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37
Thus,
256
C
≡
2101 mod 3125
.
Were the hint,
“
Try
−
4 coconuts,
”
not given, one would proceed to solve
this congruence, taking note of the fact that
(
256
,
3125
)
=
1. But
C
=−
4
is a solution of this congruence, and so
C
≡−
4 mod 3125; that is, every
number of the form 3125
k
−
4 is a solution. The minimum value for
C
is
thus 3121 coconuts.
1.98
A suspect said that he had spent the Easter holiday April 21, 1893, with
his ailing mother; Sherlock Holmes challenged his veracity at once. How
could the great detective have been so certain?
Solution.
April 21, 1893, fell on Friday, and so this date could not have
been Easter Sunday.
1.99
How many times in 1900 did the
f
rst day of a month fall on a Tuesday?
Solution.
The year
y
=
1900 was not a leap year, and
g
(
y
)
≡[
19
/
4
]−
38
≡−
34
≡
1mod7
.
We seek the number of solutions to 3
≡
1
+
j
(
m
)
+
1 mod 7; that is,
j
(
m
)
≡
1 mod 7. For the interval March through December, there is only
one such month, namely, August. But we still have to check January and
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 Fall '11
 KeithCornell

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