When 7179 and 9699 are divided by another natural number N , remainder...
To solve this problem, we need to find the values of N that will result in the same remainder when dividing both 7179 and 9699.
Let's first find the remainder when dividing 7179 and 9699 by N.
For 7179:
Let's assume the remainder is R1 when dividing 7179 by N.
So, 7179 = N * Q1 + R1
where Q1 is the quotient when dividing 7179 by N.
For 9699:
Let's assume the remainder is R2 when dividing 9699 by N.
So, 9699 = N * Q2 + R2
where Q2 is the quotient when dividing 9699 by N.
According to the problem, R1 = R2.
Now, let's simplify the equations and put them together:
N * Q1 + R1 = N * Q2 + R2
Rearranging the equation, we get:
N * (Q1 - Q2) = R2 - R1
Since R1 = R2, the right side of the equation becomes zero:
N * (Q1 - Q2) = 0
For the equation to be true, either N = 0 or (Q1 - Q2) = 0.
Now, we know that N cannot be zero because it is a natural number. So, the only possibility is (Q1 - Q2) = 0.
(Q1 - Q2) = 0 means that the quotients Q1 and Q2 are equal when dividing 7179 and 9699 by N.
To find the values of N that satisfy this condition, we need to consider the factors of 7179 and 9699.
Prime factorization of 7179: 3^2 * 797
Prime factorization of 9699: 3^2 * 1077
To have the same quotient when dividing by N, the prime factorization of N should include the common factors of 7179 and 9699, which are 3^2.
Since N is a natural number, the possible values of N are the divisors of 3^2, which are 1, 3, and 9.
Among these values, only 9 ends with one or more than one zeroes.
Therefore, the correct answer is option C) 18.