160, 162, 168, 170, 172, 174, 176, 182, 186, 190, 198, 204, 208, 216, 232, 234, 238, 240, 246, 250, 252, 270, 276, 280, 282, 288, 294, 304, 306, 310, 336, 340, 348, 360, 364, 370, 378, 390, 400, 408, 414, 418, 430, 442, 480, 492, 496, 510, 520, 522, 532, 540, 550, 552, 570, 592, 612, 630, 646, 660, 672, 682, 690, 696, 700,
702}
The product p is in the Peter Set iff there is exactly one
factorization ab=p such that a+b is an element of the Lois Set.
I obtained the Peter Set by first listing all the products ab, where
a+b is an element of the Lois Set, and then eliminating those products, p,
that can be factored two (or more) different ways p=ab=cd such that a+b
and c+d are both in the Lois Set.
Because each element, p, of the Peter Set has a unique factorization p=ab,
such that a+b is an element of the Lois Set, it's possible to partition
the Peter Set into non-overlapping subsets Peter11, Peter17,
etc. such that Petern contains elements p=ab such that a+b=n is
an element of the Lois Set. Here are the partitions of the Peter
Set:
Peter11 = {18, 24, 28}
Peter17 = {52}
Peter23 = {76, 112, 130}
Peter27 = {50, 92, 110, 140, 152, 162, 170, 176, 182}
Peter29 = {54, 100, 138, 154, 168, 190, 198, 204, 208}
Peter35 = {96, 124, 174, 216, 234, 250, 276, 294, 304, 306}
Peter37 = {160, 186, 232, 252, 270, 336, 340}
Peter41 = {114, 148, 238, 288, 310, 348, 364, 378, 390, 400, 408, 414, 418}
Peter47 = {172, 246, 280, 370, 442, 480, 496, 510, 522, 532, 540, 550, 552}
Peter53 = {240, 282, 360, 430, 492, 520, 570, 592, 612, 630, 646, 660, 672, 682, 690, 696, 700, 702}
This brings us to Statement 4, which is Lois' comment: "Now I know
what the numbers are." How could she know? All she knows
is the sum, which is one of the Lois numbers, and that the product is one
of the Peter numbers. That narrows it down for Lois to one of the
Peter subsets, listed above, but not to a particular element of the Peter
Subset. For example, if the sum given to Lois were 23, then she
would know the product is one of the elements of Peter23, which
narrows it down to 76=4*19, 112=7*16, or 130=10*13, but it doesn't narrow
it down to a particular pair of numbers. The only way Lois could
know the particular pair of numbers is if the sum she was given identifies
a Peter Subset with just one element. In other words, Peter17,
which contains just 52.
Thus the sum 17 and the product 52 must have been given to Lois and
Peter, respectively, and so the numbers must have been 4 and 13.
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