Cody

# Problem 365. Numbers with prime factors 2, 3 and 5.

Solution 1865262

Submitted on 4 Jul 2019 by Nikolaos Nikolaou
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### Test Suite

Test Status Code Input and Output
1   Pass
n = 2; A_correct = 2; assert(isequal(your_fcn_name(n),A_correct))

t = 2

2   Pass
n = 3; A_correct = [2 3]; assert(isequal(your_fcn_name(n),A_correct))

t = 2 t = 3

3   Pass
n = 4; A_correct = [2 3 4]; assert(isequal(your_fcn_name(n),A_correct))

t = 2 t = 3 t = 2 2

4   Pass
n = 5; A_correct = [2 3 4 5]; assert(isequal(your_fcn_name(n),A_correct))

t = 2 t = 3 t = 2 2 t = 5

5   Pass
n = 10; A_correct = [2 3 4 5 6 8 9 10]; assert(isequal(your_fcn_name(n),A_correct))

t = 2 t = 3 t = 2 2 t = 5 t = 2 3 t = 2 2 2 t = 3 3 t = 2 5

6   Pass
n = 17; A_correct = [2 3 4 5 6 8 9 10 12 15 16]; assert(isequal(your_fcn_name(n),A_correct))

t = 2 t = 3 t = 2 2 t = 5 t = 2 3 t = 2 2 2 t = 3 3 t = 2 5 t = 2 2 3 t = 2 7 t = 3 5 t = 2 2 2 2

7   Pass
n = 31; A_correct = [2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30]; assert(isequal(your_fcn_name(n),A_correct))

t = 2 t = 3 t = 2 2 t = 5 t = 2 3 t = 2 2 2 t = 3 3 t = 2 5 t = 2 2 3 t = 2 7 t = 3 5 t = 2 2 2 2 t = 2 3 3 t = 2 2 5 t = 3 7 t = 2 11 t = 2 2 2 3 t = 5 5 t = 2 13 t = 3 3 3 t = 2 2 7 t = 2 3 5