Aspire Faculty ID #11955 · Topic: NIMCET 2025 · Just now
NIMCET 2025

The number of all even integers between 99 and 999 which are not multiple of 3 and 5 is

Solution

Solution:

Even numbers from 100 to 998: count $= \frac{998-100}{2}+1=450$.

Exclude evens divisible by $3$ or $5$ using inclusion–exclusion:

  • Multiples of $6$ in $[100,998]$: $\lfloor 998/6 \rfloor-\lfloor 99/6 \rfloor = 166-16=150$.
  • Multiples of $10$ in $[100,998]$: $\lfloor 998/10 \rfloor-\lfloor 99/10 \rfloor = 99-9=90$.
  • Multiples of $30$ in $[100,998]$: $\lfloor 998/30 \rfloor-\lfloor 99/30 \rfloor = 33-3=30$.

Forbidden $=150+90-30=210$ ⇒ Allowed $=450-210={240}$.

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