The two-digit positive integer s is the sum of the two-digit positive

Pls Note: This is the solution from Manhattan Advanced Quant book which has been slightly modified by me for simpler and better understanding.

Try out a few randomly-chosen numbers to help you understand the question.
If m = 10 and n = 10, then s = 20 and the units digits of m and s are equal.
If m = 11 and n = 12, then s = 23 and the units digit of s is greater than the units digit of m.
If m = 19 and n = 11, then s = 30 and the units digit of s is less than the units digit of m.

Basically, if the units digits of m and n are zero, then
the units digit of s will be equal to the units digit of m (and of n), i.e. zero.
If the units digits are very small i.e. say one is 5 or less and other is 4 or less and vice versa then the units digit of s is or greater than both m and n.

On the other hand, if the units digits of m and n are large enough to cause you to “carry over” a 1 to
the tens digit, then the units digit of s will end up being smaller than the units digit of m (and of n).

So the important point to note is will there be an aforesaid "carry over"? lets analyze these two statements in this context:-

(1) SUFFICIENT: If the units digit of s is definitely less than the units digits of one of the smaller
numbers, then the “carry over” situation must apply, in which case the units digits of both m and n
must be larger than the units digit of s.

(2) SUFFICIENT: There are only two ways in which the tens digit will not equal the tens digits of the
two smaller numbers:
Case 1: The tens digits of the two smaller numbers result in a number that needs to carry over into the
hundreds digit. This is impossible for this problem because s is also a two-digit number.
Case 2: The units digits of the two smaller numbers result in a number that carries over into the tens
digit. This must be what is happening in this case. If so, then the units digit of the larger number, s,
must be smaller than the units digits of the two smaller numbers, m and n.

The correct answer is (D).

I had searched with "The two-digit positive integer s is the sum of the two-digit positive" before posting but in vain.

Maybe the search engine of gmatclub needs to be improved.

For which statement?

For (1) m + n = 70 + 10 = 80 = s, is not possible because (1) say that the units digit of s is less than the units digit of n, which does not hold for your example.

For (2) m + n = 70 + 10 = 80 = s, is not possible because (2) say that the tens digit of s is not equal to the sum of the tens digits of m and n., which does not hold for your example.

1) The units digit of s is less than the units digit of n

Unit digit of s is less than n then it is surely case of carry over.
Then definitely unit digit of m is greater than that of s
Sufficient


2) The tens digit of s is not equal to the sum of the tens digits of m and n.

Means carry over case, then in such cases resulting unit digit is less than two digits.
Means unit digit of s is less than both m and n

Sufficient

D is answer

Give kudos if it helps

Posted from my mobile device

Hi,

from the text we get that

m+n=s, let m=10a+b, n=10a'+b', then

10(a+a')+(b+b')=s

Case 1: b+b'>=10 -> (a+a') <=8
Case2: b+b'<=9 -> (a+a') <=9

We are asked if

UD(s)<UD(m)
UD(s)<b?

(1) UD(b+b')<UD(n)=b
-> b+b'>=10 -> (b,b')=(1,9),(2,8),(2,9),... which all fulfill the required

(2) This is another way of saying b+b'>=10, and following the argument in (1), we get the result again

Hence (D), no guarantees

For statement 1, what about m=49, n=39? It doesn't say that the unit digits should be unique. In that case, s = 49+39 = 88, this will be another scenario. Why are we not taking it into consideration, am I missing something?

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