A heap of stones can be made up into groups of 21. When made up into groups of 16, 20, 25 and 45, there are 3 stones left in each case. How many stones at least can there be in the heap?

Options

A. 7203
B. 2403
C. 3603
D. 4803

Answer

To solve the problem step by step, we need to find the least number of stones in the heap that meets the given conditions.

### Step 1: Understand the conditions
The problem states that:
- The stones can be grouped into 21 without any remainder.
- When grouped into 16, 20, 25, and 45, there are 3 stones left in each case.

### Step 2: Set up the equations
Let the total number of stones be \( N \). According to the problem:
- \( N \equiv 0 \mod 21 \) (N is divisible by 21)
- \( N \equiv 3 \mod 16 \)
- \( N \equiv 3 \mod 20 \)
- \( N \equiv 3 \mod 25 \)
- \( N \equiv 3 \mod 45 \)

### Step 3: Rewrite the equations
From the conditions, we can rewrite \( N \) as:
- \( N = 21k \) for some integer \( k \)
- \( N = 16m + 3 \)
- \( N = 20n + 3 \)
- \( N = 25p + 3 \)
- \( N = 45q + 3 \)

### Step 4: Solve for LCM
We need to find the least common multiple (LCM) of the numbers 16, 20, 25, and 45 to find a common base for the equations.

Calculating the LCM:
- Prime factorization:
- \( 16 = 2^4 \)
- \( 20 = 2^2 \times 5 \)
- \( 25 = 5^2 \)
- \( 45 = 3^2 \times 5 \)

Taking the highest powers of each prime:
- \( 2^4 \) from 16
- \( 3^2 \) from 45
- \( 5^2 \) from 25

Thus,
\[ \text{LCM}(16, 20, 25, 45) = 2^4 \times 3^2 \times 5^2 = 16 \times 9 \times 25 = 3600 \]

### Step 5: Write the general form of N
Since \( N \equiv 3 \mod 16, 20, 25, 45 \), we can express \( N \) in terms of the LCM:
\[ N = 3600k + 3 \quad \text{for some integer } k \]

### Step 6: Ensure divisibility by 21
Now we need \( N \) to also be divisible by 21:
\[ 3600k + 3 \equiv 0 \mod 21 \]
This simplifies to:
\[ 3600k \equiv -3 \mod 21 \]

Calculating \( 3600 \mod 21 \):
\[ 3600 \div 21 \approx 171.42857 \quad \Rightarrow \quad 21 \times 171 = 3591 \]
\[ 3600 - 3591 = 9 \quad \Rightarrow \quad 3600 \equiv 9 \mod 21 \]
Thus, we have:
\[ 9k \equiv -3 \mod 21 \]
This can be rewritten as:
\[ 9k \equiv 18 \mod 21 \]

### Step 7: Solve for k
To solve for \( k \):
Dividing both sides by 9 gives:
\[ k \equiv 2 \mod 21 \]
This means \( k \) can be expressed as:
\[ k = 21m + 2 \quad \text{for some integer } m \]

### Step 8: Substitute back to find N
Substituting back into the equation for \( N \):
\[ N = 3600(21m + 2) + 3 = 75600m + 7200 + 3 = 75600m + 7203 \]
For the minimum case, let \( m = 0 \):
\[ N = 7203 \]

### Final Answer
Thus, the least number of stones in the heap is **7203**.

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