LCM and HCF
- The product of two numbers is 2028 and their HCF is 13. The number of such pairs is
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Here, HCF = 13
Let the numbers be 13p and 13q ,where p and q are Prime to each other.
Now, 13p × 13q = 2028⇒ pq = 2028 = 12 13 × 13
Correct Option: B
Here, HCF = 13
Let the numbers be 13p and 13q ,where p and q are Prime to each other.
Now, 13p × 13q = 2028⇒ pq = 2028 = 12 13 × 13
The possible pairs are :- ( 1 , 12 ) , ( 3 , 4 ) , ( 2 , 6 )
But the 2 and 6 are not co-prime.
∴ The required no. of pairs = 2
- The HCF and LCM of two numbers are 13 and 455 respectively. If one of the number lies between 75 and 125, then, that number is :
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Given that , HCF = 13 , LCM = 455
Let the numbers be 13p and 13q , Where p and q are co-prime.
∴ LCM = 13pq
∴ 13pq = 455∴ pq = 455 13
Correct Option: B
Given that , HCF = 13 , LCM = 455
Let the numbers be 13p and 13q , Where p and q are co-prime.
∴ LCM = 13pq
∴ 13pq = 455∴ pq = 455 = 35 = 5 × 7 13
∴ Numbers are 13 × 5 = 65 and 13 × 7 = 91
- A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is :
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We know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
Here, t = Divisor – remainder = 1
t = 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1Correct Option: C
We know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
Here, t = Divisor – remainder = 1
t = 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1
∴ Required number = k - t = (L.C.M. of 10, 9, 8) – 1
Hence , Required number = 360 – 1 = 359
- The smallest number, which when divided by 12 and 16 leaves remainder 5 and 9 respectively, is :
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As we know that when a number is divided by a, b leaving remainders p, q respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a and b .
Here, t = 12 – 5 = 7, 16 – 9 = 7Correct Option: B
As we know that when a number is divided by a, b leaving remainders p, q respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a and b .
Here, t = 12 – 5 = 7, 16 – 9 = 7
∴ Required number = k - t = (L.C.M. of 12 and 16) – 7
Required number = 48 – 7 = 41
- The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is :
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LCM of 4, 6, 8, 9
∴ LCM = 2 × 2 × 3 × 2 × 3 = 72Correct Option: B
LCM of 4, 6, 8, 9
∴ LCM = 2 × 2 × 3 × 2 × 3 = 72
∴ Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13.
