1. **State the problem:** Find the number of positive integers less than or equal to 10030 that are not divisible by 3, 4, 7, or 11. 2. **Use the principle of inclusion-exclusion:** Let \(N = 10030\). - Count numbers divisible by each divisor: \(A_3 = \left\lfloor \frac{N}{3} \right\rfloor = \left\lfloor \frac{10030}{3} \right\rfloor = 3343\) \(A_4 = \left\lfloor \frac{N}{4} \right\rfloor = \left\lfloor \frac{10030}{4} \right\rfloor = 2507\) \(A_7 = \left\lfloor \frac{N}{7} \right\rfloor = \left\lfloor \frac{10030}{7} \right\rfloor = 1432\) \(A_{11} = \left\lfloor \frac{N}{11} \right\rfloor = \left\lfloor \frac{10030}{11} \right\rfloor = 911\) 3. **Count numbers divisible by intersections of pairs (lcm of pairs):** - \(LCM(3,4)=12\), count \(A_{3,4}=\left\lfloor \frac{N}{12} \right\rfloor=\left\lfloor \frac{10030}{12} \right\rfloor=835\) - \(LCM(3,7)=21\), \(A_{3,7}=\left\lfloor \frac{N}{21} \right\rfloor=\left\lfloor \frac{10030}{21} \right\rfloor=477\) - \(LCM(3,11)=33\), \(A_{3,11}=\left\lfloor \frac{N}{33} \right\rfloor=\left\lfloor \frac{10030}{33} \right\rfloor=304\) - \(LCM(4,7)=28\), \(A_{4,7}=\left\lfloor \frac{N}{28} \right\rfloor=\left\lfloor \frac{10030}{28} \right\rfloor=358\) - \(LCM(4,11)=44\), \(A_{4,11}=\left\lfloor \frac{N}{44} \right\rfloor=\left\lfloor \frac{10030}{44} \right\rfloor=228\) - \(LCM(7,11)=77\), \(A_{7,11}=\left\lfloor \frac{N}{77} \right\rfloor=\left\lfloor \frac{10030}{77} \right\rfloor=130\) 4. **Count numbers divisible by intersections of triples:** - \(LCM(3,4,7) = LCM(12,7) = 84\), \(A_{3,4,7} = \left\lfloor \frac{N}{84} \right\rfloor=\left\lfloor \frac{10030}{84} \right\rfloor=119\) - \(LCM(3,4,11) = LCM(12,11) = 132\), \(A_{3,4,11} = \left\lfloor \frac{N}{132} \right\rfloor=\left\lfloor \frac{10030}{132} \right\rfloor=76\) - \(LCM(3,7,11) = LCM(21,11) = 231\), \(A_{3,7,11} = \left\lfloor \frac{N}{231} \right\rfloor=\left\lfloor \frac{10030}{231} \right\rfloor=43\) - \(LCM(4,7,11) = LCM(28,11) = 308\), \(A_{4,7,11} = \left\lfloor \frac{N}{308} \right\rfloor=\left\lfloor \frac{10030}{308} \right\rfloor=32\) 5. **Count numbers divisible by intersection of all four:** - \(LCM(3,4,7,11) = LCM(84,11) = 924\), \(A_{3,4,7,11} = \left\lfloor \frac{N}{924} \right\rfloor = \left\lfloor \frac{10030}{924} \right\rfloor=10\) 6. **Apply inclusion-exclusion principle:** $$ \text{Count divisible} = (A_3 + A_4 + A_7 + A_{11}) - (A_{3,4} + A_{3,7} + A_{3,11} + A_{4,7} + A_{4,11} + A_{7,11}) + (A_{3,4,7} + A_{3,4,11} + A_{3,7,11} + A_{4,7,11}) - A_{3,4,7,11} $$ Plug numbers in: $$ = (3343 + 2507 + 1432 + 911) - (835 + 477 + 304 + 358 + 228 + 130) + (119 + 76 + 43 + 32) - 10 $$ Calculate each sum: $$ = 8193 - 2332 + 270 - 10 = 8193 - 2332 + 260 = 8121 $$ 7. **Final answer:** Total positive integers \(= 10030\). Non-divisible by any of the numbers are: $$ 10030 - 8121 = 1909 $$ Thus, \(\boxed{1909}\) positive integers not exceeding 10030 are not divisible by 3, 4, 7, or 11.