Find the smallest number which when divided by 20, 28, and 36 leaves 3 as remainder in each case.

Steps to Solve

1. Adjust the Remainder Condition

The smallest number we're looking for can be expressed in the form:

$$ x = k \cdot \text{LCM}(20, 28, 36) + 3 $$

where ( k ) is a non-negative integer. This means that ( x - 3 ) must be divisible by 20, 28, and 36.

2. Find the LCM of the Numbers

We need to calculate the least common multiple (LCM) of the numbers 20, 28, and 36 using their prime factorizations:

• Prime factorization of 20: ( 2^2 \cdot 5^1 )
• Prime factorization of 28: ( 2^2 \cdot 7^1 )
• Prime factorization of 36: ( 2^2 \cdot 3^2 )

The LCM takes the highest powers of all prime factors involved:

$$ \text{LCM}(20, 28, 36) = 2^2 \cdot 3^2 \cdot 5^1 \cdot 7^1 $$

3. Calculate the LCM

Calculate the LCM step by step:

• For ( 2^2 ), the max power is ( 2^2 = 4 )
• For ( 3^2 ), the max power is ( 3^2 = 9 )
• For ( 5^1 ), the max power is ( 5^1 = 5 )
• For ( 7^1 ), the max power is ( 7^1 = 7 )

Now, combine them:

$$ LCM = 4 \cdot 9 \cdot 5 \cdot 7 $$

4. Perform the Multiplication

Calculate each step:

First, calculate ( 4 \cdot 9 = 36 )
Next, ( 36 \cdot 5 = 180 )
Finally, ( 180 \cdot 7 = 1260 )

Thus,

$$ \text{LCM}(20, 28, 36) = 1260 $$

5. Find the Smallest Number

Substituting back into our expression for ( x ):

$$ x = k \cdot 1260 + 3 $$

To find the smallest number, set ( k = 0 ):

$$ x = 0 \cdot 1260 + 3 = 3 $$

However, since we need numbers that satisfy the remainder condition, we find for ( k = 1 ):

$$ x = 1 \cdot 1260 + 3 = 1263 $$