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Q9·CSAT · Prelims 2024

Finding the number of trailing zeros in a structured product sequence

NumericalNumber System (Trailing Zeros)Factual single● Hard

Question

How many consecutive zeros are there at the end of the integer obtained in the product 1^2 × 2^4 × 3^6 × 4^8 × ... × 25^50?

Options

a

50

b

55

c

100

d

200

Answer

Explanation

The number of trailing zeros in a product is determined by the exponent of 10 in its prime factorization, which equals \min(\Sigma powers of 2, \Sigma powers of 5). In any sequential sequence, factors of 5 are scarcer than 2, so the limiting factor is the count of prime factor 5.

Isolate the terms containing a base multiple of 5:

5^6 \implies contains 5 × 2 = 10, so power of 5 is 1 × 10 = 10. Wait, look at the sequence index rule: N^2N.

Let's check the terms: 1² \rightarrow power is 2 × 1 2^4 \rightarrow power is 2 × 2 3^6 \rightarrow power is 2 × 3 So term base x has exponent 2x.

Let's list the relevant terms:

5^10 \implies provides 10 factors of 5.
10^20 = (2 × 5)^20 \implies provides 20 factors of 5.
15^30 = (3 × 5)^30 \implies provides 30 factors of 5.
20^40 = (4 × 5)^40 \implies provides 40 factors of 5.
25^50 = (5²)^50 = 5^100 \implies provides 100 factors of 5.

Summing all available factors of 5: Total = 10 + 20 + 30 + 40 + 100 = 200.

For trailing zeros in sequences of the form x^2x, locate bases containing 5, capture their full exponent values, and remember that base 25 contributes double because 25 = 5².

Answer: (d).

Question details

Year

2024

Paper

CSAT

Question

Q9

Section

Numerical Ability

Sub-topic

Number System (Trailing Zeros)

Type

Factual single

Difficulty

Hard

Source hint

Consecutive zeros at the end of product 1^2 x 2^4 x 3^6 ... x 25^50

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