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SOLUTIONS FOR HOMEWORK 1
Bonus problem: 1.25. Answer: 9
,
2
,
2
. To see this, denote the ages of the
daughters, in the nonincreasing order, by
a
1
, a
2
, a
3
. We know that:
(1)
a
1
> a
2
≥
a
3
(see the reference to “the eldest daughter”).
(2)
a
1
a
2
a
3
= 36.
(3) Moreover, if
b
1
> b
2
≥
b
3
,
a
1
+
a
2
+
a
3
=
b
1
+
b
2
+
b
3
, and
b
1
b
2
b
3
= 36, then
a
1
=
b
1
,
a
2
=
b
2
, and
a
3
=
b
3
(the census taker managed to compute the ages
based on the information he had).
(4) However, there exist
b
1
=
b
2
≥
b
3
such that
a
1
+
a
2
+
a
3
=
b
1
+
b
2
+
b
3
, and
b
1
b
2
b
3
= 36 (the extra bit of info about “the eldest daughter” was needed).
Let’s look at all the factorizations
a
1
a
2
a
3
= 36, with
a
1
> a
2
≥
a
3
:
a
1
a
2
a
3
a
1
+
a
2
+
a
3
36
1
1
38
18
2
1
21
12
3
1
16
9
4
1
15
9
2
2
13
6
3
2
11
4
3
3
10
There is only one factorization
b
1
b
2
b
3
= 36 with
b
1
=
b
2
≥
a
3
:
b
1
=
b
2
= 6, and
b
3
= 1. Then
b
1
+
b
2
+
b
3
= 13. The number 13 occurs in the above table ponly
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 Fall '09
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 Math

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