Question

What is the largest value of n(natural number) such that 6n divides the product of the first 100 natural number?

Answer 1

Answer:

Y = 1×2×3×4×5×7×8×9×10×11×........×98×99×100.

Step-by-step explanation:

Let

X = product of first 100 natural numbers

That is

X = 1×2×3×4×5×6×7×8×9×10×11×........×98×99×100.

Then we can write the product of first 100 digits as product of 6 and remaining numbers.

So

X = 6Y

Where

Y = product of first 100 natural numbers except 6

That is

Y = 1×2×3×4×5×7×8×9×10×11×........×98×99×100.

So Y is actually product of 99 natural numbers.

Then it is clear that

6Y divides the X

So

Largest value of n for which the product 6n divides product of first 100 natural numbers is 6Y where Y is product of first 100 natural numbers excluding 6

That is

Y = 1×2×3×4×5×7×8×9×10×11×........×98×99×100.

So

n = Y = 1×2×3×4×5×7×8×9×10×11×........×98×99×100.

Answer 2

Answer:

y=1×2×3×4×5×6×7×8×9×10×11ו••••••••••••••••••••97×98×99×100