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# In the following correctly worked addition sum, A,B,C and D

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Director
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In the following correctly worked addition sum, A,B,C and D [#permalink]  11 Mar 2018, 08:33
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66% (01:25) correct 33% (01:18) wrong based on 6 sessions
In the following correctly worked addition sum, A,B,C and D represent different digits, and all the digits in the sum are different. What is the sum of A,B,C and D?

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A) 23
B) 22
C) 18
D) 16
E) 14
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Manager
Joined: 15 Jan 2018
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Re: In the following correctly worked addition sum, A,B,C and D [#permalink]  12 Mar 2018, 23:02
1
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You can solve this algebraically by making a lot of equations, i.e. 50 + A + 10B + C = 100D + 43, etc. But I'd avoid that mess. When you see problems like this it's usually easier to use logic or just try digits until you figure out the logic.

D looks like the easiest. It has to be 1, since there's no way you can add a 5 and a single digit to make 20 or above, which would get you a 2 or above in the hundreds place. So D is 1.

Since we need a 1 for D and a 4 in the tens place, we need to add 5 and B to get 14. Thus, B is 9.

Great. So what are A and C? They need to add to 3 so they must be 1 and 2 or 0 and 3. However, the problem says that all digits are different and we've already used a 1 and a 3. So both options are out. What's going on? Another way of making that 3 would be to make A and C add to 13. However, then a 1 would carry over so what we concluded in the last paragraph would need to be amended a bit. No big deal. So the carried over 1 and 5 and B need to add to get 14. So B is actually 8.

Double-checking to ensure that A and C add to 13 but don't duplicate anything: 9 and 4 don't work because of the 4, 8 and 5 don't work because of the 5, but 7 and 6 would work because they add to 13 and haven't duplicated anything.

So A and C make 13, B is 8, and D is 1, adding to a total of 22. So B is the answer.
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Last edited by SherpaPrep on 03 Apr 2018, 10:01, edited 1 time in total.
Manager
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Re: In the following correctly worked addition sum, A,B,C and D [#permalink]  13 Mar 2018, 23:01
I don't understand why D is 1.

SherpaPrep wrote:
You can solve this algebraically by making a lot of equations, i.e. 50 + A + 10B + C = 100D + 43, etc. But I'd avoid that mess. When you see problems like this it's usually easier to use logic or just try digits until you figure out the logic.

D looks like the easiest. It has to be 1, since there's no way you can add a 5 and a single digit to make 20 or above, which get you a 2 or above in the hundreds place. So D is 1.

Since we need a 1 for D and a 4 in the tens place, we need to add 5 and B to get 14. Thus, B is 9.

Great. So what are A and C? They need to add to 3 so they must be 1 and 2 or 0 and 3. However, the problem says that all digits are different and we've already used a 1 and a 3. So both options are out. What's going on? Another way of making that 3 would be to make A and C add to 13. However, then a 1 would carry over so what we concluded in the last paragraph would need to be amended a bit. No big deal. So the carried over 1 and 5 and B need to add to get 14. So B is actually 8.

Double-checking to ensure that A and C add to 13 but don't duplicate anything: 9 and 4 don't work because of the 4, 8 and 5 don't work because of the 5, but 7 and 6 would work because they add to 13 and haven't duplicated anything.

So A and C make 13, B is 8, and D is 1, adding to a total of 22. So B is the answer.
Director
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Re: In the following correctly worked addition sum, A,B,C and D [#permalink]  15 Mar 2018, 08:57
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gremather wrote:
I don't understand why D is 1.

SherpaPrep wrote:
You can solve this algebraically by making a lot of equations, i.e. 50 + A + 10B + C = 100D + 43, etc. But I'd avoid that mess. When you see problems like this it's usually easier to use logic or just try digits until you figure out the logic.

D looks like the easiest. It has to be 1, since there's no way you can add a 5 and a single digit to make 20 or above, which get you a 2 or above in the hundreds place. So D is 1.

Since we need a 1 for D and a 4 in the tens place, we need to add 5 and B to get 14. Thus, B is 9.

Great. So what are A and C? They need to add to 3 so they must be 1 and 2 or 0 and 3. However, the problem says that all digits are different and we've already used a 1 and a 3. So both options are out. What's going on? Another way of making that 3 would be to make A and C add to 13. However, then a 1 would carry over so what we concluded in the last paragraph would need to be amended a bit. No big deal. So the carried over 1 and 5 and B need to add to get 14. So B is actually 8.

Double-checking to ensure that A and C add to 13 but don't duplicate anything: 9 and 4 don't work because of the 4, 8 and 5 don't work because of the 5, but 7 and 6 would work because they add to 13 and haven't duplicated anything.

So A and C make 13, B is 8, and D is 1, adding to a total of 22. So B is the answer.

In this problem we are adding a pair of two digit numbers and our answer is a three digit number so we can safely assume that $$5+B$$ is 10 or more.

At this point D can be anything however we must realize that B can acquire a maximum value of $$9$$ and $$5+9 = 14$$. Also we must take into account that we may have to add any carry overs from the summation of A and C. When two single digit are added they can never exceed $$18$$ as such Our previous sum of $$5$$ and $$9$$ can have only $$1$$ as carry over so $$5+B$$ can at maximum become $$6+B$$ and If summation of two single digit number cannot exceed $$18$$ so D must be 1
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Re: In the following correctly worked addition sum, A,B,C and D   [#permalink] 15 Mar 2018, 08:57
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