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2013 CHMMC (Fall)
4
2013 CHMMC Tiebreaker 4 - sum1/2+1/2x3+ 1/2x3x5 + ....
2013 CHMMC Tiebreaker 4 - sum1/2+1/2x3+ 1/2x3x5 + ....
Source:
March 1, 2024
algebra
CHMMC
Problem Statement
Let
A
=
1
2
+
1
3
+
1
5
+
1
9
,
A =\frac12 +\frac13 +\frac15 +\frac19,
A
=
2
1
+
3
1
+
5
1
+
9
1
,
B
=
1
2
⋅
3
+
1
2
⋅
5
+
1
2
⋅
9
+
1
3
⋅
5
+
1
3
⋅
9
+
1
5
⋅
9
,
B =\frac{1}{2 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{2 \cdot 9}+\frac{1}{3 \cdot 5}+\frac{1}{3 \cdot 9} +\frac{1}{5 \cdot 9},
B
=
2
⋅
3
1
+
2
⋅
5
1
+
2
⋅
9
1
+
3
⋅
5
1
+
3
⋅
9
1
+
5
⋅
9
1
,
C
=
1
2
⋅
3
⋅
5
+
1
2
⋅
3
⋅
9
+
1
2
⋅
5
⋅
9
+
1
3
⋅
5
⋅
9
.
C =\frac{1}{2 \cdot 3 \cdot 5} + \frac{1}{2 \cdot 3 \cdot 9} + \frac{1}{2 \cdot 5 \cdot 9} +\frac{1}{3 \cdot 5 \cdot 9}.
C
=
2
⋅
3
⋅
5
1
+
2
⋅
3
⋅
9
1
+
2
⋅
5
⋅
9
1
+
3
⋅
5
⋅
9
1
.
Compute the value of
A
+
B
+
C
A + B + C
A
+
B
+
C
.
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