Gauss Solutions Worksheet With Answers - Grade 7 - University Of Waterloo - 1998 Page 10
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101998 Gauss Contest - Grade 7
Solutions
=
The best way to start is by taking
3000
˙
54 7
.
as a beginning point.
( )( )
=
<
If we try n = 54, we find 54 55
2970 3000
which is a correct estimate.
( )
(If we try n = 55 we find 55 56
=
>
. So n = 55 is not acceptable.)
3080 3000
(
)
(
)
(
)
(
)
+
+
+
+
+
+
+
+
=
This means that 1 1
2 2
3 3
...
54 54
2970
so that the longest length that Dana
completed was 54 cm. (If we had included the length 55 then we would have had a sum of 3 025 which
is too large.)
ANSWER: (C)
25. Two natural numbers, p and q, do not end in zero. The product of any pair, p and q, is a power of 10
> , the last digit of p q
(that is, 10, 100, 1000, 10 000 , ...). If p q
–
cannot be
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
Solution
If the two natural numbers p and q do not end in zero themselves and if their product is a power of 10
n
n
then p must be of the form 5
and q must be of the form 2
.
(
)
n
n
n
n
= × and 10
=
×
=
×
This is true because 10 2 5
2 5
2
5
.
The possibilities for powers of two are 2, 4, 8, 16, 32, ... and for corresponding powers of five are 5,
25, 125, 625, 3125, ... .
If we take their differences and look at the last digit of p q
–
we find the following,
p
q
last digit of p – q
5
2
3
25
4
1
125
8
7
9
625
16
and the pattern continues in groups of 4.
3125
32
3
1
15 625
64
M
M
7
9
M
Thus, the last digit of p q
–
cannot be 5.
ANSWER: (C)
9
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