Question 1
8100
17600
44100
None of these
Good Job!
Oops!
L.C. of 3, 4, 7 10 and 12 = 2*2*3*5*7 = 420 In the above factorization, 3, 5 and 7 are not in pairs, we have to multiply 420 by 3*5*7 = 420*3*5*7 = 44100
Please login to submit your explanation
You can check your performance of this question after Login/Signup
Start
Question 2
for all values of n
only for even values of n
only for odd values of n
for no values of n
Question 3
15
20
35
40
HCF of ((13863),(228138),(22863)) HCF of (75,90,165) 75=3*5*5 90=2*3*3*5 165=3*5*11 ans=3*5=15
Question 4
422
842
12723
Take LCM of 4,5,6,7. It is 420 BUt the no must leave remainder 2 in each case, so the no is of the form: 420k + 2. The smallest 4digit no is 1000. So keeping k=0,1,2,3.... We get that the largest no smaller than the smallest 4 digit no is 842
Question 5
10
12
14
Simply if divide all number with 5 than we get reminder as below :  90/5 = 18 80/5 = 16 70/5 = 14 60/5 = 12 50/5^2 = 2 40/5 = 8 30/5 = 6 20/5 = 4 10/5 = 2  So total highest power will be 9 and 1 add form 50 terms So total will be 10
Question 6
3 but not by 9
9
6
27
If a − b is divisible by 3, then a − b = 3k, for some integer k (a − b)² = (3k)² a² − 2ab + b² = 9k² a³ − b³ = (a−b) (a² + ab + b²) = (a−b) (a² − 2ab + b² + 3ab) = 3k (9k + 3ab) = 3k * 3 (3k + ab) = 9 k(3k+ab) Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9
Question 7
5
7
8
The question says how many powers of 5 are there in 30! Only the numbers 5,10,15,20, 25, 30 are divisors of 5 and 25 is divisible by 5^2 ( 5*5) Therefore the total number of five\'s are 7. Hence the greatest positive power of 5 that divides 30! is 7.
Question 8
1043
1073
1103
166 * 6 + 5 = 1001 is the smallest 4digit number gives a remainder of 5 when divided by 6. So N = 1001 + 6 * m for some m. so N = 200 * 5 + 1 + 5 m + m = (200 + m) * 5 + 1+m So Remainder when N is divided by 5 = 3 given => remainder when 1+m is divided by 5 = 3 => remainder when m is divided by 5 = 2 Smallest m is 2. Hence, N = 1001 + 6 * 2 = 1013.
Question 9
11
121
242
p=11k+7 so (p+4)(p+15)=(11k+11)(11K+22) =11(k+1)*11(k+2) =121(k+1)(k+2) here one of the terms of (k+1)and(k+2)is even and another is odd..so multiplication of odd and even no. produce a even no..so the no. is divisible by (121*2)=242
Question 10
900
1200
2500
3600
LCM (3, 4, 5, 6 ,8) = 120 120 = 2 x 2 x 2 x 3 x 5. As 2,3 and 5 are not in pair in LCM’s factor so we need to multiply 120 by 5 and 3,2 to make it a perfect square. 120 x 2 x 5 x 3 = 3,600. ∴ 3600 is the least perfect square divisible by 3, 4, 5, 6 and 8.
Please login to report
Login/Signup
Personalized Analytics only Availble for Logged in users
Analytics below shows your performance in various Mocks on PrepInsta
Your average Analytics for this Quiz
Rank

Percentile
0%
Completed
0/0
Accuracy
November 18, 2020