LCM and HCF
 The least number, which when divided by 18, 27 and 36 separately leaves remainders 5,14, and 23 respectively, is

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The difference between the divisor and the corresponding remainder is same in each case
ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13
∴ Required number = (LCM of 18, 27, and 36 ) – 13
LCM of 18, 27, and 36 = 108Correct Option: A
The difference between the divisor and the corresponding remainder is same in each case
ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13
∴ Required number = (LCM of 18, 27, and 36 ) – 13
LCM of 18, 27, and 36 = 108
Hence , Required number = 108 – 13 = 95
 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
 The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is

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LCM of 5, 10, 12, 15
∴ LCM = 2 × 3 × 5 × 2 = 60
∴ Number = 60k + 2
Now, the required number should be divisible by 7.
Now, 60k + 2 = 7 × 8k + 4k + 2
If we put k = 3 , (4k + 2) is equal to 14 which is exactly divisible by 7.Correct Option: B
LCM of 5, 10, 12, 15
∴ LCM = 2 × 3 × 5 × 2 = 60
∴ Number = 60k + 2
Now, the required number should be divisible by 7.
Now, 60k + 2 = 7 × 8k + 4k + 2
If we put k = 3 , (4k + 2) is equal to 14 which is exactly divisible by 7.
∴ Required number = 60 × 3 + 2 = 182
 Which is the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

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We have , The LCM of 12, 18, 21, 30
∴ LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260Correct Option: C
We have , The LCM of 12, 18, 21, 30
∴ LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260∴ The required number = 1260 = 630 2
 The least multiple of 7, which leaves the remainder 4, when divided by any of 6, 9, 15 and 18, is

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LCM of 6, 9, 15 and 18
∴ LCM = 2 × 3 × 3 × 5 = 90
∴ Required number = 90k + 4, which must be a multiple of 7 for some value of k.
For k = 4,Correct Option: D
LCM of 6, 9, 15 and 18
∴ LCM = 2 × 3 × 3 × 5 = 90
∴ Required number = 90k + 4, which must be a multiple of 7 for some value of k.
For k = 4,
Required Number = 90 × 4 + 4 = 364, which is exactly divisible by 7.