# What is the unit’s digit of the product of all prime numbers between $1$ and $100$?(A) $0$(B) $1$(C) $2$(D) $3$

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Hint: In the given question we are asked only to find the unit’s digit of the product of all prime numbers between $1$ and $100$. Firstly, write some starting prime numbers. We know that the product of one $2$ and one $5$ gives one zero at unit digits. Then observe the prime number to get the digit at the unit's place of the product.

Here, we have to find the unit’s digit of the product of all prime numbers between $1$ and $100$.
Now, write the prime numbers between $1$ and $100$. These are
$2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, - - - - - - - - - - - - - - - - - - - - - - - - - - - - - $97$.
Now, we have to find the product of the prime numbers between $1$ and $100$. So,
Product $= 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17$- - - - - - - - - - - - - - - - - - $\times 97$
Here, we observe that we multiply $2$ and $5$once which give one zero at the ones digit and when this is multiplied with the other numbers then also zero remains at the unit's place of the product.
So, the unit’s digit of the product is $0$.
Similar concept is applied to find the number of zero in last in the product of all natural numbers between $1$ and $100$. We have to find the total number of $2$ and $5$ in their factors and the total number of zero is equal to the number of $2$ or $5$which is less in number.