Find the Number of Zeros at the end of the product of (2⁵ * 3³ * 4⁸ * 5³ * 6⁷ * 7⁶ * 8¹² * 9⁹ * 10⁶ * 15¹² * 20¹⁴ * 22¹¹ * 25¹⁵).
Hint : (2ⁿ * 5ⁿ = 10ⁿ)
Answers
AnswEr :
• First of all the Rule to Solve this :
Any number ending with zero, will be a factor of 10 which mean it will have 2 and 5 as it factors. [Hint Provided (10ⁿ = 2ⁿ*5ⁿ)]
Here we will count number having factor 2 and 5. Whichever number is low will give you the equal number of 0.
Why Less will be the Equal No. of Zero( 0 ), Because there will no further factors of 10.
• Let's Break these in shortest form :
We can see that these numbers are in the form of (2² × 3² × 5²)
⟹ 2⁵ × 3³ × 4⁸ × 5³ × 6⁷ × 7⁶ × 8¹² × 9⁹ × 10⁶ × 15¹² × 20¹⁴ × 22¹¹ × 25¹⁵
⟹ 2⁵ × 3³ × (2²)⁸ × 5³ × (2 × 3)⁷ × (2³)¹² × (3²)⁹ × (2 × 5)⁶ × (5 × 3)¹² × (2² × 5)¹⁴ × (2 × 11)¹¹ × (5²)¹⁵
⟹ 2⁵ × 3³ × 2¹⁶ × 5³ × 2⁷ × 3⁷ × 2³⁶ × 3¹⁸ × 2⁶ × 5⁶ × 5¹² × 3¹² × 2²⁸ × 5¹⁴ × 2¹¹ × 11¹¹ × 5³⁰
⟹ (2⁵ × 2¹⁶ × 2⁷ × 2⁶ × 2³⁶ × 2²⁸ × 2¹¹) × (3³ × 3⁷ × 3¹⁸ × 3¹²) × (5³ × 5⁶ × 5¹² × 5¹⁴ × 5³⁰) × 11¹¹
⟹ 2¹⁰⁹ × 3⁴⁰ × 5⁶⁵ × 11¹¹
- And, we know that, (5ⁿ × 2ⁿ) = 10ⁿ
- So, We will Ignore (3⁴⁰ × 11¹¹)
- Now, we can see that 2¹⁰⁹ > 5⁶⁵, And Highest holding power of 2 can't contribute zero at the end of the product.
Here 65 is Less Power. That's why ;
Product of (2⁵ × 3³ × 4⁸ × 5³ × 6⁷ × 7⁶ × 8¹² × 9⁹ × 10⁶ × 15¹² × 20¹⁴ × 22¹¹ × 25¹⁵) will end up with 65 zeroes.
∴ Therefore, Correct Answer is 10⁶⁵.
Answer:
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