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Churej
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Churej

Asked: If m is the tens digit of the sum of the two-digit positive integers kr and ms, where k,m, r and s are digits of the integers, which of the following must be true?

10k + r + 10m + s = 100 + 10m + x
Case 1: r + s = x ; k = 10; Not feasible
Case 2: r + s = 10 + x > 9; k =9; Feasible

I. k = 9
II. m < 9
III. r + s >9

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

IMO D
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HI Kinshook , I understood the case 1 but I dint understand why wrote like that in case 2 and concluded from there that the answer is D  ( the highlighted part ) . Kindly explain. Please.
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Churej

Asked: If m is the tens digit of the sum of the two-digit positive integers kr and ms, where k,m, r and s are digits of the integers, which of the following must be true?

10k + r + 10m + s = 100 + 10m + x
Case 1: r + s = x ; k = 10; Not feasible
Case 2: r + s = 10 + x > 9; k =9; Feasible

I. k = 9
II. m < 9
III. r + s >9

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

IMO D
­
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gmatophobia , I  dint understand the case 2. WIll you please explain in simpler terms ? Thank you !!   :)
gmatophobia

Churej
If m is the tens digit of the sum of the two-digit positive integers kr and ms, where k,m, r and s are digits of the integers, which of the following must be true?

I. k = 9
II. m < 9
III. r + s >9

A. II only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
\(\quad kr\)
+
\(\quad ms\)
--------------------
\(\quad xmy\)

Given: tens digit of the sum = \(m\)

As \(m\) is a single-digit integer, the unit digit of the other number must be zero for the unit digit of the sum of both digits to be equal to \(m\).

Case 1: \(k\) can be = 0 ⇒ This is not possible as \(kr\) is a two digit number.

Case 2: The sum of \(r\) and \(s\) resulted in a value greater than 9 and 1 was carried over to the tens place. Therefore the sum \(k + 1\) resulted in 10 10 ⇒ If this was the case, 10 + m will have a unit digit of m.

For this to happen \(k + 1 = 10, k = 9\) and we should have a carry from the sum of \(r\) and \(s\). Therefore \(r + s > 9\).


Option D
­
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­This question showcases the power of drawing accurate inferences to streamline your problem-solving process.
Watch this solution to see how making smart deductions from the given information can significantly reduce your solving time and boost accuracy.

­
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There are two number kr and ms. It is given that on summing these two number, the ten's place digit of the sum is m.

a) To simplify, take an example, 98 + 32, if I want to get 3 in ten's place of the sum, somehow I need to add 10 to 3 of 32. Which will result is 13 and hence 3 at tens place.
So, it is clear to get m at the tens place of the sum. we need to add 10.

b) Maximum carryover possible from sum of unit digit is (9+9 = 18) is 1.

Based on above two we can conclude that,

k has to be equal to 9
and r+s > 9, so we get a carry over of 1.
making (9+1) = 10.

Hence, Option D is correct.
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In this case, is it fair to assume that if m can equal 9, we do not care if the variables have the same values? I ruled option II out because if k = 9, I thought m must have a different value for k since they are separate variables. But in the GMAT, I am assuming that variables in this circumstance can carry the same value unless explicitly stated otherwise?
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Mr_domore
In this case, is it fair to assume that if m can equal 9, we do not care if the variables have the same values? I ruled option II out because if k = 9, I thought m must have a different value for k since they are separate variables. But in the GMAT, I am assuming that variables in this circumstance can carry the same value unless explicitly stated otherwise?
­Unless explicitly stated otherwise, different variables CAN represent the same number.
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Nothing to do but see how m remains the same as the addition happens, and then the rest is obvious:

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Churej
If m is the tens digit of the sum of the two-digit positive integers kr and ms, where k,m, r and s are digits of the integers, which of the following must be true?

I. k = 9
II. m < 9
III. r + s >9

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

This is an alphametic and you must write it as that:

....k r
+ m s
------------
...m ...


The only hint we have is that when k and m are added, we get a sum with units digit as m.
This is possible in 2 cases:
When k = 0. But kr is a two digit number so k cannot be 0
When k = 9 and there is a carry over from r+s. Hence this is the only possibility then.

So statements I and III must be true.
It is necessary that m < 9? No. m can be 9 too. We are not given that digits are distinct. Hence II is not necessarily true.

Answer (D)

Alphametics are discussed here:
https://anaprep.com/puzzles-corner-solving-alphametics/
https://anaprep.com/puzzles-corner-mult ... phametics/
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Suppose we're adding k r and m s
In order for m to be 10s digit of the result there should be a carryover of 1 from adding r&s and k should be 9 inorder to sum up to 10, so m becomes 10s digit
Hence r+s > 9 and k=9
But m need not be less than 9
Hence option D
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