Find the Smallest Number Which Leaves Remainder 7 When Divided by 35, 56 and 91
Video Explanation
Watch the video below to understand the complete solution step by step:
Solution
Question: What is the smallest number that, when divided by 35, 56 and 91, leaves remainders of 7 in each case?
Step 1: Subtract the Common Remainder
If a number leaves remainder 7 when divided by 35, 56 and 91, then the number minus 7 is exactly divisible by all three numbers.
Required number − 7 is divisible by 35, 56 and 91
Step 2: Find the LCM of 35, 56 and 91
Prime factorisation:
35 = 5 × 7
56 = 23 × 7
91 = 7 × 13
LCM = 23 × 5 × 7 × 13
LCM = 3640
Step 3: Find the Required Smallest Number
Required number = LCM + 7
Required number = 3640 + 7
Required number = 3647
Final Answer
∴ The smallest number that leaves remainder 7 when divided by 35, 56 and 91 is 3647.
Conclusion
Thus, by subtracting the common remainder and finding the LCM of the given divisors, we obtain the required smallest number as 3647.