To find the least number $N$ which, when divided by 16, 28, 40, and 77, leaves a remainder of 5 in each case, follow these steps:

Step 1: Represent the condition mathematically

The number $N$ can be expressed as: $N = LCM(16, 28, 40, 77) \cdot k + 5$ where $k$ is an integer and $LCM$ is the least common multiple.

Step 2: Compute the LCM of 16, 28, 40, and 77

We find the prime factorizations:

• $16 = 2^4$
• $28 = 2^2 \cdot 7$
• $40 = 2^3 \cdot 5$
• $77 = 7 \cdot 11$

The $LCM$ is the product of the highest powers of all primes involved: $LCM = 2^4 \cdot 5 \cdot 7 \cdot 11 = 16 \cdot 5 \cdot 7 \cdot 11 = 6160$

Step 3: Construct the number $N$

The least number $N$ is obtained when $k = 0$: $N = 6160 \cdot 0 + 5 = 5$

The next such number is: $N = 6160 \cdot 1 + 5 = 6165$

Thus, the least number is 6165.

Verification

Check if $6165$ leaves a remainder of 5 when divided by 16, 28, 40, and 77:

• $6165 \mod 16 = 5$
• $6165 \mod 28 = 5$
• $6165 \mod 40 = 5$
• $6165 \mod 77 = 5$

The solution satisfies all conditions.

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Would you like me to clarify any step or solve related problems?

Related Questions:

1. What is the difference between LCM and GCD?
2. How do you calculate the LCM of more than two numbers quickly?
3. Can you explain modular arithmetic in more detail?
4. How would the solution change if the remainder were different, e.g., 3?
5. Why does adding the remainder work for finding $N$?

Tip:

Always verify