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It is an integer which, when divided by 2, leaves a remainder of 1.
Matlab:for i=50:70if mod(i,3) 3fprintf('%d\n',i);endend58.
The answer is 301
23457 divided by 23456 leaves a remainder of 1, and it is divisible by 7. So that is one of infinitely many possible answers.
12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.