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If A, B and C represent different digits in the multiplication, then A

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If A, B and C represent different digits in the multiplication, then A [#permalink]

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New post 29 Dec 2017, 03:58
2
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (03:05) correct 40% (02:32) wrong based on 58 sessions

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If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

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It does take time to solve this problem.
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Re: If A, B and C represent different digits in the multiplication, then A [#permalink]

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New post 29 Dec 2017, 05:28
chesstitans wrote:
If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

9
12
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It does take time to solve this problem.


Questions like these are best approached with a mix of trial and error and logic.
This combines Alternative and Logical approaches.

First, the ones digit of AAB*B is B. This means that B*B is B (because the ones digit of the product is determined only by the ones digits of the factors!).
Only three numbers fit the pattern - 0,1, 5 and 6. 0 and 1 are clearly irrelevant so B must be 5 or 6.
Next, the tens digits AAB*B is 5 meaning that AB*B has a tens digit of 5.
If B=5 then A5*5 has a tens digit of 5 --> 50*A + 25 has a tens of 5 --> 5*A +2 has a ones digit of 5 --> 5*A has a ones digit of 3.
This is impossible so B must equal 6.
Let's verify:
If B=6 then A6*6 has a tens digit of 5 --> 60*A + 36 has a tens of 5 --> 6*A + 3 has a ones digit of 5 --> 6*A has a ones digit of 2 --> A equals 2 or 7.

So, our problem is now 226*6 = C656 or 776*6 = C656
Let's solve the first problem as it looks easier. 226*6 = 1200+120+36= 1356. But then B is both 3 and 6. Impossible! Then A must equal 7.
776*6 = 4200 + 420 + 36 = 4656 and C = 4. In this case A+B+C = 7+6+4=17 and (E) is our answer.
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If A, B and C represent different digits in the multiplication, then A [#permalink]

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New post 29 Dec 2017, 05:55
chesstitans wrote:
If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

9
12
14
15
17

It does take time to solve this problem.



STEP by step approach will help you get answer and that too within decent time..

STEP I - UNIT's digit - B
when can be B*B lead to B as units digit..
a) \(0*0 = 0\)... But entire answer will become 0
b) \(1*1 = 1\)... BUT the product will remain AAB
c) \(5*5 = 25\)..... But \(5*A+2 = x5\).... Not possible.. the digit cannot be 5, it will be "5*even+2=x2" OR "5*odd+2 = 7"
d) \(6*6=36\) ..... \(6*A+3=x5\)... Possible when A is 2 or A is 7
no other possibility, so B is 6 and A is 2 or 7

STEP II - A?
our product is \(AA6*6=C656\)...
if A is 2, it becomes \(226*6 = 1356\) Not same as C656

so A is 7..
\(776*6 = 4656...YES\)
so \(A+B+C = 7+6+4 = 17\)

E
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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If A, B and C represent different digits in the multiplication, then A   [#permalink] 29 Dec 2017, 05:55
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