three and four digit numbers

alex.gellatly wrote:The numbers x and y are three-digit positive integers, and x+y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x<y, which of the following must be true?

I. The units digit of x + y is greater than the units digit of either x or y.
II. The tens digit of x + y equals 2.
III. The hundreds digit of y is at least 5.

A. II only
B. III only
C. I and II
D. I and III
E. II and III

Thanks

x and y are both between 500 and 999. The problem is using the phrase 'must be true' so even one instance where a statement does not hold true is sufficient to rule it out.

(I) is ruled out because when the units place of both x and y is 0, the units place of (x + y) is also 0. For example: x= 570 and y =650
(II) is ruled for this example: x = 576 and y = 656 because x + y will have 3 has the tens digit in this case

No need to check (III) because none of the options says that none of the statements could be true.

[spoiler](B)[/spoiler] is correct.

aneesh.kg wrote: x and y are both between 500 and 999.

Hi!

Is there a reason why we couldn't have x=170 and y=850?

In order for x+y to be a 4 digit number, at least one of the two must be greater than 500 (and the hundreds digits must sum to at least 9), but both don't need to be greater than 500.

Fortunately, that issue doesn't cause any problems with these particular statements and the rest of your reasoning is excellent!

Stuart Kovinsky wrote:

aneesh.kg wrote: x and y are both between 500 and 999.

Hi!

Is there a reason why we couldn't have x=170 and y=850?

In order for x+y to be a 4 digit number, at least one of the two must be greater than 500 (and the hundreds digits must sum to at least 9), but both don't need to be greater than 500.

Fortunately, that issue doesn't cause any problems with these particular statements and the rest of your reasoning is excellent!

Ah no, My bad! I was fortunate indeed.
Thanks Stuart for pointing it out.