Some children were sharing oranges. If each child took 3 oranges, there would be 2 oranges left over. But if each child took 4 oranges, there would be 2 oranges short. How many oranges were there?
I have a money pouch Rs.700.there are equal number of 25 pais,50 paise,1 one Rupee coins.how many of each is there?
PLS HELP!!!
Some children were sharing oranges. If each child took 3 oranges, there would be 2 oranges left over. But if each child took 4 oranges, there would be 2 oranges short. How many oranges were there?
To solve this problem, let's represent the number of children as "c" and the total number of oranges as "o."
According to the problem, if each child took 3 oranges, there would be 2 oranges left over. This means that the number of oranges is 2 more than a multiple of 3. We can express this as:
o ≡ 2 (mod 3) -- Equation 1
Similarly, if each child took 4 oranges, there would be 2 oranges short. This means that the number of oranges is 2 less than a multiple of 4. We can express this as:
o ≡ -2 (mod 4) -- Equation 2
Now, we can use these two congruence equations to find the number of oranges. To do this, we need to find the least common multiple (LCM) of 3 and 4.
The LCM of 3 and 4 is 12.
By adding Equation 1 and Equation 2 together, we can find the number of oranges:
o ≡ 2 (mod 3)
o ≡ -2 (mod 4)
--------------------
o + o ≡ 2 + (-2) (mod 3) + (mod 4)
2o ≡ 0 (mod 12)
Simplifying the congruence equation:
2o ≡ 0 (mod 12)
Since the coefficient of "o" is 2, we need to find the inverse modulo 12:
2 * 6 ≡ 12 (mod 12)
Therefore, the inverse of 2 modulo 12 is 6.
Multiplying both sides of the congruence equation by the inverse of 2 modulo 12:
(2o * 6) ≡ (0 * 6) (mod 12)
12o ≡ 0 (mod 12)
Reducing the equation:
12o ≡ 0 (mod 12)
o ≡ 0 (mod 12)
This means that the number of oranges is a multiple of 12.
To find the least positive solution, we can substitute different values for "o" into each equation and look for a solution that satisfies both equations.
If we substitute o = 12 into the equations, we get:
Equation 1: 12 ≡ 2 (mod 3) -- This is true because 12 divided by 3 leaves a remainder of 0, which is equivalent to 2 (mod 3).
Equation 2: 12 ≡ -2 (mod 4) -- This is not true because 12 divided by 4 leaves a remainder of 0, not -2.
If we substitute o = 24 into the equations, we get:
Equation 1: 24 ≡ 2 (mod 3) -- This is not true because 24 divided by 3 leaves a remainder of 0, not 2.
Equation 2: 24 ≡ -2 (mod 4) -- This is not true because 24 divided by 4 leaves a remainder of 0, not -2.
If we substitute o = 36 into the equations, we get:
Equation 1: 36 ≡ 2 (mod 3) -- This is not true because 36 divided by 3 leaves a remainder of 0, not 2.
Equation 2: 36 ≡ -2 (mod 4) -- This is true because 36 divided by 4 leaves a remainder of 2, which is equivalent to -2 (mod 4).
Therefore, the least positive solution for "o" that satisfies both equations is o = 36.
Hence, there were 36 oranges.