M40-88

Question

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

Bunuel · Answer

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

Vikramaditya00 · Answer

I think this is a  high-quality  question and the explanation isn't clear enough, please elaborate. Why AAAA=1111 why it cannot be 2222.3333,4444.......

Bunuel · Answer

This is explanted in the solution: 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.

The sum of two three-digit integers cannot be more than 999 + 999 = 1,998.

lnyngayan · Answer

Could you explain more on why B cannot be less than 9?
Why have to be 9+2 =11, but not 8+3 or 7+4, etc?

Many thanks.

Bunuel · Answer

If B = 8, then we would have 118 + 88C, and the sum would be less than 1111 for any value of C. Therefore, B must be 9 for the sum to reach 1111.

Sujithz001 · Answer

I think this is a  high-quality  question and I agree with explanation. Wonderful question.

MBAToronto2024 · Answer

KarishmaB  any frameworks how to solve it?

KarishmaB · Answer

.. A A B + B B C ---------- A A A A When two 3 digit numbers are added together, can we get 2000 or more? No! Never. The sum will always be less than 2000 because each number is less than 1000. This means that the 4 digit number AAAA begins with 1. So A = 1 When I add a small three digit number 11x and I want a sum above 1000, the other number should be a big number. So try B = 9 .. 1 1 9 + 9 9 C ---------- 1 1 1 1 If C = 2, then the sum is valid. Answer (C) Alphametics are discussed here: https://anaprep.com/puzzles-corner-solving-alphametics/ https://anaprep.com/puzzles-corner-mult ... phametics/

PrameshLuitel123 · Answer

I don’t quite agree with the solution. Isn't there a possibility that AAB=664 AND BBC = 447 which would yeild 1111?

Bunuel · Answer

If A = 1, then AAB cannot be  66 4. The digits A, B, and C must be used consistently throughout the equation. Please review the solution carefully. Your example doesn't follow the given constraints.

Apoorv0194 · Answer

I did it a bit differently - we have AAB + BBC
this can be written as - 100A + 10A + B + 100B +10B + C = 110A + 111B + C
We know its equal to AAAA = 1000A + 100A + 10A + A

Equating,
110A + 111B + C = 1000A + 100A + 10A + A
110A + 111B + C = 1111A
111B + C = 1001A

since A,B and C are single-digit numbers, only value satisfying would be 
A=1, B=9 and C=2

janvibatra.50 · Answer

I like the solution - it’s helpful.

ravi1522 · Answer

Hello ,
AAB+BBC=AAAA
this can be written as
100A+10B+C+100B+10B+C=1000A+100A+10A+A
100(A+B)+10(A+B)+B+C=1000A+100A+10A+A
 by comparing we can write

A+B=10A
11A=B+C

solving this equation 
we get C=2A

So A can be from 1to 9
so C will be 2,4,6,8

Hence option C is correct

Is this approach is correct ?

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