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Integers r, s and t

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Integers r, s and t

by yass20015 » Fri Aug 21, 2015 12:34 am
The integers r, s ans t all have the same remainder when divided by 5. What is the value of t ?

1) r+s=t
2) 20 <= t <= 24

any method to solve this one ? Thanks
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by GMATGuruNY » Fri Aug 21, 2015 4:22 am
yass20015 wrote:The integers r, s and t all have the same remainder when divided by 5. What is the value of t ?

1) r+s=t
2) 20 ≤ t ≤ 24
Since r, s, and t all have a remainder of R when divided by 5, they must ALL be contained in ONE of the following lists:
Case 1: R=0
0, 5, 10, 15, 20, 25...
Case 2: R=1
1, 6, 11, 16, 21, 26...
Case 3: R=2
2, 7, 12, 17, 22, 27...
Case 4: R=3
3, 8, 13, 18, 23, 28...
Case 5: R=4
4, 9, 14, 19, 24, 29...

Statement 1: r+s = t
Only Case 1 is viable.
No combination of values from the remaining cases will satisfy the constraint that r+s = t.
In Case 1, it's possible that r=5. s=5, and t=10.
In Case 1, it's possible that r=5, s=10, and t=15.
Since t can take on different values, INSUFFICIENT.

Statement 2: 20 ≤ t ≤ 24
In Case 1, it's possible that t=20.
In Case 2, it's possible that t=21.
Since t can take on different values, INSUFFICIENT.

Statements combined:
Only one value in Case 1 satisfies the constraint that t is between 20 and 24, inclusive: t=20.
SUFFICIENT.

The correct answer is C.
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by Max@Math Revolution » Fri Aug 21, 2015 8:20 pm
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 equations ensures a solution.


The integers r, s ans t all have the same remainder when divided by 5. What is the value of t ?

1) r+s=t
2) 20 <= t <= 24

In the original condition, we have 3 variables (r,s,t) and 1 equation (that the remain is the same) thus we need 2 more equations to match the number of variables. Since we have 1 in each 1) and 2), C is likely the answer.

In actual calculation, r=s= multiple of 5, and t=20 which makes them unique and thus the conditions are sufficient. Therefore the answer is C



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by Matt@VeritasPrep » Sun Aug 23, 2015 12:20 pm
Let's say that the remainder is x. This means that

r = 5*something + x
s = 5*something else + x
t = 5*something else again + x

We can write this algebraically as

r = 5k + x
s = 5m + x
t = 5n + x

where k, m, and n are integers whose values we don't care about.

S1 tells us that

r + s = t, or
5k + x + 5m + x = 5n + x, or
5*(k + m) + 2x = 5n + x

In other words, 5*(something) + 2x = 5*(something else) = x, which means that 2x and x represent the same remainder! If x = 0, this works, but if x = 1, 2, 3, or 4, it doesn't. (2*1 = remainder 2, 2*2 = remainder 4, 2*3 = remainder 1, and 2*4 = remainder 3.)

So we know that r, s, and t all have remainder 0 when divided by 5; in other words, they're all multiples of 5. This is close, but no cigar: NOT SUFFICIENT.

S2:: By itself, obviously unhelpful.

Together, we know that t is a multiple of 5 between 20 and 24, inclusive, so t must be 20. Sufficient!
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