Problem on Numbers - GeeksforGeeks

04 Aug, 2021
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  
 
Shyam Mahato
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