Mrs. Jones has a prime number of children in her class. When she divides the children into teams of 7, one team has 6. When she divides the children into teams of 6, one team has 5. How many children are in the class?
x/7 = A 6/7 = (7 A + 6)/7
x/6 = B 5/6 = (6 B + 5)/6
so x = 7 A + 6
and x= 6 B + 5
7 A +6 = 6 B + 5
6 B - 7 A = 1
what multiples of 7 and 6 differe by one?
35 and 36 or 48 and 49 ?
try 35 and 36
A = 5 and B = 6
x = 6 B + 5 = 41
or
x = 7 A + 6 = 41 sure enough
To solve this problem, let's work through it step by step.
Let's assume the prime number of children is denoted by "x."
According to the given information, when Mrs. Jones divides the children into teams of 7, one team has 6 children. This implies that the total number of children in her class, "x," leaves a remainder of 6 when divided by 7.
Similarly, when the children are divided into teams of 6, one team has 5 children. This means that the total number of children in Mrs. Jones' class, "x," leaves a remainder of 5 when divided by 6.
Now we can set up a system of congruences to represent the given information:
x ≡ 6 (mod 7)
x ≡ 5 (mod 6)
To find the value of "x," we can solve this system of congruences using the Chinese Remainder Theorem.
One approach is to use trial and error to find a number that satisfies both congruences simultaneously. We can start by trying small values of "x" and incrementing until we find a solution.
Let's start with x = 6 and check if it satisfies the congruences:
6 ≡ 6 (mod 7) --> True
6 ≡ 5 (mod 6) --> False
Since 6 does not satisfy both congruences simultaneously, let's try the next value, x = 13:
13 ≡ 6 (mod 7) --> True
13 ≡ 5 (mod 6) --> False
We continue this process until we find a value that satisfies both congruences simultaneously.
Let's try x = 20:
20 ≡ 6 (mod 7) --> True
20 ≡ 5 (mod 6) --> True
Since 20 satisfies both congruences simultaneously, we have found a solution. Therefore, the number of children in Mrs. Jones' class is 20.