Ars Mathematica
Blog para los que les gustan las matemáticas.
Intentaré proponer un problema cada lunes y una solución cada viernes. Pero tu puedes aportar la tuya.
miércoles, 30 de diciembre de 2015
Solución al problema 58
Enunciado del problema 58
1
⋅
2
⋅
3
⋅
…
⋅
99
⋅
100
(
1
⋅
3
⋅
5
⋅
…
⋅
99
)
(
1
⋅
2
⋅
3
⋅
…
⋅
50
)
=
3
⋅
4
⋅
6
⋅
…
⋅
98
⋅
100
1
⋅
2
⋅
3
⋅
…
⋅
49
⋅
50
=
2
50
(
1
⋅
2
⋅
3
⋅
…
⋅
49
⋅
50
)
1
⋅
2
⋅
3
⋅
…
⋅
49
⋅
50
=
2
50
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