for a prime number greater than 10, which are the possible digits in one's place
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2
For a prime number greater than 10 , the possible digits in 1's place:
=>Cannot be an even number or 0 (0,2,4,6,8)
=>Cannot be 5.
The possible numbers are
1 (eg.11)
3 (eg.13)
7 (eg.17)
9 (eg.19)
=>Cannot be an even number or 0 (0,2,4,6,8)
=>Cannot be 5.
The possible numbers are
1 (eg.11)
3 (eg.13)
7 (eg.17)
9 (eg.19)
Answered by
0
it can be
1,3,5,7,9 at one's place but not(always.yeah but mostly)
...so why not (2,4,6,8,0) ? because they are even or most apparently they are divided by 2 which is breaking prime number's law.......and also 5 can not be there because then it would be divisible by 5
1,3,5,7,9 at one's place but not(always.yeah but mostly)
...so why not (2,4,6,8,0) ? because they are even or most apparently they are divided by 2 which is breaking prime number's law.......and also 5 can not be there because then it would be divisible by 5
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