Find the number of zeros.

Find the number of zeros in:


1.   Trick: In any product, zeros = minimum of (number of 2s, number of 5s).Super simple rule : Count 2’s , Count 5’s ,


     Zeros = Smaller number


2.   Rule to count 5s (very important) : Zero is made by 2 × 5, But 2s are always more, so we only count 5s.


3.   Simple rule: Divide the number by 5, 25, 125 … until division is 0.


Ex : Count 5s in 100



  • 100 ÷ 5 = 20

  • 100 ÷ 25 = 4

  • 100 ÷ 125 = 0


Total 5s = 20 + 4 = 24


Rule to count 2s: Divide the number by 2, 4, 8, 16… until division is 0.


Ex: Count 2s in 16



  • 16 ÷ 2 = 8

  • 16 ÷ 4 = 4

  • 16 ÷ 8 = 2

  • 16 ÷ 16 = 1


Total 2s = 8 + 4 + 2 + 1 = 15


Final Zero Rule: Number of zeros = smaller of (count of 2s, count of 5s) 


But in 99% questions:



  • 2s are more

  • So count only 5s


Simple Way to Remember


Count 5s.



  • 25 gives 2 fives.

  • 125 gives 3 fives.

  • 625 gives 4 fives…


Ex:


1) Zeros in 100!



  • Multiples of 5 → 100 ÷ 5 = 20

  • Multiples of 25 → 100 ÷ 25 = 4


Total = 20 + 4 = 24


2) Zeros in 75!



  • 75 ÷ 5 = 15

  • 75 ÷ 25 = 3


Total = 18


3) Zeros in 50!



  • 50 ÷ 5 = 10

  • 50 ÷ 25 = 2


Total = 12


Most Simple Memory Trick


Keep dividing by 5 until the answer becomes 0. Add them all.


Find Zero Ex :-


1. Find the number of zeros 2222 x 5555 = ?


(A)  222  (B) 555  (C) 777  (D) 333


Rule


Zeros come from 10 = 2 × 5
So we count how many pairs of (2, 5) can be formed.



  • Number of 2s = 222

  • Number of 5s = 555


To make a pair, we need one 2 and one 5.


So the number of pairs = smaller number → 222


Ans: B. 222 zeros


2. Find the number of zeros in: 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50


Trick: In any product, zeros = minimum of (number of 2s, number of 5s). Here, 5s = 11.


Count numbers ending in 5 or 0 → gives 5s


5, 10, 15, 20, 25, 30, 35, 40, 45, 50
→ 10 fives


Extra fives from 25 and 50 → +2


Total 5s = 12


Count even numbers → gives 2s


Only even: 10, 20, 30, 40, 50 → These give total 8 twos


Zeros = smaller number


Smaller of (5s = 12 , 2s = 8) = 8


Ans = 8


3. Find the number of trailing zeros in: 2 × 4 × 6 × 8 × … × 200


Options: (A) 49 (B) 24 (C) 25 (D) 50


Given: The product is:


2×4×6×…×200 = 2^100 × (1×2×3×…×100)


Trailing zeros depend on number of 5s in the factors (because 2s are already plenty).


So we only count number of 5s in 1 to 100:


Simple Trick: Count the number of 5s in 1 to 100.


Divide 100 by 5


100 ÷ 5 = 20


Divide 100 by 25


100 ÷ 25 = 4


Add both: 20 + 4 = 24


Ans: B. 24


4. Find the number of trailing zeros in:1×3×5×7 × … × 99×64


Options: (A) 23 (B) 6 (C) 0 (D) 5


Simple Trick : Trailing zeros come from pairs of (2 × 5).


Count factors of 5


Only odd numbers from 1 to 99 are used.


Multiples of 5 among odd numbers: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95
→ 10 numbers


Now count multiples of 25 (give extra 5): 25, 75
→ 2 extra 5s


Total 5s: 10+2=12


Count factors of 2


Only 64 has factors of 2 :- 64 = 2^6 ⇒ 6 twos


Trailing Zeros


Zeros = min(12, 6) = 6


Why only 64 has 2 ?


Ans : All numbers from 1 to 99 are odd, so they do not contain factor 2. Only 64 is even, and: 64 = 2^6


So only 64 has 2 - factors (six of them).


Ans: B. 6


5. Find the number of trailing zeros in: 20 × 40 × 7600 × 600 × 300 × 1000


Options: (A) 15 (B) 11 (C) 2 (D) 16


Short Trick:


Count the zeros at the end of each number → add them.


20 → 1 zero, 40 → 1 zero, 7600 → 2 zeros, 600 → 2 zeros, 300 → 2 zeros, 1000 → 3 zeros


Add them: 1 + 1 + 2 + 2 + 2 + 3 = 11


Ans: B. 11


6. Find the number of zeros in: 100! + 200!


Options: A. 24  B. 25  C. 49  D. 73


Trick


Trailing zeros depend on number of 5s in the factorial.


But here we have: 100! + 200!


And we know:



  • 200! has more trailing zeros than 100!

  • When adding two numbers, the number of trailing zeros equals the smaller number of trailing zeros.


So the answer is simply:


Trailing zeros = zeros in 100!


Find trailing zeros in 100!


100/5 = 20


100/25 = 4


Total zeros: 20 + 4 = 24


Ans: A. 24


7. Find the number of zeros in: 100! x 200!


Options: A. 24  B. 25  C. 49  D. 73


Options: A. 24 B. 25 C. 49 D. 73


Trick : Trailing zeros = count of 5s


Just count 5s in 100! and 200!, then add.


5s in 100!


→ 100÷5 = 20
→ 100÷25 = 4


Total = 24


5s in 200!


→ 200÷5 = 40
→ 200÷25 = 8
→ 200÷125 = 1


Total = 49


Add them : 24 + 49 = 73


Ans: 73


8. The number of zeroes at the end of 21 × 35 × 625 × 8 × 165 is:


Option: (A) 1 (B) 3 (C) 5 (D) 7


Soln


Trailing zero = 2 × 5


Count 5s in all numbers: 35 → 1, 625 → 4, 165 → 1 → total 6


21 → 3 × 7
35 → 5 × 7
625 → 5 × 5 × 5 × 5
8 → 2 × 2 × 2
165 → 5 × 3 × 11


Count 2s in all numbers: 8 → 3 → total 3


Count 5s and 2s


5s: 35(1) + 625(4) + 165(1) = 6
2s: 8(3) = 3 


Take smaller of 2s and 5s → 3 zeroes


Ans: B) 3


Super short trick: Count 5s & 2s → smaller number = zeroes.