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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

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[ACTION NAME]

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# Your action step 3

...

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system

Let \( N \) be the total number of students in the school before the New Year, among which \( M \) are boys, making up \( k \) percent of the total. This means \( M = \frac{k}{100} N \), or \( 100M = kN \).

After the New Year, the number of boys became \( M+1 \), and the total number of students became \( N+3 \). If the boys now make up \( \ell \) percent (with \( \ell < 100 \) since there are definitely still some girls in the school), then:

\[ 100(M+1) = \ell(N+3) \]

Recalling the equality \( 100M = kN \), we find that:

\[ \ell N + 3\ell = 100M + 100 = kN + 100 \]

Thus, \( 100 - 3\ell = (\ell - k)N \). If \( 3\ell < 100 \) (that is, if girls make up less than one-third), then \( N \) is a natural divisor of the positive number \( 100 - 3\ell < 100 \), and therefore \( N \) is less than 100. If \( 3\ell \) is greater than 100 (it clearly cannot be equal to 100), then \( (k - \ell)N = 3\ell - 100 \leq 3 \cdot 99 - 100 = 197 \), and the number of students from the previous year is at most 197.

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