Question

What is the largest value of n such that 10^n divides the product 2^5 * 3^3 * 4^8 * 5^3 * 6^7 * 7^6 * 8^12 * 9^9 * 10^6 * 15^12 * 20^14 * 22^11 * 25^15 ?

Answer 1

Here the answer is actually the no. of zeros by which the product  2⁵ × 3³ × 4⁸ × 5³ × 6⁷ × 7⁶ × 8¹² × 9⁹ × 10⁶ × 15¹² × 20¹⁴ × 22¹¹ × 25¹⁵  ends!

Find the no. of zeros at the end of the product. The power of 10 ending in the same no. of zeros can divide this product, and so this divisor will be the largest power of 10 by which the product is divisible too.

That's why the largest value of  'n'  is the no. of zeros at the end of the product.

It is true that the no. of zeros at the end of a number is the least no. among its factors either 2 or 5.

E.g.: Consider the number 4500. Its prime factorization is 2² × 3² × 5³. Here 2 twos and 3 fives are multiplied. Since the no. of twos is lesser than that of fives here, we can say that 4500 ends in '2' zeros.

But consider 360. Its prime factorization is 2³ × 3² × 5. Here the no. of fives is 1, which is lesser than the no. of twos, i.e., 3. So 360 ends in '1' zero.

Now let's come to the question.

First we consider the even factors multiplied in the product given.

2⁵ × 4⁸ × 6⁷ × 8¹² × 10⁶ × 20¹⁴ × 22¹¹

Here the factors other than 2⁵ should be simplified to get them as powers of 2 or product of power of 2. Means,

    2⁵ × 4⁸ × 6⁷ × 8¹² × 10⁶ × 20¹⁴ × 22¹¹

⇒  2⁵ × (2²)⁸ × (2 × 3)⁷ × (2³)¹² × (2 × 5)⁶ × (2² × 5)¹⁴ × (2 × 11)¹¹

⇒  2⁵ × 2¹⁶ × 2⁷ × 3⁷ × 2³⁶ × 2⁶ × 5⁶ × 2²⁸ × 5¹⁴ × 2¹¹ × 11¹¹

We can ignore the powers other than those of 2 here.

    2⁵ × 2¹⁶ × 2⁷ × 2³⁶ × 2⁶ × 2²⁸ × 2¹¹

⇒  2¹⁰⁹

So 109 twos are multiplied in the product.

Now we consider factors which are powers of 5 multiplied in the product.

5³ × 10⁶ × 15¹² × 20¹⁴ × 25¹⁵

Even  10⁶  and  20¹⁴  are considered in the case of even factors earlier, it doesn't matter.

As we did earlier, we have to simplify each factors other than 5³ to get them as powers of 5 or products of power of 5. Means,

    5³ × 10⁶ × 15¹² × 20¹⁴ × 25¹⁵

⇒  5³ × (5 × 2)⁶ × (5 × 3)¹² × (5 × 2²)¹⁴ × (5²)¹⁵

⇒  5³ × 5⁶ × 2⁶ × 5¹² × 3¹² × 5¹⁴ × 2²⁸ × 5³⁰

Here also we can ignore powers other than those of 5 here.

    5³ × 5⁶ × 5¹² × 5¹⁴ × 5³⁰

⇒  5⁶⁵

So 65 fives are multiplied in the product.

Now compare the nos. of twos and fives each. Here  65 < 109.

So the product  2⁵ × 3³ × 4⁸ × 5³ × 6⁷ × 7⁶ × 8¹² × 9⁹ × 10⁶ × 15¹² × 20¹⁴ × 22¹¹ × 25¹⁵  ends in '65' zeros.

So this product can be divided by 10⁶⁵ which is the highest power of 10 by which it's divisible.

Hence answer is  n = 65.

Answer 2

Answer:

You can See all Numbers are in the form 2² * 3² * 5²

➟ 2⁵ × 3³ × 4⁸ × 5³ × 6⁷ × 7⁶ × 8¹² × 9⁹ × 10⁶ × 15¹² × 20¹⁴ × 22¹¹ × 25¹⁵

➟ 2⁵ × 3³ x (2²)⁸ × 5³ x (2 × 3)⁷ × (2³)¹² × (3²)⁹ x (2 × 5)⁶ × (5 × 3)¹² x (2² × 5)¹⁴ × (2 × 11)¹¹ × (5²)¹⁵

➟ 2¹⁰⁹ * 5⁶⁵ * 3⁴⁰

But we know , (5ⁿ x 2ⁿ) = 10ⁿ

So , We will Ignore 3⁴⁰

Now . 2¹⁰⁹ > 5⁶⁵

Here 65 is Less . So Product of 2⁵ × 3³ × 4⁸ × 5³ × 6⁷ × 7⁶ × 8¹² × 9⁹ × 10⁶ × 15¹² × 20¹⁴ × 22¹¹ × 25¹⁵ will end with 65 zeroes .

★ So Correct Answer is 10⁶⁵