AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
(A) 1
(B) 3
(C) 7
(D) 9 answer
(E) Cannot be determined
digit issue
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Since it is said AB, C D are 2 digit numbers, the highest can be 198 without any restrictions (99+99).francoisph wrote:AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
(A) 1
(B) 3
(C) 7
(D) 9 answer
(E) Cannot be determined
They have given some restrictions like A,B,C,D should be distinct.
So on any count AAA will be 111.as 111<198
Try few options for B,C,D (hit & trial) to get 111 as ur total.
So finally u wil see B=3, C=9, D=8 and A=1
AAA=111.
So C is 9...Move on!!
- albatross86
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The maximum value of AB + CD is = 98 + 76 = 174
Thus A must be 1, as 111 is the only 3 digit number that could have all 3 digits equal
So, 1B + CD = 111
Now consider B = 1 => The number AB is 11, which makes the number CD = 100 which is impossible
Thus B must be atleast 2, yielding AB as 12 and CD as 99
The maximum value of B is 9, yielding AB as 19 and CD as 92
In all possible values of AB, the value of CD is in the nineties. Thus C must be equal to 9
Pick D
Thus A must be 1, as 111 is the only 3 digit number that could have all 3 digits equal
So, 1B + CD = 111
Now consider B = 1 => The number AB is 11, which makes the number CD = 100 which is impossible
Thus B must be atleast 2, yielding AB as 12 and CD as 99
The maximum value of B is 9, yielding AB as 19 and CD as 92
In all possible values of AB, the value of CD is in the nineties. Thus C must be equal to 9
Pick D
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The sum of two 2 digit integers cannot be greater than 198, as gmatmachoman also mentioned.francoisph wrote:hi
how do you know that AAA = 111 please?
AAA is a 3 digit number that has all 3 digits the same. Thus, it must be of the form 111, 222 and so on
As you can see, the only possible value is 111.
- kvcpk
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B +10A +D + 10c = 111Afrancoisph wrote:AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
(A) 1
(B) 3
(C) 7
(D) 9 answer
(E) Cannot be determined
B+D+10C = 101A
B+D = A or B+D = 10 +A [Last digits of the sum]
if B+D = A, then c=10A from above.. not possible..Since C is single digit +ve Integer.
So B+D = 10+A [Since of sum of two digits cannot be more than 18]
10+10C = 100A
C+1 = 10A
so A has to be 1 because C+1 cannot exceed 10
Therefore c= 9
Answer D
- cricketsteve5
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Hi All,
I have a doubt regarding this problem..
Why cant it be like 98 +13 so that C is 1?
1 is also one of the options right?
Hope am right am clear..
Thanks,
cricketsteve
I have a doubt regarding this problem..
Why cant it be like 98 +13 so that C is 1?
1 is also one of the options right?
Hope am right am clear..
Thanks,
cricketsteve
- albatross86
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Hi,cricketsteve5 wrote:Hi All,
I have a doubt regarding this problem..
Why cant it be like 98 +13 so that C is 1?
1 is also one of the options right?
Hope am right am clear..
Thanks,
cricketsteve
Remember that it the equation is AB + CD = AAA
In your scenario, (98+13) where C = 1, A here = 9. which means the equation would be 98 + 13 = 999 which is not true.
Thus, since the 3 digit number MUST be 111, A must be 1.
Hope that clears your doubt.
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
- albatross86
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Hi ngkontos,
Actually your example is perfectly correct but you have arranged the data incorrectly, it should be 17 + 94 = 111 where C = 9 (i.e. D, which is the correct and only answer)
If you arrange it as 94 + 17 = 111 you are suggesting that in the equation AB + CD = AAA, A is 9 in the LHS and 1 in the RHS which is, incorrect.
No matter what combinations you choose, A must be 1 and C will ALWAYS be = 9.
Hope that makes sense.
Actually your example is perfectly correct but you have arranged the data incorrectly, it should be 17 + 94 = 111 where C = 9 (i.e. D, which is the correct and only answer)
If you arrange it as 94 + 17 = 111 you are suggesting that in the equation AB + CD = AAA, A is 9 in the LHS and 1 in the RHS which is, incorrect.
No matter what combinations you choose, A must be 1 and C will ALWAYS be = 9.
Hope that makes sense.
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
- albatross86
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ngkontos,
You're missing the fact that A must be = 1. When you say 94 + 17 = 111, you are suggesting that A = 9, you should write it as 17 + 94 = 111 which means that A = 1 and C = 9 ... i.e the answer is (d)
The question is worded perfectly. Give it some thought
You're missing the fact that A must be = 1. When you say 94 + 17 = 111, you are suggesting that A = 9, you should write it as 17 + 94 = 111 which means that A = 1 and C = 9 ... i.e the answer is (d)
The question is worded perfectly. Give it some thought
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
Yeah I actually realised where the mistake was but couldnt be bothered to reply again
Anyway, I say this question is not worded well because if on gmat the numbers would have been placed one on top of the other so it would have been much clearer. This can be a bit confusing but this is just a formatting issue in this case.
Anyway, I say this question is not worded well because if on gmat the numbers would have been placed one on top of the other so it would have been much clearer. This can be a bit confusing but this is just a formatting issue in this case.
- albatross86
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Oh I see what you mean now Glad you solved it!
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
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