The concept of Zeros
Q1–10 of 54The numbers 1, 3, 5, ........, 25 are multiplied together. The number of zeros at the right end of the product is
Explanation: All terms are odd numbers, meaning there are no factors of $2$. Since there are no pairs of $2 \times 5$, zero trailing zeros are formed.
Answer: (a) 0
The numbers 1, 2, 3, 4, ........, 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be
Count of 5s in 1000!:
Answer: (d) 249
First 100 multiples of 10 i.e. 10, 20, 30, ........, 1000 are multiplied together. The number of zeros at the end of the product will be
Product: $10 \times 20 \times 30 \times \dots \times 1000 = 10^{100} \times (1 \times 2 \times 3 \times \dots \times 100) = 10^{100} \times 100!$
Zeros from $10^{100}$: $100$ zeros
Zeros from $100!$: $\lfloor 100/5 \rfloor + \lfloor 100/25 \rfloor = 20 + 4 = 24$ zeros
Total Zeros: $100 + 24 = 124$
Answer: (c) 124
The number of zeros at the end of the product 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 is
Product: $5 \times 10 \times 15 \times \dots \times 50 = 5^{10} \times (1 \times 2 \times 3 \times \dots \times 10) = 5^{10} \times 10!$
Count of 2s in $10!$: $\lfloor 10/2 \rfloor + \lfloor 10/4 \rfloor + \lfloor 10/8 \rfloor = 5 + 2 + 1 = 8$
Count of 5s: $10 + 2 = 12$
Trailing Zeros: $\min(8, 12) = 8$
Answer: (c) 8
The number of zeros at the end of 60! is
Count of 5s in 60!:
Answer: (b) 14
The numbers 1, 3, 5, 7, ........, 99 and 128 are multiplied together. The number of zeros at the end of the product must be
Product: $(1 \times 3 \times 5 \times \dots \times 99) \times 128$
Count of 5s in odd product (up to 99): $5, 15, 25(5^2), 35, 45, 55, 65, 75(5^2), 85, 95 \implies 12$ factors of $5$
Count of 2s in 128: $128 = 2^7 \implies 7$ factors of $2$
Trailing Zeros: $\min(12, 7) = 7$
Answer: (b) 7
The numbers 2, 4, 6, 8, ........, 98, 100 are multiplied together. The number of zeros at the end of the product must be
Product: $2 \times 4 \times 6 \times \dots \times 100 = 2^{50} \times 50!$
Count of 5s in 50!: $\lfloor 50/5 \rfloor + \lfloor 50/25 \rfloor = 10 + 2 = 12$
Answer: (c) 12
Let S be the set of all prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeros will the product end?
Primes < 100: $2, 3, 5, 7, 11, \dots$
Factors: Contains exactly one $2$ and one $5$.
Trailing Zeros: $1$
Answer: (a) 1
Find the number of zeros at the end of the result 3 × 6 × 9 × 12 × 15 × ........ × 99 × 102.
Product: $3 \times 6 \times 9 \times \dots \times 102 = 3^{34} \times 34!$
Count of 5s in 34!: $\lfloor 34/5 \rfloor + \lfloor 34/25 \rfloor = 6 + 1 = 7$
Answer: (c) 7
The number of zeros at the end of the product 11×12×13×14×15 is
Product: $11 \times 12 \times 13 \times 14 \times 15$
Factors of 5: $15 = 3 \times 5 \implies 1$ five
Factors of 2: $12, 14$ give plenty of 2s
Answer: (a) 1