If some marbles are divided in the groups of 32-32, 10 marbles are left behind and if divided in the 40-40 groups, 18 marbles are left behind. If these are divided in groups of 7272, 50 marbles are left behind. The minimum marbles are

Options

A. 1450
B. 1440
C. 1418
D. 1412

Answer

To solve the problem, we need to find the minimum number of marbles (let's call it \( N \)) that satisfies the following conditions:

1. When divided by 32, a remainder of 10 is left.
2. When divided by 40, a remainder of 18 is left.
3. When divided by 72, a remainder of 50 is left.

We can express these conditions mathematically as follows:

1. \( N \equiv 10 \mod 32 \)
2. \( N \equiv 18 \mod 40 \)
3. \( N \equiv 50 \mod 72 \)

### Step 1: Rewrite the congruences

From the first condition, we can express \( N \) in terms of \( k \):

\[ N = 32k + 10 \]

From the second condition, we can express \( N \) in terms of \( m \):

\[ N = 40m + 18 \]

From the third condition, we can express \( N \) in terms of \( p \):

\[ N = 72p + 50 \]

### Step 2: Set the equations equal to each other

We can set the first two equations equal to each other:

\[ 32k + 10 = 40m + 18 \]

Rearranging gives:

\[ 32k - 40m = 8 \]

Dividing the entire equation by 8 simplifies it to:

\[ 4k - 5m = 1 \quad \text{(Equation 1)} \]

### Step 3: Solve for \( k \) and \( m \)

We can express \( k \) in terms of \( m \):

\[ 4k = 5m + 1 \implies k = \frac{5m + 1}{4} \]

For \( k \) to be an integer, \( 5m + 1 \) must be divisible by 4. This means:

\[ 5m + 1 \equiv 0 \mod 4 \]

Calculating \( 5 \mod 4 \):

\[ 5 \equiv 1 \mod 4 \implies 1m + 1 \equiv 0 \mod 4 \implies m \equiv -1 \equiv 3 \mod 4 \]

Let \( m = 4n + 3 \) for some integer \( n \).

### Step 4: Substitute back to find \( k \)

Substituting \( m \) back into the equation for \( k \):

\[ k = \frac{5(4n + 3) + 1}{4} = \frac{20n + 15 + 1}{4} = \frac{20n + 16}{4} = 5n + 4 \]

### Step 5: Substitute \( k \) and \( m \) back into \( N \)

Now substituting \( k \) back into the equation for \( N \):

\[ N = 32(5n + 4) + 10 = 160n + 128 + 10 = 160n + 138 \]

### Step 6: Use the third condition

Now we need to satisfy the third condition:

\[ 160n + 138 \equiv 50 \mod 72 \]

Calculating \( 160 \mod 72 \):

\[ 160 \div 72 = 2 \quad \text{(remainder 16)} \implies 160 \equiv 16 \mod 72 \]

Thus, we have:

\[ 16n + 138 \equiv 50 \mod 72 \]

Simplifying gives:

\[ 16n \equiv 50 - 138 \equiv -88 \equiv -16 \mod 72 \]

This simplifies to:

\[ 16n \equiv 56 \mod 72 \]

### Step 7: Solve for \( n \)

Dividing through by 8 gives:

\[ 2n \equiv 7 \mod 9 \]

Finding the multiplicative inverse of 2 modulo 9, we see that \( 2 \times 5 \equiv 1 \mod 9 \). Therefore, we multiply both sides by 5:

\[ n \equiv 35 \mod 9 \implies n \equiv 8 \mod 9 \]

Let \( n = 9p + 8 \) for some integer \( p \).

### Step 8: Substitute back to find \( N \)

Substituting \( n \) back into \( N \):

\[ N = 160(9p + 8) + 138 = 1440p + 1280 + 138 = 1440p + 1418 \]

### Step 9: Find the minimum \( N \)

For the minimum \( N \), set \( p = 0 \):

\[ N = 1418 \]

### Conclusion

Thus, the minimum number of marbles is:

\[ \boxed{1418} \]