What is the remainder when $51\times 27\times 35\times 62\times 75$is divided by 100?

Correct Option

50

To determine the remainder when $51\times 27\times 35\times 62\times 75$ is divided by 100, we need to evaluate the product modulo 100. The number 100 can be factored as $4\times 25$.

Let the given product be $P=51\times 27\times 35\times 62\times 75$.

We can express some of the factors in terms of their prime components or components related to 100:

• $35=5\times 7$
• $62=2\times 31$
• $75=3\times 25$

Substitute these into the product expression:

$P=51\times 27\times (5\times 7)\times (2\times 31)\times (3\times 25)$

Rearrange the terms to group factors that form multiples of 100 or 50:

$P=(5\times 2\times 25)\times (51\times 27\times 7\times 31\times 3)$

$P=(10\times 25)\times (51\times 27\times 7\times 31\times 3)$

$P=250\times (51\times 27\times 7\times 31\times 3)$

Let $K=51\times 27\times 7\times 31\times 3$. All the individual factors (51, 27, 7, 31, 3) are odd numbers. The product of any number of odd integers is always an odd integer. Therefore, K is an odd integer.

Now, we need to find the remainder of $250\times K$ when divided by 100. We know that $250\equiv 50(\mathrm{mod}100)$.

So, the product $P\equiv 50\times K(\mathrm{mod}100)$.

Since K is an odd number, it can be written in the form $2m+1$ for some integer m.

$P\equiv 50\times (2m+1)(\mathrm{mod}100)$

$P\equiv (100m+50)(\mathrm{mod}100)$

$P\equiv 50(\mathrm{mod}100)$

Thus, the remainder when the given product is divided by 100 is 50.

Incorrect Options:

The options 25, 5, and 1 are incorrect because the rigorous application of modular arithmetic, as demonstrated above, yields a remainder of 50. Any other remainder would contradict the derived result. For instance, a remainder of 25 would imply that the product is congruent to 25 modulo 100, which is not the case. Similarly, remainders of 5 or 1 are inconsistent with the calculated value of 50.