The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, coprimes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Q. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Given numbers are 43, 91, and 183.
Subtract smallest number from both the highest numbers.
We have three cases:
183 > 43; 183 > 91 and 91 > 43
Therrefore,
183 – 43 = 140
183 – 91 = 92
91 – 43 = 48
Now, we have three new numbers: 140, 48 and 92.
HCF (140, 48 and 92) = 4
The highest number that divides 183, 91, and 43 and leaves the same remainder is 4.
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
N = greatest number that will divide 105, 4665 and 6905 leaving the same remainder in each case
⇒ N = H.C.F. of (4665  1305), (6905  4665) and (6905  1305)
⇒ N = H.C.F. of 3360, 2240 and 5600 = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
Greatest number of 4digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999  399) = 9600.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10, and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
The Greatest Common Divisor of 1.08, 0.36 and 0.9 is:
Given numbers are 1.08 , 0.36 and 0.90
G.C.D. i.e. H.C.F of 108, 36 and 90 is 18
Therefore, H.C.F of given numbers = 0.18
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
⇒ ab = 12.
Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1
Now, the coprimes with product 12 are (1, 12) and (3, 4).
⇒ The required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
∴ Required number = (90 x 4) + 4 = 364.
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