Smallest Number with Remainders 7 when Divided by 35, 56, and 91

To find the smallest number that satisfies the given conditions, we need to determine the least common multiple (LCM) of 35, 56, and 91. The LCM is the smallest multiple that is divisible by all three numbers.

Step 1: Prime Factorization
- Prime factorize each of the given numbers:

35 = 5 × 7
56 = 2^3 × 7
91 = 7 × 13

Step 2: Determine the Highest Power of Each Prime Factor
- Identify the highest power of each prime factor among the three numbers:

The highest power of 2 is 3 (from 56).
The highest power of 5 is 0 (from 35).
The highest power of 7 is 1 (from 35, 56, and 91).
The highest power of 13 is 1 (from 91).

Step 3: Calculate the LCM
- Calculate the LCM by multiplying the highest powers of each prime factor:

LCM = 2^3 × 5^0 × 7^1 × 13^1
= 8 × 1 × 7 × 13
= 728

Therefore, the smallest number that when divided by 35, 56, and 91 leaves remainders of 7 in each case is 728.

Explanation
The LCM of the three given numbers ensures that when we divide the LCM by any of these numbers, it will leave a remainder of 0. However, since we are looking for a remainder of 7, we need to add 7 to the LCM.

For example, when we divide 728 by 35, the remainder is 7. Similarly, dividing 728 by 56 or 91 will also result in a remainder of 7. Any number smaller than 728 will not satisfy the condition.

The prime factorization helps us find the highest powers of all prime factors present in the given numbers. By multiplying these highest powers, we obtain the LCM.

In conclusion, the smallest number that leaves remainders of 7 when divided by 35, 56, and 91 is 728.