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In the correctly worked addition problem shown, where the [#permalink]

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19 Oct 2007, 05:19

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AB +BA ----------- AAC

In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?

In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?

A. 9 B. 6 C. 3 D. 2 E. 0

Ans.is E

Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0

In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?

In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?

A. 9 B. 6 C. 3 D. 2 E. 0

Ans.is E

Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0

And why is that? Could you elaborate more on this please? Thank you very much sir.

In the correctly worked addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?

A. 9 B. 6 C. 3 D. 2 E. 0

Ans.is E

Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0

And why is that? Could you elaborate more on this please? Thank you very much sir.

Cheers J

The sum of 2 two-digit integers cannot be greater than 200 (actually it cannot be greater than 99+99=198).

Thus from: AB +BA ___ AAC

It follows that A=1. So, we have that 1B +B1 ___ 11C

If B is less than 9, AB+BA will not be a three-digit integer (consider 18+81=99), thus B=9 --> 19+91=110 --> C=0.

Re: In the correctly worked addition problem shown, where the [#permalink]

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19 Apr 2015, 03:15

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In the correctly worked addition problem shown, where the [#permalink]

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24 Sep 2015, 06:54

10a + b +10b + a= 100a+10a+c 11(b-9a)=c what gives us this information? C cannot be a multiple of 11 it can take only values between 0 - 9 so c must equal 0 and a=1,b=9 I'm just looking for the quickiest way to solve such kind of problems. Experts do you think such representation is a reliable option ? Thanks in advance.

BTW it's no way a 600 Level question --> 600-700
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Re: In the correctly worked addition problem shown, where the [#permalink]

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27 Sep 2015, 02:08

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It is fairly simple if you dont get afraid and block the logic. As posted above the largest number that we are able to get adding two-digit numbers is 198 (99+99).

Looking at the clue, AAC. We have already solved half of the task. Because A=must be 1 in order to satisfy the condition above. So in this case we should find a number that has a structure of

11_

Because we have AB+ BC= AAC.

Applying the logic we have a number 1B+B1= 11C. So AB is a number smaller than 20 because the tens digit is 1. In order to get a number above or equal to 11C.

The number BA must be greater than 90. Because we have already A=1. The only solution left is B=9. Final step:

19+91= 110 .... The solution E

Hope I helped and did not cause additional confusion.

In the correctly worked addition problem shown, where the [#permalink]

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21 Jan 2016, 03:15

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I solved this problem in the following way:

When two digits are interchanged and added, they are always a multiple of 11. So AB + BA = Multiple of 11. The first 3 digit multiple of 11 is 110. With this, you can use the clue given in the question that the first 2 digits are same so I don't need to consider 121, 132, 144 & so on. Now if you want to double check your answer then input numbers which can equal to 110, starting small 91+19 =110.

Also, Subtraction of two interchanged digits is always a multiple of 9. For example 54 - 45 = 9.

In the correctly worked out addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits. what is the units digit of the integer AAC?

AB + BA CCA

A) 9

B) 6

C) 3

D) 2

E) 0

I got it that:

10A + B + 10B + A = AAC

11A + 11B = 110A + C

-99A + 11B = C

11 * (-9A + B) = C

But I didn't know what else to do.

I looked up the solution, it says:

C is divisible by 11, and 0 is the only digit that is divisbe by 11.

How comes that 0 is divisible by 11? Shouldn't 0 divided by 11 be equal to 0??

In the correctly worked out addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits. what is the units digit of the integer AAC?

AB + BA CCA

A) 9

B) 6

C) 3

D) 2

E) 0

I got it that:

10A + B + 10B + A = AAC

11A + 11B = 110A + C

-99A + 11B = C

11 * (-9A + B) = C

But I didn't know what else to do.

I looked up the solution, it says:

C is divisible by 11, and 0 is the only digit that is divisbe by 11.

How comes that 0 is divisible by 11? Shouldn't 0 divided by 11 be equal to 0?? I hope someone knows how to explain this.

Thanks for any help!

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

In the correctly worked out addition problem shown, where the sum of the two-digit positive integers AB and BA is the three-digit integer AAC, and A, B, and C are different digits. what is the units digit of the integer AAC?

AB + BA CCA

A) 9

B) 6

C) 3

D) 2

E) 0

I got it that:

10A + B + 10B + A = AAC

11A + 11B = 110A + C

-99A + 11B = C

11 * (-9A + B) = C

But I didn't know what else to do.

I looked up the solution, it says:

C is divisible by 11, and 0 is the only digit that is divisbe by 11.

How comes that 0 is divisible by 11? Shouldn't 0 divided by 11 be equal to 0??

Problem Solving question that i was not able to understand [#permalink]

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10 Aug 2016, 15:50

Hi everyone,

My name is Marcos Ramalho and i am a Portuguese Industrial Engineer.

Yesterday after dinner and during my study time, i was faced with the following PS question that i was not able to understand besides book's proposed resolution.

Whats the logic behind the problem as a hole? Whats the logic behind the proposed equations? Is there any other way to achieve the solution?

My name is Marcos Ramalho and i am a Portuguese Industrial Engineer.

Yesterday after dinner and during my study time, i was faced with the following PS question that i was not able to understand besides book's proposed resolution.

Whats the logic behind the problem as a hole? Whats the logic behind the proposed equations? Is there any other way to achieve the solution?