If m, n, and p are three-digit integers and m + n = p

Abib, both \(K\) and \((B + E)\) are integers and according to the question, \(K\) can take a maximum value of \(9\) as \(K\) is just a single digit.

So, the maximum value of \((B + E)\) will be \(8\).

Also, if \(C + F < 10\), then only \(B + E \ge K\) will be true. But we already know that \(B + E < K\).

So, \(C + F \ge 10\).

Since, \(B + E \le 8\) and \(C + F \ge 10\),

\(C+F \ge B+E+2\) will always be true.

Hope that helps.

Thanks, eshan429.

It helped me to figure out my mistake.

Posted from my mobile device

Hi Mike,

I have one doubt in your explanation. The question asks whether the sum of the units digits of m and n at least 2 more than the sum of the tens digits of m and n.Letter K doesn't represent sum of B and E. Let us take 137 and 129 as two numbers. sum of 137 and 129 is 266, which satisfies both the conditions.
sum of units digits=7+9=16 doesn't mean units digit of 266 i.e. 6
sum of tens digits=3+2=5 doesn't mean tens digit of 266 i.e. 6

we are asked about sum of the digits and can't place condition on tens or units digit of p.
Can you please help me understand your solution.

Took almost 6 minutes to arrive at A
I solved it exactly the way you showed in your solution but i could not solve it under 5 minutes leave alone 3 minutes.
I took examples rather then making the inequality.
Statement 2 is a bit easier.
How should i approach these kinda questions ?

This has to be a 780 level question.
Your thoughts?
mikemcgarry

hi mikemcgarry
Took almost 6 minutes to arrive at A
I solved it exactly the way you showed in your solution but i could not solve it under 5 minutes leave alone 3 minutes.
I took examples rather then making the inequality.
Statement 2 is a bit easier.
How should i approach these kinda questions ?

This has to be a 780 level question.
Your thoughts?

Dear stonecold,

I'm happy to respond.

First of all, this is a hard question, a really brilliantly designed question. Of course, there is no "trick" for solving hard questions, in part because each hard question is hard in its own way. You have to have excellent number sense, which involves investigating and playing with all kinds of patterns of numbers on your own. You will find a link to an fun app on that number sense blog. I also will suggest this blog article:
How to do GMAT Math Faster
One way to build intuition for problem solving is to pursue a ton of difficult practice questions. For that, I would recommend:
Nova’s GMAT Math Prep Course Book

Finally, while I agree that this is a challenging question, I am extremely skeptical about the process of assigning numerical score-values to individual questions. See:
Is this a 700+ level GMAT question?

I hope all this helps.
Mike

I'll try to outline my thought process on this one when I just tried it for the first time. The solution above is great, so I won't rewrite it - but here are the actual steps I took in order to get this one in 2 minutes.

1. Thought: this is asking whether the sum of the units digits is at least two more than the sum of the tens digits. That means that a case like 129 + 348 would give us a 'yes', since 9+8 is much larger than 2+4. A case like '111+111' would give us a 'no', since 1+1 and 1+1 are the same.

2. Go to statement 1. If the tens digit of P is greater than the sum of the tens digits of M and N, that means that something must have been 'carried over' from the units place when we did the addition. If nothing was carried over, that'd mean the tens digit of P would be equal to the sum of the tens digits of M and N.

3. What does it tell us, if something was carried over? That just means that the sum of the units digits was 10 or higher.

4. Okay, it seems pretty simple to find a case where the sum of the units digits is much higher than the sum of the tens digits, then. We'll just make the units digits as large as possible, and the tens digits as small as possible. A case like that would give us a 'yes', so now we need a 'no'. The trick is going to be finding a case where the sum of the units digits isn't at least two higher than the sum of the tens digits.

5. Can we come up with a case where the sum of the units digits is just one higher than the sum of the tens digits? That would give us a 'no'. Suppose that the sum of the units digits is 10, and the sum of the tens digits is 9. In that case, we'd carry over a 1, add it to 9, and end up with P having a tens digit of... 0. Uh-oh. It's no longer greater than the sum of M and N's tens digits, so we just disobeyed the statement. That's an invalid case.

What if the sum of the units digits is 17, and the sum of the tens digits is 16? Wait, that wouldn't work either. If the sum of the tens digits is that large, then the actual tens digit of P wouldn't be greater than the sum of the tens digits of M and N. For instance, if the sum of the tens digits of M and N was 16, like in this case, the tens digit of P would be 6 or 7, which is much smaller than 16. Okay, so the case where the sums are 9 and 10 didn't work, and the case where the sums are bigger than that doesn't work either. There's no good case at all, so this statement seems sufficient.

At this point, I wrote 'AD/BCE' on my paper and crossed off B, C, and E.

6. On to statement 2. This looks simpler, so I'm going to try cases. It seems like what we really care about is whether the sum of the units digits is greater than 10 - that affects whether a 1 gets 'carried over'. So, I'll try a really simple case first: 111 + 111 = 222. That case gives us an answer of 'no'.

Note: At this point, I could actually stop. That's because the statements, in a real GMAT problem, will never directly contradict each other. It's actually impossible to have one sufficient statement that gives you a 'yes', and another sufficient statement that gives you a 'no'. So, the second statement must be insufficient! But, I'm going to keep going just to illustrate the rest of the problem.

7. Now, I want a case where the sum of the units digits is big, and the sum of the tens digits is small. Suppose the units digits sum to 16, like 8+8. Then the units digit of P would be 6. Can we make the tens digit of P equal to 6, also, to fulfill the statement? Yeah, by making the tens digits of M and N equal to 3 and 2, so that adding the 1 that gets carried over gives us 6.

At this point, I jotted this down on my paper, to double check:

_ 3 8
_ 2 8
_ 6 6

8+8 is at least two greater than 3+2, so we have our 'yes' case, and the statement is insufficient. Cross off D and pick A!

HI,

although Q is asked from someone else and may be someone would have already unanswered the way I am going to go about, I ll still put in my bit..

all are three digit numbers so p < 1000...
let the number be m = xyz and n = abc, also p = rst...
Q asks us if \(z+c\geq{y+b+2}\)

lets see the I statement-

(1) The tens digit of p is greater than the sum of the tens digits of m and n.
MAIN point -- possible if and only if s<10
so \(s\geq{y+b+1}\)..

and TENS of p can be greater ONLY if the UNITS of m and n add up to 10 or more than 10, lets take the MINIMUM value of 10, so z+c=10....(i)

Also by statement \(s\g{y+b}\), lets take the max value of y+b possible that is s=y+b+1....
Also lets take the max possible of s, which is<10, so 9...
thus we get 9=y+b+1 or y+b=8....(ii)

subtract (ii) from (i)...
(z+c)-(y+b)=10-8.......z+c=2+y+b..
Thus z+c or the sum of UNITS digit of m and n is ATLEAST 2 greater than y+b or sum of TENS digit of m and n..

If m, n, and p are three-digit integers and m + n = p, is the sum of the units digits of m and n at least 2 more than the sum of the tens digits of m and n?
 
(1) The tens digit of p is greater than the sum of the tens digits of m and n.
 
(2) The tens and units digits of p are equal.

Here is my take:

Option A ) Lets try to find out scenarios when this can be possible. When can a single digit number be greater than the sum of its digit when you add them- ONLY in those situations where you can borrow a left over "1" from the unit digit numbers being added. So what is the maximum sum of numbers we can get by adding the tens digit is 8, as if you chose 9 , then the condition won't hold true as 9+1=10, the digit in tens place becomes 0 and that is an illegal case, as 0 can't be bigger than any number. So lets come back to the case where the digits total up to 8. Now we have a 3 digit number P that has a tens digit of 8. Now in what cases can you get a carryover of 1 from addition of unit digits. You can get the carry over in all the below mentioned cases:

1. 5+5 +10
2. 5+6= 11
3. 5+7=12......

and so on. You will find innumerable ways to find this. But you really don't have to worry about those giving a sum of more than 10, as they are already at least 2 units greater than tens place of N. Concentrate on case, where the sum adds upto 10. BINGO, If this case can give us the minimum difference possible, then there is no need to look further. Thus the difference between the sum of units and tens digit can either be equal to 2 or greater than that. HENCE SUFFICIENT

Option B) Look at these two cases- this one is easy.

149
+139
2 8 8

Sum of units digit = 18
Sum of tens digit= 7
Yes, the difference is greater than 2

144
+122
266

Sum of unit digits : 6
Sum of tens digit : 6
No, the difference is not greater than 2

Hence Insufficient

Your answer is A

OFFICIAL EXPLANATION FROM MANHATTAN

It may help to write the sum in a traditional (columnar) format.  Let m have digits ABC, n have digits DEF, and p have digits GHJ:


 
The question asks whether C + F is at least 2 more than B +E, or is C + F > 2 + B +E?
 
(1) SUFFICIENT: Statement 1 says that H > B + E. (You can see the advantage of writing letters for the digits!) If that is true, then two other things must be true:
 
1. The sum of C and F must “carry over” to the tens place. Without such a carryover, H = B + E if B + E < 9, and H < B + E if B + E > 10. (If you’re not sure, test some real numbers to see how this works.)
 2. The sum of B and E (even with the “carried-over” unit) must NOT “carry over” to the hundreds place. If such a carryover occurs, then H = B + E if at least one of B and E is 9, and H < B + E otherwise. (Again, test some numbers if you’re not sure.)
 
The first fact guarantees that C + F is at least 10; the second guarantees that B + E is at most 8. The difference must be at least 2, so the statement is sufficient.
 
(2) NOT SUFFICIENT: Statement 2 says that H = J; in other words, the last two digits of p (the sum) are the same.
 
With this statement, the answer to the question can be either yes or no. For instance, in the sum 128 + 127 = 255, the sum of the units digits is 11 more than the sum of the tens digits. In the sum 172 + 183 = 355, on the other hand, the sum of the units digits is 10 less than the sum of the tens digits. The statement is not sufficient to answer the question.
 
The correct answer is (A).

OFFICIAL EXPLANATION:

It may help to write the sum in a traditional (columnar) format. Let m have digits ABC, n have digits DEF, and p have digits GHJ:
ABC
DEF
----------
GHJ

The question asks whether C + F is at least 2 more than B +E, or is C + F ≥ 2 + B +E?
(1) SUFFICIENT: Statement 1 says that H > B + E. (You can see the advantage of writing letters for the digits!) If that is true, then two other things must be true:
1. The sum of C and F must “carry over” to the tens place. Without such a carryover, H = B + E if B + E ≤ 9, and H < B + E if B + E ≥ 10. (If you’re not sure, test some real numbers to see how this works.)

2. The sum of B and E (even with the “carried-over” unit) must NOT “carry over” to the hundreds place. If such a carryover occurs, then H = B + E if at least one of B and E is 9, and H < B + E otherwise. (Again, test some numbers if you’re not sure.)
The first fact guarantees that C + F is at least 10; the second guarantees that B + E is at most 8. The difference must be at least 2, so the statement is sufficient.

(2) NOT SUFFICIENT: Statement 2 says that H = J; in other words, the last two digits of p (the sum) are the same.

With this statement, the answer to the question can be either yes or no. For instance, in the sum 128 + 127 = 255, the sum of the units digits is 11 more than the sum of the tens digits. In the sum 172 + 183 = 355, on the other hand, the sum of the units digits is 10 less than the sum of the tens digits. The statement is not sufficient to answer the question.
The correct answer is (A).

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