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28 Mar 2024, 01:42
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
69% (02:22) correct
31%
(02:31)
wrong
based on 382
sessions
History
Date
Time
Result
Not Attempted Yet
\(AAB + BBC = AAAA\)
A. 0
B. 1
C. 2
D. 7
E. 9
05 Apr 2024, 03:45
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Official Solution:
\(AAB + BBC = AAAA\)
In the correctly worked addition problem shown, where the sum of the three-digit positive integers AAB and BBC is the four-digit integer AAAA, and A, B, and C are different digits, what is the units digit of the integer BBC?
A. 0
B. 1
C. 2
D. 7
E. 9
Given that AAB and BBC are three-digit integers, their sum can give us only one four-digit integer of a kind of AAAA: 1111. Hence, A = 1, and we have:
\(11B + BBC = 1111\)
Next, observe that B cannot be less than 9, because if it is, the sum won't reach 1111. Thus, B = 9, and we have:
\(119 + 99C = 1111\)
The sum of 9 and C to produce the result with the units digit of 1, C must be 2.
Answer: C
\(AAB + BBC = AAAA\)
A. 0
B. 1
C. 2
D. 7
E. 9
Given that AAB and BBC are three-digit integers, their sum can give us only one four-digit integer of a kind of AAAA: 1111. Hence, A = 1, and we have:
\(11B + BBC = 1111\)
\(119 + 99C = 1111\)
Answer: C
General Discussion
28 Mar 2024, 02:26
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Maximum value of 2 3 digit number addition would be 1998.
So, A has to be 1
Since, A is 1 so B would be 9
Subtracting 1111 - 119 would give us 992
So, the value for C would be 2.
So, A has to be 1
Since, A is 1 so B would be 9
Subtracting 1111 - 119 would give us 992
So, the value for C would be 2.
