remainder data sufficiency question

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.

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.