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AB + CD = AAA, where AB and CD are two-digit numbers
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07 Nov 2022, 00:39
07 Nov 2022, 00:39
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45% (02:19) correct
54% (01:18) wrong based on 24 sessions
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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
(E) Cannot be determined
(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined
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AB + CD = AAA, where AB and CD are two-digit numbers
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07 Nov 2022, 22:55
07 Nov 2022, 22:55
2
Thanks for the tough problem. I solved it by choosing numbers.
AB+CD = AAA
The sum has all the same digits so let's try A=2
2B+CD = 222
Well, that doesn't work. The maximum value C could be is 9, then if B and D are 8 and 7 we have:
28 + 97 = 125 -----> This is far off from 222. ====> Let's try A =1
1B + CD = 111 ---->C has to be 9 as it is the only way we can make the sum have a one in the hundreds ===> Answer choice D
AB+CD = AAA
The sum has all the same digits so let's try A=2
2B+CD = 222
Well, that doesn't work. The maximum value C could be is 9, then if B and D are 8 and 7 we have:
28 + 97 = 125 -----> This is far off from 222. ====> Let's try A =1
1B + CD = 111 ---->C has to be 9 as it is the only way we can make the sum have a one in the hundreds ===> Answer choice D
AB + CD = AAA, where AB and CD are two-digit numbers
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21 May 2024, 19:04
21 May 2024, 19:04
2
Great problem! I solved this algebraically.
I wrote out the numbers in terms of 10.
AB = 10(A) + B
CD = 10(C) + D
AAA = 100(A) + 10(A) + A
10(A) + B + 10(C) + D = 100(A) + 10(A) + A
I subtracted 10A from both sides.
B + 10C + D = 100A + A
B + 10C + D = 101A
At this point, I solved it logically.
If A, B, C, and D are single digit numbers, the right-hand side of this equation can only equal 101. (If A = 2, then the right hand side would be 201 and this is too big). Therefore A = 1.
After realizing the entire equation equals 101, I maximized C = 9 so 10(C) = 90 (Anything less than 90 would be too small since the numbers have to be single digits). B + D has to be either 8 or 3 since 90 + 8 + 3 = 101.
At this point you can stop since you found the value of C, but to check your answer:
Going back to the original stem...13 + 98 = 121 does not match the format of the equation (AB + CD = AAA) however 18 + 93 = 111 does.
The answer is D
I wrote out the numbers in terms of 10.
AB = 10(A) + B
CD = 10(C) + D
AAA = 100(A) + 10(A) + A
10(A) + B + 10(C) + D = 100(A) + 10(A) + A
I subtracted 10A from both sides.
B + 10C + D = 100A + A
B + 10C + D = 101A
At this point, I solved it logically.
If A, B, C, and D are single digit numbers, the right-hand side of this equation can only equal 101. (If A = 2, then the right hand side would be 201 and this is too big). Therefore A = 1.
After realizing the entire equation equals 101, I maximized C = 9 so 10(C) = 90 (Anything less than 90 would be too small since the numbers have to be single digits). B + D has to be either 8 or 3 since 90 + 8 + 3 = 101.
At this point you can stop since you found the value of C, but to check your answer:
Going back to the original stem...13 + 98 = 121 does not match the format of the equation (AB + CD = AAA) however 18 + 93 = 111 does.
The answer is D
