Counting Questions with Solutions — Set 4 (Q31-Q40)

This set contains counting questions with solutions — Set 4 (Q31-Q40) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

31. k is the product of first 100 multiples of 15. Find the number of trailing zeros in k.

32. Given that
k = 1! + 2! + 3! + 4! +5! + ……….. + 120!. Then which one of the following is true
(i) k is odd
(ii) k is even
(iii) k is even or odd, can not be determined
(iv) (k+3) is even.

33. Find the smallest perfect square number divisible by 11!.

34. Consider a number x.(x + 1).(x + 2).(x + 3)………till x terms. If above number is completely divisible by all two digit prime numbers then what is the minimum value of x.

35. Find the number of trailing zeros in (1! + 2! + 3! + 4! +5!)(6! + 7! + 8! + 9! + 10!)…………(96! + 97! + 98! + 99! + 100!).

36. An expression is given below:
N = 1! + 2! + 3! + 4! + 5! + ………. + 99! + 100!
(i) Find the unit digit of N
(ii) Find the last two digit of N
(iii) What will be the remainder when N is divided by 7.

37. If p = 𝑥!𝑦!𝑧! & p is a single digit number then find the possible value of p.

38. If p = 10! + 20! + 30! + 40! + ………… + 150!. then find the number of trailing zeros in pᵖ.

39. If p = x¹⁰⁰⁰(1000!) & p is divisible by 10¹⁰⁰⁰. then what would be the minimum number of zeros at the end of p.

40. If k! is not divisible by 1155 then find the maximum value of k
(i) 11
(ii) 17
(iii) 10
(iv) 7

31. k is the product of first 100 multiples of 15. Find the number of trailing zeros in k.
Solution:-
k = (15×1).(15×2).(15×3)…………(15×100)
k = 15¹⁰⁰ × 100!
now highest power of 5 in 100! ⇒

Hence Number of trailing zeros in k = minimum of(Highest power of 5 in k & Highest power of 2 in k)
= 97 Answer

32. Given that 
        k = 1! + 2! + 3! + 4! +5! + ……….. + 120!. Then which one of the following is true
(i) k is odd
(ii) k is even
(iii) k is even or odd, can not be determined
(iv) (k+3) is even.

Solution:- Since 1! = 1 & all factorials starting from 2 till ∞ are even numbers.
Hence

& we know that odd + even = odd
Hence k is an odd number      option(1)✔        Answer

now since k ia an odd number. Hence k + 3 i.e. odd + odd = even
Hence (k+3) is even number option (iv) ✔       Answer

33.  Find the smallest perfect square number divisible by 11!.
Solution:-

11! = 11×10×9×8×7×6×5×4×3×2×1
= 11×2×5×3²×2³×7×2×3×5×2×2×3×2
= 11×7×5²×3⁴×2⁸

Hence smallest perfect square number will be
=11²×7²×5²×(3²)²×(2⁴)²
= 11!×11×7
= 307359300                 Answer

34. Consider a number x.(x + 1).(x + 2).(x + 3)………till x terms. If above number is completely divisible by all two digit prime numbers then what is the minimum value of x.
Solution:-

Since largest two digit prime number is = 97 & Since above number is completely divisible by all two digit prime numbers.
Then, let the highest term of the expression be 97 then expression ⇒
49×50×51×52×…………..till 49 terms.
= 49×50×51×52×………….×97
Hence every prime number from 49 till 97 will divide this expression completely.
& prime numbers which are less than  49 ⇒

Let 47
Then 47×2 = 94 ✔
Let 43
Then 43×2 = 86 ✔
⋮
⋮
⋮
⋮
Let 2
Then 2×25 = 50 ✔
Let 13
Then 13×5 = 65 ✔

Hence when x = 49 then all two digits prime number divide this expression.
Hence minimum value of x = 49      Answer

35. Find the number of trailing zeros in (1! + 2! + 3! + 4! +5!)(6! + 7! + 8! + 9! + 10!)…………(96! + 97! + 98! + 99! + 100!).
Solution:-

(1! + 2! + 3! + 4! + 5!) ➝ as factorial number goes on increasing in every term.
Hence in every term number of trailing zeros will be the number of trailing zeros in the term’s lowest factorial number.
Hence
(1! + 2! + 3! + 4! + 5!) ➝ 0
(6! + 7! + 8! + 9! + 10!) ➝ 1
(11! + 12! + 13! + 14! + 15!) ➝ 2
(16! + 17! + 18! + 19! + 20!)➝ 3
(21! + 22! + 23! + 24! + 25!) ➝ 4
(26! + 27! + 28! + 29! + 30!) ➝ 6
(31! + 32! + 33! + 34! + 35!) ➝ 7
(36! + 37! + 38! + 39! + 40!) ➝ 8
(41! + 42! + 43! + 44! + 45!) ➝ 9
(46! + 47! + 48! + 49! + 50!)➝ 10
(51! + 52! + 53! + 54! + 55!) ➝12
(56! + 57! + 58! + 59! + 60!) ➝ 13
(61! + 62! + 63! + 64! + 65!) ➝ 14
(66! + 67! + 68! + 69! + 70!) ➝15
(71! + 72! + 73! + 74! + 75!) ➝ 16
(76! + 77! + 78! + 79! + 80!) ➝ 18
(81! + 82! + 83! + 84! + 85!) ➝ 19
(86! + 87! + 88! + 89! + 90!) ➝ 20
(91! + 92! + 93! + 94! + 95!) ➝ 21
(96! + 97! + 98! + 99! + 100!) ➝ 22

Now since expression is multiplication of these terms. Hence for value of expression all trailing zeros of individual terms will be added & this will be the final number of trailing zeros in the expression which is = 220       Answer

36. An expression is given below:
N = 1! + 2! + 3! + 4! + 5! + ………. + 99! + 100!
(i) Find the unit digit of N
(ii) Find the last two digit of N
(iii) What will be the remainder when N is divided by 7.

Solution:-
(i)

Hence U𝒹 of N = 3 + 0 + 0 = 3 Answer

(ii) Last two digit of N = ?

Hence last two digit of N = 13 + 00
= 13 Answer

(iii) Remainder when N is divided by 7 = ?

Since \(\frac{{7!}}{7}\) ➜ R = 0
\(\frac{{8!}}{7}\) ➜ R = 0

          \(\frac{{9!}}{7}\) ➜ R = 0
⋮
⋮
⋮

          \(\frac{{100!}}{7}\) ➜ R = 0
So final remainder will be
Remainder of \(\left( {\frac{{1! + 2! + 3! + 4! + 5! + 6!}}{7}} \right)\)

= Remainder of \(\left( {\frac{{873}}{7}} \right)\)

= 5          Answer

37. If p = \(\boldsymbol{x{!^{y{!^{z!}}}}}\) & p is a single digit number then find the possible value of p.
Solution:-

➠ if x = 0 or 1 ⇒ x! = 1 then y & z can take any value & value of p will be = 1.

➠ if x = 2 ⇒ x! = 2
if y = 0 or 1 ⇒ y! = 1 then z can take any value & value of p will be 2.
if y = 2 ⇒ y! = 2 then z may take value 0 or 1 then p = 4

➠ if x = 3 ⇒ x! = 6 then y can be 0 or 1 & z can take any value & value of p will be = 6

Hence Possible values of p = 1,2,4,6             Answer

38. If p = 10! + 20! + 30! + 40! + ………… + 150!. then find the number of trailing zeros in pᵖ.
Solution:-

Number of trailing zeros in ➞

So number of trailing zeros in p = 2

Hence number of trailing zeros in pᵖ = 2p        Answer

39. If p = x¹⁰⁰⁰(1000!) & p is divisible by 10¹⁰⁰⁰. then what would be the minimum number of zeros at the end of p.
Solution:-

First find number of trailing zeros in 1000!

Hence 1000! has 249 trailing zeros.
∴ 1000! = 2⁹⁹⁴ × 5²⁴⁹ × (other factors)
now 10¹⁰⁰⁰ has 1000 zeros after 1 (i.e. 1000 trailing zeros) & 1000! has 249 trailing zeros.
So for p to be completely divisible by 10¹⁰⁰⁰ , remaining zeros should come from “x¹⁰⁰⁰” term & x¹⁰⁰⁰ will have trailing zeros only when x is a multiple of 2 & 5.
So to find minimum zeros at the end of p take the least value of x as 2×5

x = 2×5 = 10

∴ p should have 1249 zeros at the end             Answer

40. If k! is not divisible by 1155 then find the maximum value of k
(i) 11
(ii) 17
(iii) 10
(iv) 7

Solution:- 
Since 1155 = 3×5×7×11
if we take k as 11 or above then k! will be completely divisible by 1155 because

So for maximum value take k as 11 – 1 = 10
∴ k = 10 option(iii)           Answer

✅ Well done on completing Set 4!

Now revisit the Counting Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering counting concept for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Factorial on Wikipedia before attempting the next set.

This page is part of our complete series of counting concept questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core counting concept before your exam day.