4 bells toll together ay 9 am. they toll after 7,8 11,12 seconds resp. how many times will they toll together again in the next 3 hrs.?
a) 3
b)4
c)5
d)6
The least common multiple of 2^6-1 and 2^9-1.
Hi Nilesh !
A = (26 - 1) = ((23)3 -1) = ( 23 - 1 ) ( 23 + 1 )
B = (29 - 1) = (23 -1) ( 23 + 1 + 26 )
HCF (A, B) = ( 23 - 1 )
We know,
Product of two numbers = HCF x LCM
Hence, LCM (A, B) = [(26 - 1) (29 -1)]/ [23 -1]
LCM (A, B) = (23 +1) (29 -1)
= 212 - 23+ 29 -1
= 212 - 23 (26-1) -1
= 212 + 23 × 63 - 1.
B is multiplied by 25, LCM (A,B) remains unchanged so A has 5^3 in its prime factorization .
When A is multiplied by 8 ,LCM (A,B) remains unchanged so B has 2^4 in its prime factorization .
And A can have 2^0 or 2^1 => 2 cases
for each case exponent of 7 can range from 0 to 4 => 5 cases
Total : 2 x 5 = 10 cases
H.C.F of how many distinct pairs of factors of 18000 is 75?
75k and 75n are the factors , where k and n are coprime to each other .
=> k,n are the coprime factors of 18000/75= 240.
240 = 24 x 3 x 5
k = 2a1 3b1 5c1
n = 2a2 3b2 5c2
one of a1 , a2 has to be 0
=> ( 5 x 5) – ( 4 x 4) = 9 cases
one of b1 , b2 has to be 0
=> ( 2 x 2 ) – ( 1x 1 )= 3 cases
one of c1 , c2 has to be 0
=> ( 2 x 2 ) – ( 1 x 1 )= 3 cases
Hence , total number of ordered pair solution : 9 x 3 x 3 = 81 cases
Unordered pairs 82/2 = 41
What is the HCF of 3n+23 and n+8? (Please explain with long division method)
Find the largest 3 digit number which gives same remainder when divided by 13 or 17?
A.998
B.996
C.896
D.884
How can we do this without options?
Largest possible remainder among 13 and 17 is 12 .since remainder has to be lesser than divisor
So , n [ LCM (13,17)] + 12
221n + 12 < 1000
largest possible n : 4
Hence 221 x 4 + 12 = 896.
How many pairs of natural numbers exist such that their greatest common divisor is 45 and product is 21600?
45a x 45b = 21600
ab = 21600/45
Which is not possible as a, and b are positive integers .
Three runners running around a circular track can complete one revolution in 2,4and 5.5 hours respectively. When will they meet at the starting point?
First runner reaches the starting point every two hour
Second runner reaches the starting point every four hour
The third runner reaches the starting point every 5.5
So just take LCM of 2 , 4 and 5.5 🙂
A rectangular cloth measuring 54 cm *90cm has tobe cut into equal squares such that no cloth is wasted. What is the least number of squares that can be made?
Hi Nancy
Take the HCF of 54 and 90.
HCF ( 54 and 90) = 18
So we can put 54/18 i.e 3 square tiles in one direction and 90/18 i.e 5 square tiles in another direction .
Total : 5 x 3 = 15 square tiles
Alternate Approach :
For least number of tiles side of the square tiles should be maximum.
side of the square = HCF (54, 90) = 18
Number of squares = (Area of the floor )/ (Area of a tile) = (54 x 90)/ ( 18 x 18) = 15
A number N is divisible by 10,90,98,882 but it is not divisible by 50or 270 or 686 or 1764. It os also known that N is a factor of 9261000. What is N?
Use option - elimination.
P and Q are two disticnt whole no. And p+1,p+2,p+3,.......p+7 are integral multiples of q+1,q+2,q+3,.......q+7 Respectively. What is the minimum value of P?
For minimum p, q should be minimum that is 0.
so p +1 will be divisible by 1 ,
p + 2 will be divisible by 2 , p + 3 will be divisible by 3 and so on ....
Hence , just take LCM of 1, 2 , 3, 4....., 7 i.e 420
