Question

After distributing the sweets equally among 25 children, 8 sweets remain. Had the number of children been 28, 22 sweets would have been left after equally distributing. What is the smallest possible total number of sweets ?

• A. 328
• B. 348
• C. 358
• D. Data inadequate

Answer 1

The Correct Option is C

Let the total number of sweets be \( x \). When \( x \) sweets are distributed among 25 children, each child gets the same number of sweets, and 8 sweets remain. Therefore, we can express this situation mathematically as:

\[ x \equiv 8 \pmod{25} \]

This means that when \( x \) is divided by 25, the remainder is 8.

Similarly, when \( x \) sweets are divided among 28 children, 22 sweets remain, which can be expressed as:

\[ x \equiv 22 \pmod{28} \]

This means that when \( x \) is divided by 28, the remainder is 22.

We now have two simultaneous congruences:

\[ x \equiv 8 \pmod{25} \]

\[ x \equiv 22 \pmod{28} \]

We need to find the smallest positive \( x \) that satisfies both congruences. First, express these congruences in form of equations:

\[ x = 25k + 8 \]

\[ x = 28m + 22 \]

Equating the two expressions for \( x \):

\[ 25k + 8 = 28m + 22 \]

Rearranging terms gives:

\[ 25k - 28m = 14 \]

This is a linear Diophantine equation, which we can solve for integer solutions \( k \) and \( m \). We simplify the equation by rewriting it as:

\[ 25k = 28m + 14 \]

Trial and error or systematic searching for integer solutions gives \( k = 6 \), \( m = 5 \), thus:

\[ x = 25 \times 6 + 8 = 158 \]

\[ x = 28 \times 5 + 22 = 162 \]

Adjusting for the smallest positive solution, observe the general solution:

\[ x = 175n + 158 \]

\[ x = 175n + 22 \]

Solving for consistency across values of \( x \) yields \( x \equiv 22 \pmod{28} \). Hence:

If \( x = 358 \), checking the conditions:

\[ 358 \equiv 8 \pmod{25} \] and \[ 358 \equiv 22 \pmod{28} \]

Both conditions satisfy, hence, the smallest possible total number of sweets is:

\[ \boxed{358} \]