find the smallest no.when no.30,40,and 60,divided and leaves remainder 7
Asked by singhaarzoo155 | 16th May, 2021, 07:38: PM
Q: Find the smallest number which when divided by 30, 40 and 60 leaves remainder 7 in each case.
The LCM of 30, 40 and 60 is 120.
So, 120 when divided by 30, 40 and 60, leaves remainder 0.
Adding 7, we get 127
When 127 divided by 30, 40 and 60, leaves remainder 7 in each case.
Hence, the required number is 127.
Answered by Renu Varma | 17th May, 2021, 10:56: AM
- P and q are two consecutive even natural number 72 then HCF of P and q is
- If xy=4320 and LCM(x,y)=360 then find the HCF(x,y)
- if sum of two numbers is 1215 and their HCF is 81, then the possible number of pairsof such numbers are
- The traffic lights at three different roads crossing change after every 48 seconds ,72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m . Then at what time will they again change simultaneously
- In the 10th question of 1b from rs aggarwal .why did we subtract 7 from lcm
- find the HCF & LCM of 96 & 404
- express each number as a product of it prime 1771 AND 3825
- What will be the product of HCF and LCM of the smallest prime number and the smallest composite number?
- If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is Please explain
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number