Answer:
[tex]N = 6[/tex] so that number becomes divisible by 3, 6 and 9.
Step-by-step explanation:
In Number Theory there is a rule of thumb which states that sum of digits of a multiple of 3 equal 3 or a multiple of three. If we know that [tex]n = 154N38[/tex], then its sum of digits is:
[tex]x = 1 + 5+4+N+3+8[/tex]
[tex]x = 21+N[/tex] (Eq. 1)
We have to determine which digits corresponds to multiples of three, there are four digits:
N = 0
[tex]x = 21+0[/tex]
[tex]x = 21[/tex] ([tex]3\times 7 = 21[/tex])
N = 3
[tex]x = 21+3[/tex]
[tex]x = 24[/tex] ([tex]3\times 8 = 24[/tex])
N = 6
[tex]x = 21+6[/tex]
[tex]x = 27[/tex] ([tex]3\times 9 = 27[/tex])
N = 9
[tex]x = 21+9[/tex]
[tex]x = 30[/tex] ([tex]3\times 10 = 30[/tex])
We get the following four distinct options: 154038, 154338, 154638, 154938. Now we find which number is divisible by 6 and 9 by factor decomposition:
[tex]154038 = 2\times 3\times 25673[/tex]
[tex]154338 =2\times 3\times 29 \times 887[/tex]
[tex]154638 = 2\times 3\times 3\times 11\times 11\times 71[/tex]
[tex]154938 = 2\times 3\times 7\times 7\times 17\times 31[/tex]
It is quite evident that [tex]N = 6[/tex] so that number becomes divisible by 3, 6 and 9.
