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14 Oct 2025, 00:27
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
59% (01:16) correct
41%
(01:16)
wrong
based on 154
sessions
History
Date
Time
Result
Not Attempted Yet
How many trailing zeroes are there in 26! - 25! ?
A. 0
B. 2
C. 4
D. 6
E. 8
A. 0
B. 2
C. 4
D. 6
E. 8
14 Oct 2025, 01:33
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26! = 26 × 25!.
26! - 25! = 26 × 25! - 25! = 25!(26 - 1) = 25! × 25.
Now, let's find the number of trailing zeroes in 25! × 25.
Trailing zeroes come from multiples of 10, which is 2 × 5.
For 25!:
Number of factors of 5 in 25! is:
[25/5] = 5
[25/25] = 1
Total = 6
Number of factors of 2 in 25! is much higher, so the limiting factor is 5.
25 adds 2 more factors of 5 (since 25 = 5^2).
Total factors of 5 = 6 + 2 = 8.
Hence, there are 8 trailing zeroes in 26! - 25!.
26! - 25! = 26 × 25! - 25! = 25!(26 - 1) = 25! × 25.
Now, let's find the number of trailing zeroes in 25! × 25.
Trailing zeroes come from multiples of 10, which is 2 × 5.
For 25!:
Number of factors of 5 in 25! is:
[25/5] = 5
[25/25] = 1
Total = 6
Number of factors of 2 in 25! is much higher, so the limiting factor is 5.
25 adds 2 more factors of 5 (since 25 = 5^2).
Total factors of 5 = 6 + 2 = 8.
Hence, there are 8 trailing zeroes in 26! - 25!.
14 Oct 2025, 06:53
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Bunuel
The trailing zeroes are determined by the number of (2*5)s present in a number.
26! - 25! = 25! * (26 - 1)
= 25! * 25
The number of 5s in 25! are obtained by successive division of 5, which gives (5+1) = 6.
The number of 5s in 25 = 5^2 . That’s 2.
The number of trailing zeroes = (6+2) = 8 zeroes.
Option E
