Paper I · 9 Logical Reasoning & Analytical Ability · 9.1 Number system and simplification
9.1.9 Factorials: trailing zeros and the highest power of a prime in n!
Count how many times a prime goes into n! by repeated division (n/p + n/p² + …), and every question on zeros, powers and divisibility of factorials becomes a short sum.
- APPSC Paper I · Paper I
- Standard level
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Core facts
- n! = 1 × 2 × 3 × … × n. 0! = 1 and 1! = 1.
- Values: 4! = 24, 5! = 120, 6! = 720, 7! = 5,040, 8! = 40,320, 10! = 36,28,800.
- Highest power of a prime p in n! (Legendre): ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … (stop when pᵏ > n).
- Short form: divide n by p, then divide the quotient by p again, and add all the quotients.
- Trailing zeros of n! = highest power of 5 in n! (2s are always more plentiful than 5s in n!).
- Composite divisor: split it into prime powers, find each limit, and take the smallest.
- From 5! onwards every factorial ends in 0; from 10! onwards in 00.
Example 1. How many zeros are at the end of 70!? Solution. 70/5 = 14; 14/5 = 2 (whole part). 14 + 2. Answer: 16
Example 2. How many zeros are at the end of 200!? Solution. 200/5 = 40; 40/5 = 8; 8/5 = 1. 40 + 8 + 1. Answer: 49
Example 3. Find 10!/7!. Solution. 10 × 9 × 8. Answer: 720
Explained
Type 1: highest power of a prime.
Example 4. Highest power of 3 that divides 80!. (a) 3²⁶ (b) 3³⁴ (c) 3³⁶ (d) 3⁴⁰ Solution. 80/3 = 26; 26/3 = 8; 8/3 = 2. 26 + 8 + 2 = 36. Answer: (c) 3³⁶
Example 5. Highest power of 7 in 100!. Solution. 100/7 = 14; 14/7 = 2. Answer: 16
Type 2: composite divisor. Count each prime, divide by its exponent in the divisor, take the smallest.
Example 6. Highest power of 12 that divides 30!. Solution. 12 = 2² × 3. Power of 2 in 30!: 15 + 7 + 3 + 1 = 26, giving ⌊26/2⌋ = 13 copies of 2². Power of 3: 10 + 3 + 1 = 14. Smaller is 13. Answer: 13
Example 7. Highest power of 15 in 75!. Solution. 15 = 3 × 5. Power of 5: 15 + 3 = 18. Power of 3: 25 + 8 + 2 = 35. The 5s are scarcer. Answer: 18
Type 3: zeros in a product of factorials. Add the zeros of each factorial (2s are plentiful in each).
Example 8. Zeros at the end of 25! × 30!. Solution. 25!: 5 + 1 = 6. 30!: 6 + 1 = 7. Answer: 13
Type 4: "exactly k zeros". The count jumps at each multiple of 25 (two 5s), 125 (three), and so on.
Example 9. For how many values of n does n! end in exactly 6 zeros? Is there an n! with exactly 5 zeros? Solution. 24! has 4 zeros; 25! has 5 + 1 = 6. The count stays 6 for 25!, 26!, 27!, 28!, 29! and becomes 7 at 30!. It jumps from 4 straight to 6, so no factorial ends in exactly 5 zeros. Answer: 5 values (25 to 29); no
Type 5: remainders and unit digits of factorial sums. Once k! is divisible by the divisor, all later terms drop out.
Example 10. Remainder when 1! + 2! + 3! + … + 100! is divided by 12. Solution. 4! = 24 and every later factorial is divisible by 12. So only 1 + 2 + 6 = 9 is left. Answer: 9
Speed tips.
- For trailing zeros, divide by 5 repeatedly in your head: 200 → 40 → 8 → 1.
- For a composite divisor, the largest prime is usually the scarce one, but check: in 30!, 12 = 2² × 3 is limited by 2², not by 3.
Common traps
- ✗ "Zeros in 200! = 200/5 = 40." ✓ Keep dividing: 40 + 8 + 1 = 49.
- ✗ "Power of 3 in 80! = 80/3 = 26." ✓ Add 8 and 2 as well: 36.
- ✗ "Highest power of 12 in 30! = power of 3 = 14." ✓ 2² needs two 2s per copy: ⌊26/2⌋ = 13, which is smaller.
- ✗ "Every whole number of zeros occurs for some n!." ✓ The count jumps at 25, 50, 75, 125, …; no n! ends in exactly 5 zeros.
- ✗ "In any product, zeros = number of 5s." ✓ That is true for n!, not for every product; count both 2s and 5s.
- ✗ "0! = 0." ✓ 0! = 1.
Quick revision
- Power of prime p in n! = ⌊n/p⌋ + ⌊n/p²⌋ + …
- Zeros of n! = power of 5 in n!.
- Composite divisor: limit for each prime power, take the smallest.
- 0! = 1; 5! = 120; 7! = 5,040; 10! = 36,28,800.
- 5! onwards ends in 0; 10! onwards ends in 00.
- 24! has 4 zeros, 25! has 6: no factorial has exactly 5.
- 250! has 62 zeros (249! has 59): the count jumps by at least 3 at each multiple of 125.
- Factorial sums: drop every term divisible by the divisor.
- Products of powers: count 2s and 5s separately.
Go deeper: Deeper (beyond what APPSC Paper I usually asks)
Deeper Advanced
Example 11. Smallest n for which n! ends in at least 60 zeros. (a) 245 (b) 249 (c) 250 (d) 255 Solution. 245! and 249! both have 49 + 9 + 1 = 59 zeros. 250 = 2 × 5³ adds three 5s at once: 250! has 50 + 10 + 2 = 62. Answer: (c) 250
Example 12. Zeros at the end of 5⁵ × 10¹⁰ × 15¹⁵ × … × 50⁵⁰. Solution. Here the 2s, not the 5s, are scarce, so count both.
- 5s: each term (5k)^(5k) gives 5k for every 5 in 5k. Sum of 5k for k = 1 to 10 is 275; 25 (k = 5) and 50 (k = 10) give an extra 25 + 50. Total 5s = 350.
- 2s: only even 5k give 2s: 10¹⁰ (10), 20²⁰ (40), 30³⁰ (30), 40⁴⁰ (120), 50⁵⁰ (50). Total 2s = 250.
Answer: 250
Example 13. Last two digits of 1! + 2! + … + 10!. Solution. 10! ends in 00. 1 + 2 + 6 + 24 + 120 + 720 + 5,040 + 40,320 + 3,62,880 = 4,09,113. Answer: 13
Example 14. Highest power of 2 in 60!, and the number of zeros at the end of 120!/60!. Solution. Power of 2 in 60!: 30 + 15 + 7 + 3 + 1 = 56. 120! has 24 + 4 = 28 zeros and 60! has 12 + 2 = 14, and 120!/60! still has more 2s than 5s (116 − 56 = 60), so it ends in 28 − 14 = 14 zeros. Answer: 56; 14
- A check by expansion: 25! = 15,511,210,043,330,985,984,000,000 (26 digits, 6 trailing zeros), a quick check for Legendre counts.
- Factorials in base other than 10: zeros of n! in base b = highest power of b in n! (same composite-divisor method).
Practise this topic. Questions on Numerical ability, with answers explained.
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