We need to find the greatest prime factor of the expression $(3^{199} – 3^{196})$. First, factor out the common term, which is the lowest power of 3, $3^{196}$.
$3^{199} – 3^{196} = 3^{196} (3^{199-196} – 1)$
Now, simplify the term inside the parenthesis:
$3^{196} (3^3 – 1)$
Calculate $3^3$:
$3^3 = 3 \times 3 \times 3 = 27$
Substitute this back into the expression:
$3^{196} (27 – 1) = 3^{196} \times 26$
To find the greatest prime factor, we determine the prime factors of the simplified expression $3^{196} \times 26$.
The prime factors of the entire expression $(3^{199} – 3^{196})$ are therefore 3, 2, and 13.
Comparing the prime factors {2, 3, 13}, the largest number is 13.
Thus, the greatest prime factor of $(3^{199} – 3^{196})$ is 13.
Find the value of \(\sqrt{2025}\) .
How many times does the number 5 occur in the range of numbers from 1 to 100?
A. 21
B. 22
C. 20
D. 19
A prime number
A. is not a positive integer.
B. has no divisor at all.
C. has only 1 and itself as divisors.
D. has more than two divisors.
__________ are twin prime number.
A. (4, 9)
B. (2, 3)
C. (4, 6)
D. (3, 5)A factory produced 18,58,509 cassettes in the month of January, 7623 more cassettes in the of February and owing to short supply of electricity produced 25,838 less cassettes in March than in February. Find the total production in all?
A. 55,57,312
B. 59,83,245
C. 55,64,935
D. 56,08,988