LCM and HCF
 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.
 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
 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
 What is the smallest number which leaves remainder 3 when divided by any of the numbers 5, 6 or 8 but leaves no remainder when it is divided by 9 ?

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We find LCM of 5, 6 and 8
5 = 5
6 = 3 × 2
8 = 2^{3}
LCM of 5, 6 and 8 = 2^{3} × 3 × 5 = 8 × 15 = 120
Required number = 120K + 3
when K = 2,Correct Option: D
We find LCM of 5, 6 and 8
5 = 5
6 = 3 × 2
8 = 2^{3}
LCM of 5, 6 and 8 = 2^{3} × 3 × 5 = 8 × 15 = 120
Required number = 120K + 3
when K = 2,
∴ Required number = 120 × 2 + 3 = 243
It is completely divisible by 9
 What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?

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LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
∴ Required number = 1936 – 1897Correct Option: C
LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
∴ Required number = 1936 – 1897 = 39