Five different integers are selected at random from the numbers 1 to 1

Question

Five different integers are selected at random from the numbers 1 to 100. Each of the selected numbers is then divided by integer n. What is the value of n ?

(1) The greatest possible remainder for any selected integer is 4.

(2) For some selected integer, k, the quotient when k is divided by n is 9, the quotient when k is divided by n + 2 is 6, and the difference of the remainders is 3.

Aks111 · Answer

We need to know what is the value of Integer n.

(1) The greatest possible remainder for any selected integer is 4.
=> If an Integer is divided by 5 the possible remainders are 0,1,2,3,4. Hence, we know n = 5
Statement 1 alone is sufficient.

(2) For some selected integer, k, the quotient when k is divided by n is 9, the quotient when k is divided by n + 2 is 6, and the difference of the remainders is 3.
=> Let remainder when k is divided by n = a and remainder when k is divided by 6 = b
k = 9n + a and k = 6(n+2)+b
=> 9n+a = 6n+12+b
=> 3n = 12 + (b-a)

Also, given difference of remainders = 3, but we don't know whether a>b
=> So, either b-a=3 or a-b=3 
=> 3n = 15 or 3n = 9
i.e., 'n' could be 3 or 5.

e.g., Statement 2 is valid in both cases:
if n=3 and k=30 (a=3;b =0) or if n=5 and k=45 (a=0; b=3)
Statement 2 alone is not sufficient.

Answer: A

Aks111 · Answer

I noticed where I went wrong. The difference of remainders is 3. If n=3 we cannot have remainder as 0 and 3.
The maximum difference of remainders when n=3 is 2.
Statement 2 alone is sufficient as we get unique value for n (n=5).

Answer: D

RAHUL208 · Answer

I didn't understand the explanation on A)
How is A Alone sufficient ? - 100 / 96 also give 4 as remainder.

Bunuel · Answer

Statement (1) says the  greatest possible  remainder is 4, not that one example gives a remainder of 4.

If n = 96, the greatest possible remainder would be 95, if one of the selected numbers were 95, not 4.

Since the greatest  possible  remainder is always n - 1, we get n - 1 = 4, so n = 5.

tcocarogallart · Answer

Bunuel, is there a better way to discard the option that ellicits n=3 (not possible if we have a remainder of 3) before doing the math as was done in the answer above?

Bunuel · Answer

For n = 3, the remainder when k is divided by n can be at most 2. Therefore, the difference between that remainder and the other remainder can be at most 2, not 3.

So n = 3 is impossible, and n = 5.

egmat · Answer

Hi tcocarogallart,

Yes, there is a cleaner shortcut, and the good news is you've already spotted the right principle:  a remainder when dividing by d must satisfy 0 ≤ r < d.  You just need to know  where  in Aks111's algebra to deploy it, so you never have to hunt for a k at all.

Where to insert the bound check

Look at Aks111's line that splits the cases:

3n =  12  + (b - a)  with  |b - a| =  3   - either b - a =  3  (giving n =  5 ) or a - b =  3  (giving n =  3 ).

That second branch is the one to kill on sight. Here's the move:

The branch demands a - b =  3 , so  a ≥ 3  (since b ≥  0 ). But on this branch n =  3 , and a is the remainder when dividing by n - so a is forced into  {0, 1, 2} . The two requirements  a ≥ 3  and  a ≤ 2  can't both hold. The branch dies right there.

No constructing k =  30 , no checking mod  5  =  0 , no extra arithmetic - one inequality comparison and you're done.

The general DS habit

Whenever a remainder problem produces a candidate divisor, immediately check:

"Does this candidate force the remainder to be ≥ the divisor itself?"

If yes, kill the branch. Two tiny rehearsals:

- "Remainder  3  when divided by  3 " - impossible (max remainder is  2 ).
- "Remainder  5  when divided by  3 " - impossible (max remainder is  2 ).

Same principle, easy to apply in two seconds.

One caveat

You can't skip the algebra  entirely  - you still need 3n =  12  + (b - a) to expose the two candidates and, crucially, which side of the difference is the bigger remainder. That sign is what makes the bound check possible in the first place. But once the candidates appear, you can dismiss n =  3  with the inequality alone.

So Statement (2) cleanly gives  n = 5 , matching Statement (1).

Answer: D

Five different integers are selected at random from the numbers 1 to 1

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GMAT Data Sufficiency (DS) Questions

Five different integers are selected at random from the numbers 1 to 1

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