there are fewer than 6 dozen eggs in a basket. If I count them by 2's there is 1 left over. If I count them by 3's there are 2 left over
Three are left if I count by 4s. Four are left if I count by 5s. how many egs are there
To solve this problem, we can start by setting up a system of equations based on the given information.
Let's assume the number of eggs in the basket is "x".
According to the first condition, if we count them by 2's, there is 1 left over. This means that when we divide the number of eggs by 2, the remainder is 1:
x mod 2 = 1
According to the second condition, if we count them by 3's, there are 2 left over. This means that when we divide the number of eggs by 3, the remainder is 2:
x mod 3 = 2
According to the third condition, if we count them by 4's, three are left over. This means that when we divide the number of eggs by 4, the remainder is 3:
x mod 4 = 3
Lastly, according to the fourth condition, if we count them by 5's, four are left over. This means that when we divide the number of eggs by 5, the remainder is 4:
x mod 5 = 4
Now, we can solve this system of equations step-by-step using a technique called "Chinese remainder theorem."
Step 1: Solve the first pair of equations (x mod 2 = 1 and x mod 3 = 2) by finding the least common multiple (LCM) of the two divisors, in this case, 2 and 3.
The LCM of 2 and 3 is 6, so we need to find the multiple of 6 that satisfies both equations.
The multiples of 6 are 6, 12, 18, 24, 30, ...
Checking the remainders, we find that when x = 6, both equations are satisfied:
6 mod 2 = 0 (satisfies the first equation)
6 mod 3 = 0 (satisfies the second equation)
Therefore, the solution for the first pair of equations is x = 6.
Step 2: Solve the next pair of equations (x mod 4 = 3 and x mod 5 = 4) by finding the LCM of the two divisors, in this case, 4 and 5.
The LCM of 4 and 5 is 20, so we need to find the multiple of 20 that satisfies both equations.
The multiples of 20 are 20, 40, 60, ...
Checking the remainders, we find that when x = 23, both equations are satisfied:
23 mod 4 = 3 (satisfies the third equation)
23 mod 5 = 3 (satisfies the fourth equation)
Therefore, the solution for the second pair of equations is x = 23.
Step 3: Finally, we need to find the LCM of the two previous solutions, 6 and 23.
The LCM of 6 and 23 is 138.
Thus, there are 138 eggs in the basket.
To solve this problem, we can use the method of finding the least common multiple (LCM) of the numbers 2, 3, 4, and 5.
First, let's analyze the information given:
1. If we count the eggs by 2's, there is 1 left over. This means the total number of eggs is not divisible evenly by 2.
2. If we count the eggs by 3's, there are 2 left over. This means the total number of eggs is not divisible evenly by 3.
3. If we count the eggs by 4's, there are 3 left over. This means the total number of eggs is not divisible evenly by 4.
4. If we count the eggs by 5's, there are 4 left over. This means the total number of eggs is not divisible evenly by 5.
Now, let's find the LCM of 2, 3, 4, and 5:
The prime factorization of each number is as follows:
- 2 = 2^1
- 3 = 3^1
- 4 = 2^2
- 5 = 5^1
To find the LCM, we take the highest power of each prime factor that appears in any of the given numbers:
- The highest power of 2 is 2^2 = 4.
- The highest power of 3 is 3^1 = 3.
- The highest power of 5 is 5^1 = 5.
The LCM of 2, 3, 4, and 5 is obtained by multiplying the highest powers of each prime factor:
LCM = 2^2 * 3^1 * 5^1 = 4 * 3 * 5 = 60.
Therefore, there are 60 eggs in the basket.
" If I count them by 2's there is 1 left over"
---- all odd numbers , including 59
" If I count them by 3's there are 2 left over "
--- 5 8 11 14 17 20 23 ....59 62 65 68 71
"Three are left if I count by 4s"
-- 7 11 15 19 23 27 31 .... 59 63 67 71
"Four are left if I count by 5s"
--- 9 14 19 24 29 34 39 44 49 55 59 65 69
looks like 59 is in all 4 groups