# Problem on Numbers

**Question 1 :** Find the number of zeroes in 155!

**Solution:** Multiplication of 2×5 results into 10. So number of zeroes depend on the number of pairs of 2 and 5.

In any factorial, number of 5’s is lesser than the number of 2’s. So, we need to count the maximum power of 5 in 155!

[155/5] + [155/5^{2}] + [155/5^{3}]

=31 + 6 +1

=38

Hence, number of zeroes is **38**.

**Question 2: ** Find the number of factors in 8820.

**Solution: ** Number of factors can be calculated by finding the prime factors.

8820= 2^{2}x3^{2}x5^{1}x7^{2}

Number of factors= (2 + 1)(2 + 1)(1 + 1)(2 + 1)

= 3 x 3 x 2 x 3

= **54**

**Question 3: **Find the sum of all the factors of 576.

**Solution: ** Prime factorization of 576= 2^{6}x3^{2}

Sum of all the factors= (2^{0} + 2^{1}+ 2^{2} + 2^{3} + 2^{4} + 2^{5} + 2^{6})*(3^{0} + 3^{1} + 3^{2})

= 127*13

= **1651**

**Question 4: **Find the product of all the factors of 600.

**Solution:** Prime factorization of 600= 2^{3}x3^{1}x5^{2}

Number of factors= (3+1)(1+1)(2+1)

= 4 * 2 * 3

= 24

Product of all the factors= (2^{3}x3^{1}x5^{2})^{24/2}

= (2^{3}x3^{1}x5^{2})^{12}

**Question 5: **Find the highest power of 126 which divides 366!.

**Solution:** 126=2×3^{2}x7

We need to check among 2, 3 and 7 which appears least number of times in 366!.

If we check factorial (1 x 2 x 3 x 4 x 5 x 6 x **7** x 8 x 9 x 10 x 11 x 12 x 13 x **14 **…….) then we got to know 7 appears less frequently than 2 and 3.

So, need to count only number of 7’s in 366!

[366/7] + [366/7^{2}] + [366/7^{3}]

=52 + 7 + 1

=**60**.

**Question 6: ** Find the maximum value of n such that 671! is perfectly divisible by 45n

**Solution:** Prime Factor of 45= 3^{2}x5

We will count the number of 32 and 5 in 671!, and which one is lesser in number would be the answer.

No of 3’s= 671/3 + 671/9 + 671/27 + 671/81 + 671/243

= 223 + 74 + 24 + 8 + 2

= 331

No of 3^{2}= 331/2 = 165

No of 5= 671/5 + 671/25 + 671/125 + 671/625

= 134 + 26 + 5 + 1

= 166

**165** will be the answer because 3^{2} is lower in number than 5.

**Question 7 : **Find the remainder (359 x 471)/11.

**Solution:** Divide individually both numbers and put remainder.

359/11 gives remainder 7

471/11 gives remainder 9

Put (7 x 9)/11 = 63/11 = 8 is the remainder.

**Question 8 : **Find the number of zeroes in the product:

1^{1}x2^{2}x3^{3}x4^{4}x5^{5}………55^{5}

**Solution:** Number of zeroes will be given by counting the number of 5’s.

5^{5}, 10^{10}, 15^{15}, 20^{20}, 25^{25}, 30^{30}, 35^{35}, 40^{40}, 45^{45}, 50^{50}, 55^{55}

The number of 5’s in these values

5 + 10 + 15 + 20 + **50** + 30 + 35 + 40 + 45 + **100** + 55 = **405**

**Question 9: ** Which of the following fraction is the smallest 7/6, 7/9, 4/5, 5/7 ?

**Solution:**

**Step #1**

Compare first two fractions 7/6 and 7/9

Cross multiply 63 > 42

7/9 fraction is smaller

**Step #2** Compare the smaller one and the next fraction.

7/9 4/5 Cross multiply 35 < 36 7/9 is smaller

**Step #3** Compare smaller one and next and repeat cross multiplication.

7/9 5/7 cross multiply 49 > 45

Here **5/7** is smallest fraction among all.

**Question 10: ** If 19^{200} is divided by 20, the remainder is

**Solution: ** We can write it as (20 – 1)^{200}

Hence apply binomial theorem,

20^{200} (-1)^{0} + 20^{199} (-1)^{1} +………….. 20^{0} (-1)^{200}

Remainder always comes from the last term

20^{0} (-1)^{200} / 20 = 1/20= **1**

**Question 11: ** If tripling a number and adding 10 to the result gives the same answer as multiplying the number by 4 and taking away 20 from the product, the number is :

**Solution: ** Let x be the number

Acc. to question

3x + 10 = 4x – 20

**x = 30**

**Question 12: ** Some students decided to go on campaign and planned to spend Rs 150 on eatables.Five friends did not turn up. As a consequence, each one of the remaining had to contribute Rs 5 extra. The number of students who attend campaign is

**Solution: ** Let the number of students in the beginning is x.

Acc. to question

150/(x – 5) – 150/x = 5

150x – 150x + 750/ x(x – 5) = 5

x^{2} – 5x -150 = 0

x^{2} – 15x + 10x – 150 = 0

x(x – 15) + 10(x – 15) = 0

x = 15, -10

Hence the number of students in the beginning is 15.

**Question 13: ** The value of ( + ) is

**Solution: ** It can be write as

(0.43434343….. + 0.54272727….)

After addition it will be

( 0.9770707070…) or ()

**Question 14: ** Find 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56.

**Solution: ** It can be write as

1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) + 1/(6*7) + 1/(7*8)

= (1/2 – 1/3) + (1/3 – 1/4)+ (1/4 – 1/5) + (1/5 – 1/6) + (1/6 – 1/7)+ (1/7 – 1/8)

= 1/2 – 1/8

= (4 – 1)/8

= **3/8**

## Recommended Posts:

- Problem on HCF and LCM
- Problem on Pipes and Cisterns
- Problem on Time Speed and Distance
- Problem on Trains, Boat and streams
- Numbers
- TCS Coding Practice Question | LCM of 2 Numbers
- TCS Coding Practice Question | HCF or GCD of 2 Numbers
- TCS Coding Practice Question | Average of 2 Numbers
- TCS Coding Practice Question | Greatest of 3 Numbers
- TCS Coding Practice Question | Swap two Numbers
- Quantitative Aptitude: Maths Tricks | Set 1 (Squares of numbers)
- TCS Coding Practice Question | Prime Numbers upto N
- Age of AI-based recruitment... What to expect?
- Placement 100 Course Walk-Through

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.