We need to find the smallest natural number, let's call it $N$, such that when $N$ is divided by 20, 42, or 76, the remainder is always 7.
This can be expressed mathematically as:
This implies that $N - 7$ must be exactly divisible by 20, 42, and 76. Therefore, $N - 7$ must be a common multiple of these three numbers.
To find the smallest such natural number $N$, we need $N - 7$ to be the least common multiple (LCM) of 20, 42, and 76.
First, find the prime factorization of each number:
The LCM is found by taking the highest power of each prime factor present in any factorization:
$\text{LCM}(20, 42, 76) = 2^2 \times 3 \times 5 \times 7 \times 19$
Calculating the value:
$\text{LCM}(20, 42, 76) = 4 \times 3 \times 5 \times 7 \times 19 = 12 \times 5 \times 7 \times 19 = 60 \times 7 \times 19 = 420 \times 19 = 7980$
Now we know that $N - 7 = \text{LCM}(20, 42, 76)$:
$N - 7 = 7980$
Solving for $N$:
$N = 7980 + 7$
$N = 7987$
Thus, the smallest natural number that leaves a remainder of 7 when divided by 20, 42, or 76 is 7987.
$A$ is an ($n \times n$) matrix. Consider the following two statements
Statement 1: Columns of matrix $A$ are linearly independent
Statement 2: Inverse of matrix $A$ exists
Which one of the following statements is TRUE?