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remainder data sufficiency question

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remainder data sufficiency question

by alsergi » Tue Sep 23, 2008 3:48 am
Please, could you help me with this question?


If n is the sum of three consecutive nonnegative integers p, s and t,
with p<s<t, and n is also the product of three consecutive nonnegative integers
x, y and z, with x<y<z

What is the remainder when n is divided by 5?

1. The remainder when p is divided by 5 is 1
2. The remainder when x is divided by 5 is 1


Thanks in advance
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by siddarth » Tue Sep 23, 2008 4:43 am
The Answer is D, the remainder will be 1 in both cases. Try using numbers like 6+7+8 or 6*7*8, 11+12+13 or 11*12*13. Since we know that p & x are the smallest.
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by mals24 » Tue Sep 23, 2008 4:50 am
@ siddarth

I don't think you can use 6,7,8 or 11,12,13 because the question says

p+q+r=n
xyz=n

6+7+8=21
But 6*7*8=336

I guess the only set of numbers that work is 1,2,3 for both pqr and xyz.

1+2+3=6
1*2*3=6

But the answer is still D
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by siddarth » Tue Sep 23, 2008 5:09 am
As usual i got excited, thanks so much for correcting. This will be my biggest drawback, got to really work on this .
mals24 wrote:@ siddarth

I don't think you can use 6,7,8 or 11,12,13 because the question says

p+q+r=n
xyz=n

6+7+8=21
But 6*7*8=336

I guess the only set of numbers that work is 1,2,3 for both pqr and xyz.

1+2+3=6
1*2*3=6

But the answer is still D
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by Ian Stewart » Tue Sep 23, 2008 5:55 am
mals24 wrote:
I guess the only set of numbers that work is 1,2,3 for both pqr and xyz.

1+2+3=6
1*2*3=6

But the answer is still D
The answer is D, but you've misunderstood the question. n is equal to the product of three consecutive integers (x, y and z), and is also equal to the sum of three consecutive integers (p, s and t), but there is no reason to think that x, y and z are the same numbers as p, s and t. It could be, for example, that n = 24, which is the product of three consecutive numbers (2*3*4 = 24) and the sum of three consecutive numbers (7+8+9 = 24).

Statement 1: We know that n = p + s + t, but we also know that p, s and t are increasing consecutive numbers. So s = p+1, and t = p+2, and n = 3p + 3. If the remainder is 1 when p is divided by 5, that means p is one greater than some multiple of 5:

p = 5k + 1

So n = 3p + 3 = 3(5k+1) + 3 = 15k + 6 = 5(3k + 1) + 1, and n is one greater than a multiple of 5. So the remainder will be 1; the statement is sufficient.

Statement 2: As above, x = 5k + 1 for some k, and y = 5k+2, z = 5k+3 because x, y and z are consecutive. Thus

n = (5k+1)(5k+2)(5k+3)

and if you were to multiply this out, every term will be a multiple of 5 except the last- 1*2*3 = 6. So the number is six larger than some multiple of 5, and again the remainder must be 1 when n is divided by 5.
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by mals24 » Tue Sep 23, 2008 6:05 am
@ Ian

I have a small doubt the statements say that p and x when divided by 5 have a remainder of 1. But when you divide 2 and 7 by 5 the remainder is 2 not 1. That's why i chose 1,2,3 because they seemed to satisfy all the conditions. Did I misread something???
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by Ian Stewart » Tue Sep 23, 2008 8:55 am
mals24 wrote:@ Ian

I have a small doubt the statements say that p and x when divided by 5 have a remainder of 1. But when you divide 2 and 7 by 5 the remainder is 2 not 1. That's why i chose 1,2,3 because they seemed to satisfy all the conditions. Did I misread something???
When I gave the example of 2*3*4 = 7+8+9 = 24, I wasn't using the statements, just so I could give an example with small numbers. The smallest positive numbers we can use besides 1, 2 and 3, if we want both statements to be true, are:

6*7*8 = 111+112+113 = 336

That is, 336 can be written as the product of three consecutive integers, and also as the sum of three consecutive integers, in each case beginning with an integer which gives a remainder of 1 when divided by 5. You can also make larger examples using 11*12*13, or 16*17*18, and so on.
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by mals24 » Tue Sep 23, 2008 9:07 am
Finally I got the logic
Thanks a ton Ian for making the concept clear :)
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by alsergi » Thu Sep 25, 2008 3:56 am
Yes, the answer is D, thank you all !! You totally Rock :mrgreen:
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