The smallest number which when divided by 30, 40, and 60 leaves a remainder of 7 can be found using the concept of LCM (Least Common Multiple).
- Step 1: Find the LCM of 30, 40, and 60
To find the LCM, list down the prime factors of each number:
30 = 2 x 3 x 5
40 = 2 x 2 x 2 x 5
60 = 2 x 2 x 3 x 5
Multiply the highest power of each prime factor to find the LCM:
LCM = 2 x 2 x 2 x 3 x 5 = 120
- Step 2: Subtract the common remainder from the LCM
Subtract the common remainder (7) from the LCM:
120 - 7 = 113
- Step 3: Check if the number satisfies the condition
Now, check if 113 is divisible by 30, 40, and 60 with a remainder of 7:
113 ÷ 30 = 3 with a remainder of 23
113 ÷ 40 = 2 with a remainder of 33
113 ÷ 60 = 1 with a remainder of 53
- Step 4: Find the smallest number that satisfies the condition
Since 113 does not satisfy the condition, we need to keep adding the LCM (120) until we find a number that satisfies the condition:
113 + 120 = 233
233 satisfies the condition:
233 ÷ 30 = 7 with a remainder of 23
233 ÷ 40 = 5 with a remainder of 33
233 ÷ 60 = 3 with a remainder of 53
Therefore, the smallest number that when divided by 30, 40, and 60 leaves a remainder of 7 is 233.