LCM and HCF
- A number p is divisible by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of p is:
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We get LCM of 8, 12 and 16 = 48
∴ Required number = 48a + 3 which is divisible by 7.
∴ p = 48a + 3 = (7 × 6a) + (6a + 3) which is
divisible by 7.
i.e. 6a + 3 is divisible by 7.
When a = 3, 6a + 3 = 18 + 3 = 21 , which is divisible by 7.Correct Option: D
We get LCM of 8, 12 and 16 = 48
∴ Required number = 48a + 3 which is divisible by 7.
∴ p = 48a + 3 = (7 × 6a) + (6a + 3) which is
divisible by 7.
i.e. 6a + 3 is divisible by 7.
When a = 3, 6a + 3 = 18 + 3 = 21 , which is divisible by 7.
∴ p = 48 × 3 + 3 = 144 + 3 = 147
Hence , The least value of p is 147 .
- The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 is
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First of all , we find the LCM of 12 , 18 , 21 and 28
∴ LCM = 2 × 2 × 3 × 3 × 7 = 252
The largest 4-digit number = 9999
Here , Remainder = 171
∴ Required number = The largest 4-digit number - RemainderCorrect Option: A
First of all , we find the LCM of 12 , 18 , 21 and 28
∴ LCM = 2 × 2 × 3 × 3 × 7 = 252
The largest 4-digit number = 9999
Here , Remainder = 171
∴ Required number = The largest 4-digit number - Remainder
Hence , Required number = 9999 – 171 = 9828
- Let p be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of p is
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We find LCM of 5, 6, 7 and 8 = 840
∴ LCM = 2 × 5 × 3 × 7 × 4 = 840
∴ Required number = 840p + 3 which is divisible by 9 for a certain least value of p.
Now, 840p + 3 = 93p × 9 + 3p + 3
∴ 3p + 3 , is divisible by 9 for p = 2Correct Option: C
We find LCM of 5, 6, 7 and 8 = 840
∴ LCM = 2 × 5 × 3 × 7 × 4 = 840
∴ Required number = 840p + 3 which is divisible by 9 for a certain least value of p.
Now, 840p + 3 = 93p × 9 + 3p + 3
3p + 3 , is divisible by 9 for p = 2
∴ Required number = 840 × 2 + 3 = 1680 + 3 = 1683
∴ Sum of digits = 1 + 6 + 8 + 3 = 18
- The number between 3000 and 4000 which is exactly divisible by 30, 36 and 80 is
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Firstly, we find the LCM of 30, 36 and 80.
∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720Correct Option: D
Firstly, we find the LCM of 30, 36 and 80.
∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
∴ Required number = Multiple of 720 = 720 × 5 = 3600 ; because 3000 < 3600 < 4000
- HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is :
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Let the numbers be 7p and 7q where p and q are co-prime.
Now, LCM of 7p and 7q = 7pq
∴ 7pq = 140⇒ pq = 140 = 20 7
Now, required values of p and q whose product is 20 and are co-prime, will be 4 and 5.
Correct Option: C
Let the numbers be 7p and 7q where p and q are co-prime.
Now, LCM of 7p and 7q = 7pq
∴ 7pq = 140⇒ pq = 140 = 20 7
Now, required values of p and q whose product is 20 and are co-prime, will be 4 and 5.
∴ Numbers are 28 and 35 which lie between 20 and 45.
∴ Required sum = 28 + 35 = 63.
