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6) ¿Qué valor no puede tener " d " en \( \mathrm{N}=453 \mathrm{~d} \) para que sea divisible por 3 ? \( \begin{array}{lllll}\text { a) } 0 & \text { b) } 3 & \text { c) } 6 & \text { d) } 7 & \text { e) } 9\end{array} \)

Ask by Schofield Bryant. in Venezuela
Feb 03,2025

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El valor que no permite que \( \mathrm{N} = 453d \) sea divisible por 3 es **7**.

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Bonus Knowledge

Para que un número sea divisible por 3, la suma de sus dígitos debe ser divisible por 3. En el caso de \( N = 453d \), sumamos los dígitos 4, 5 y 3, obteniendo 12. Ahora, al sumar \( d \), obtenemos \( 12 + d \). Para determinar qué valores hacen que \( 12 + d \) no sea divisible por 3, podemos analizar los diferentes valores de \( d \) de acuerdo a las opciones dadas. Si probamos con \( d = 0 \) (12+0=12), \( d = 3 \) (12+3=15), y \( d = 6 \) (12+6=18), todos son divisibles por 3. Con \( d = 7 \) (12+7=19), y \( d = 9 \) (12+9=21), notamos que 19 no es divisible por 3, mientras que 21 sí lo es. Así que la respuesta es **d) 7**.

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