Seven different numbers are selected from the integers 1 to

The trick here is to know that remainder is always non-negative integer less than divisor \(0\leq{r}<d\), so in our case \(0\leq{r}<7\).

So the remainder upon division of any integer by 7 can be: 0, 1, 2, 3, 4, 5, or 6 (7 values).

(1) The range of the seven remainders is 6 --> if we pick 6 different multiples of 7 (all remainders 0) and the 7th number 6 (remainder 6) then the range would be 6 and the sum also 6. But if we pick 7 consecutive integers then we'll have all possible remainders: 0, 1, 2, 3, 4, 5, and 6 and their sum will be 21. Not sufficient.

(2) The seven numbers selected are consecutive integers --> ANY 7 consecutive integers will give us all remainders possible: 0, 1, 2, 3, 4, 5, and 6. It does not matter what the starting integer will be: if it's say 11 then the remainder of 7 consecutive integers from 11 divided by 7 will be: 4, 5, 6, 0, 1, 2, and 3 and if starting number is say 14 then the remainder of 7 consecutive integers from 14 divided by 7 will be: 0, 1, 2, 3, 4, 5 and 6. So in any case sum=0+1+2+3+4+5+6=21. Sufficient.

Answer: B.

Similar question to practice: n-consecutive-integers-are-selected-from-the-integers-131349.html

Hope it helps.