Skill Subsist Impulse Ltd Interview Question for Analysts


Country: India
Interview Type: In-Person




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There are infinitely many prime numbers, so there is no 'largest prime'

- D PRAVEEN KUMAR September 04, 2017 | Flag Reply
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Prime numbers are unlimited.

- adiya.tb@gmail.com September 04, 2017 | Flag Reply
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Prime numbers are unlimited.

- Adiya September 04, 2017 | Flag Reply
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Prime numbers are unlimited.

- adiya.tb September 04, 2017 | Flag Reply
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in decadic system. digits at the end are not. 2,4,6,8,0 because these numbers can be divided by 2 and 2 is divider of 10. (10 represents decadid system), not 5 because 5 if also divider of 10.

numbers 1,3,7,9 and no matter that 1 is not a prime number in decadic system are able to generate each number in a last position using following system a * b mod(10) = c
where a = {1,2,3...9, 10} b={1,3,7,9} generates c= {1,2,3,4,5,6,7,8,9} with occurence 1.

example:

1 * 3 = 3;
2 * 3 = 6;
3 * 3 = 9;
4 * 3 = 12 mod 10 = 2;
5 * 3 = 15 mod 10 = 5;
6 * 3 = 18 mod 10 = 8;
7 * 3 = 21 mod 10 = 1;
8 * 3 = 24 mod 10 = 4;
9 * 3 = 27 mod 10 = 7;
10 * 3 = 30 mod 10 = 0;

So 1,3,7,9 with probability 25%.

... and if I am wrong then sorry ;)

- Mike September 04, 2017 | Flag Reply
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0
of 0 vote

in decadic system. digits at the end are not. 2,4,6,8,0 because these numbers can be divided by 2 and 2 is divider of 10. (10 represents decadid system), not 5 because 5 if also divider of 10.

numbers 1,3,7,9 and no matter that 1 is not a prime number in decadic system are able to generate each number in a last position using following system a * b mod(10) = c
where a = {1,2,3...9, 10} b={1,3,7,9} generates c= {1,2,3,4,5,6,7,8,9} with occurence 1.

example:

1 * 3 = 3;
2 * 3 = 6;
3 * 3 = 9;
4 * 3 = 12 mod 10 = 2;
5 * 3 = 15 mod 10 = 5;
6 * 3 = 18 mod 10 = 8;
7 * 3 = 21 mod 10 = 1;
8 * 3 = 24 mod 10 = 4;
9 * 3 = 27 mod 10 = 7;
10 * 3 = 30 mod 10 = 0;

So 1,3,7,9 with probability 25%.

... and if I am wrong then sorry ;)

- Mike September 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

in decadic system. digits at the end are not. 2,4,6,8,0 because these numbers can be divided by 2 and 2 is divider of 10. (10 represents decadid system), not 5 because 5 if also divider of 10.

numbers 1,3,7,9 and no matter that 1 is not a prime number in decadic system are able to generate each number in a last position using following system a * b mod(10) = c
where a = {1,2,3...9, 10} b={1,3,7,9} generates c= {1,2,3,4,5,6,7,8,9} with occurence 1.

example:

1 * 3 = 3;
2 * 3 = 6;
3 * 3 = 9;
4 * 3 = 12 mod 10 = 2;
5 * 3 = 15 mod 10 = 5;
6 * 3 = 18 mod 10 = 8;
7 * 3 = 21 mod 10 = 1;
8 * 3 = 24 mod 10 = 4;
9 * 3 = 27 mod 10 = 7;
10 * 3 = 30 mod 10 = 0;

So 1,3,7,9 with probability 25%.

... and if I am wrong then sorry ;)

- Michal Havelec September 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

in decadic system. digits at the end are not. 2,4,6,8,0 because these numbers can be divided by 2 and 2 is divider of 10. (10 represents decadid system), not 5 because 5 if also divider of 10.

numbers 1,3,7,9 and no matter that 1 is not a prime number in decadic system are able to generate each number in a last position using following system a * b mod(10) = c
where a = {1,2,3...9, 10} b={1,3,7,9} generates c= {1,2,3,4,5,6,7,8,9} with occurence 1.

example:

1 * 3 = 3;
2 * 3 = 6;
3 * 3 = 9;
4 * 3 = 12 mod 10 = 2;
5 * 3 = 15 mod 10 = 5;
6 * 3 = 18 mod 10 = 8;
7 * 3 = 21 mod 10 = 1;
8 * 3 = 24 mod 10 = 4;
9 * 3 = 27 mod 10 = 7;
10 * 3 = 30 mod 10 = 0;

So 1,3,7,9 with probability 25%.

... and if I am wrong then sorry ;)

- Mike September 05, 2017 | Flag Reply


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