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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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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? Attachment:
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A) 23 B) 22 C) 18 D) 16 E) 14




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Re: In the following correctly worked addition sum, A,B,C and D [#permalink]
12 Mar 2018, 23:02
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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. Doublechecking 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.



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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.
Doublechecking 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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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.
Doublechecking 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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Re: In the following correctly worked addition sum, A,B,C and D
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15 Mar 2018, 08:57





