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] + [155/5²] + [155/5³] 
=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= 2²x3²x5¹x7²
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= 2⁶x3²
Sum of all the factors= (2⁰ + 2¹+ 2² + 2³ + 2⁴ + 2⁵ + 2⁶)*(3⁰ + 3¹ + 3²) 
= 127*13 
= 1651

Question 4: Find the product of all the factors of 600. 
Solution: Prime factorization of 600= 2³x3¹x5²
Number of factors= (3+1)(1+1)(2+1) 
= 4 * 2 * 3 
= 24 
Product of all the factors= (2³x3¹x5²)^24/2
= (2³x3¹x5²)¹²

Question 5: Find the highest power of 126 which divides 366!. 
Solution: 126=2×3²x7 
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] + [366/7²] + [366/7³] 
=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= 3²x5 
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 3²= 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 3² 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: 
1¹x2²x3³x4⁴x5⁵………55⁵
Solution: Number of zeroes will be given by counting the number of 5’s. 
5⁵, 10¹⁰, 15¹⁵, 20²⁰, 25²⁵, 30³⁰, 35³⁵, 40⁴⁰, 45⁴⁵, 50⁵⁰, 55⁵⁵
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 19²⁰⁰ is divided by 20, the remainder is 
Solution: We can write it as (20 – 1)²⁰⁰
Hence apply binomial theorem, 
20²⁰⁰ (-1)⁰ + 20¹⁹⁹ (-1)¹ +………….. 20⁰ (-1)²⁰⁰
Remainder always comes from the last term 
20⁰ (-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 
x² – 5x -150 = 0 
x² – 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 written 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 written 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

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