For example, {14, 21, 28, 35, 42, 49, 55} --> remainders {0, 0, 0, 0, 0, 0, 6} --> range 6 --> the sum of the remainders 6.

Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?

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 could be: 0, 1, 2, 3, 4, 5, or 6 (7 values).

(1) The range of the remainders is 6 --> if we pick 6 different multiples of 7 (all remainders 0) and the 7th number is 20 (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 integers are consecutive. 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 questions to practice:
seven-different-numbers-are-selected-from-the-integers-1-to-99943.html
n-consecutive-integers-are-selected-from-the-integers-131349.html

Hope it helps.
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My answer will be B. Here is the explanation.

Clue 1 - Range is the difference between the largest and the smallest numbers. Possible remainders when a number is divided by 7 is 0 thru 6. SO range is 6. This can be derived from the question itself. So Clue 1 is the same derivation made is explicit. With this we cannot say what numebers are choosen and what would be ther remainders and in turn the sum. S0 Clue 1 is INSUFFICIENT.

Clue 2 - It says that the numbers are consecutive. This clearly tells that one number is divisible by 7. So remainder is 0. And the rest of the 6 numbers should leave the remainders from 1 thru 6. From this we can get the sum of the remainders. So Clue 2 is sufficient.

Any OA?

first step:
xi = 7*ki+ri where i=1 to 7.
sum of ri=?
(1) max-min (r1,....,r6)=6
(2) xi is such that x(i+1) = xi+1
So,
x1 = 7k1+r1
x2 = x1+1=7k1+r1+1
x3 = x2+1 = 7k1+r1+2
x7 = 7k1+r1+6.
sum of remainders = 7r1+1+2+3+4+5+6 = 7r1+21
one remainder is zero. so r1 can be solved for. hence B.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Seven integers, x1, x2, x3, x4, x5, x6, and x7, are picked at random from the set of all integers between 10 and 110, inclusive. If each of these integers is divided by 7 and the 7 remainders are all added together, what would be the sum of the 7 remainders?

(1) The range of the remainders is 6.
(2) The seven integers are consecutive.

There are 7 variables, so we need 7 equations in order to solve the problem; only 2 equations are given, so there is high chance (E) will be the answer.
When we look at the conditions together, the remainders become 1,2,3,4,5,6,0 so the sum becomes 1+2+3+4+5+6=21, a unique answer. The condition is sufficient, so the answer seems to be (C), but this is a question with commonly made mistakes in 4(A) (Common Mistake Type 4 4(A)).
If we look at the conditions separately,
Condition 1 gives 1,2,3,4,5,6,7==> 1+2+3+4+5+6+0=21, but this also works for
7,14,21,35,42,48,49==>0+0+0+0+0+6+0=6, so this condition is not unique, and insufficient by itself.
Condition 2, on the other hand, the sum of the remainders becomes 1+2+3+4+5+6+0=21, so this condition is unique and therefore sufficient. So the answer becomes (B).

This type of question is a common type in today GMAT math

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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St 1 doesn't tell us anything that isn't already contained in St 2 because

11,12,13,14,15,16,16 will yield the same sum as

20,21,22,23,24,25,26


B

1. If x1 were a two-digit integer, then it would be explicitly stated: x1 is a tw-digit integer where x is a tens digit and 1 is an units digit.
2. On the real test subscripts would be clearly visible, not to create confusion: \(x_1\), \(x_2\), \(x_3\), \(x_4\), \(x_5\), \(x_6\), and \(x_7\), ... Edited above.
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