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705-805 (Hard)|   Arithmetic|      
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Bunuel
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If A, B and C represent different digits in the multiplication, AAB * 22 Feb 2022, 04:45
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75% (hard)

Question Stats:

64% (03:08) correct 36% (02:50) wrong based on 411 sessions

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

A. 9
B. 12
C. 14
D. 15
E. 17


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chetan2u
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GMAT Focus 1: 735 Q90 V89 DI81
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If A, B and C represent different digits in the multiplication, AAB * 23 Feb 2022, 00:21
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Bunuel
If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

A. 9
B. 12
C. 14
D. 15
E. 17
Show more


Step 1: Value of B.
If we look at the units digit of the numbers being multiplied and the product, we get B*B=_B.
Thus, B can only be 0, 1, 5 and 6.
If 0, then product will be 0. Discard
If 1, then product will be AA1. Discard
If 5, then AA5*5=C555. But product of two multiples of 5 should end in 00, 25, 50 or 75. Discard
So, B=6.

Step 2: Value of C.
So, the product is => AA6*6=C656
Now, C656 should be a multiple of 6, so C can be 1, 4 or 7.
If C is 1, then \(\frac{1656}{6}=276\neq AA6\)
If C is 4, then \(\frac{4656}{6}=776= AA6\)
So, C=4 and A=7

SUM = 7+4+6 = 17
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aryanrgupta
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If A, B and C represent different digits in the multiplication, AAB * 22 Sep 2024, 21:47
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chetan2u
Bunuel
If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

A. 9
B. 12
C. 14
D. 15
E. 17


Step 1: Value of B.
If we look at the units digit of the numbers being multiplied and the product, we get B*B=_B.
Thus, B can only be 0, 1, 5 and 6.
If 0, then product will be 0. Discard
If 1, then product will be AA1. Discard
If 5, then AA5*5=C555. But product of two multiples of 5 should end in 00, 25, 50 or 75. Discard
So, B=6.

Step 2: Value of C.
So, the product is => AA6*6=C656
Now, C656 should be a multiple of 6, so C can be 1, 4 or 7.
If C is 1, then \(\frac{1656}{6}=276\neq AA6\)
If C is 4, then \(\frac{4656}{6}=776= AA6\)
So, C=4 and A=7

SUM = 7+4+6 = 17
Show more
Can you please explain how did the values of C boil down to 1, 4, 7 ?
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If A, B and C represent different digits in the multiplication, AAB * 23 Sep 2024, 00:06
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aryanrgupta
chetan2u
Bunuel
If A, B and C represent different digits in the multiplication, AAB * B = CB5B, then A + B + C = ?

A. 9
B. 12
C. 14
D. 15
E. 17


Step 1: Value of B.
If we look at the units digit of the numbers being multiplied and the product, we get B*B=_B.
Thus, B can only be 0, 1, 5 and 6.
If 0, then product will be 0. Discard
If 1, then product will be AA1. Discard
If 5, then AA5*5=C555. But product of two multiples of 5 should end in 00, 25, 50 or 75. Discard
So, B=6.

Step 2: Value of C.
So, the product is => AA6*6=C656
Now, C656 should be a multiple of 6, so C can be 1, 4 or 7.
If C is 1, then \(\frac{1656}{6}=276\neq AA6\)
If C is 4, then \(\frac{4656}{6}=776= AA6\)
So, C=4 and A=7

SUM = 7+4+6 = 17
Can you please explain how did the values of C boil down to 1, 4, 7 ?
Show more

C656 is a multiple of 6, so it must also be a multiple of 3. The sum of the digits of a multiple of 3 must also be a multiple of 3. Since C + 6 + 5 + 6 = 17 + C, for this to be a multiple of 3, C must be 1, 4, or 7.

Hope it's clear.
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If A, B and C represent different digits in the multiplication, AAB * 29 Sep 2024, 05:37
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AAB
x B
--------
CB5B
since B x B gives unit place as B. B is either 1, 5 or 6
now assume B as 1
we have AA1 x 1 = CB5B (not possible)
try B = 5
AA5 x 5 = C555
take A = 1 ten 115x5 tens place becomes 7 hence not possible
any value of A does not gives tens digit as 5
hence A = 5 is rejected

not take B = 6
look for value of A such that B x A +3 gives 5
B = 6 then A could be 2, 7
try A = 2
226 x 6 = 1356 (hundred place is B which is 3 here in answer. hence rejected
then A must be 7
just verify to confirm
776 x 6 = 4656 (everything matches)

A = 7 , B = 6 and C = 4 therefore A+B+C =17
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If A, B and C represent different digits in the multiplication, AAB * 08 Jan 2026, 14:09
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AAB
x B
------------- this is the given multiplication problem , need to find A+B+C=?
CB5B
check the units digit B, only choices which give the same digit on multiplication to itself are 1,5 & 6.
B=1 can be discarded at once as if even A is 9, it will not result in the thousands digit.
B=5 can also be discarded as 5*5=25 carry is 2 , and 5 multiplied with any digit A will result in either 0 or 5 which on addition to carry can never result in 5 at the tens place.
so correct choice for B is 6
now carry for the tens place is 3, and 6 multiplied by 2 or 7 result in 12 or 42 which on accumulation of carry will result in 5 at the tens place of the result.
however A=2 can be discarded as multiplication again B*A=12 plus carry of 1 will never result in B=6 at the hundreds place of the result.
hence A=7 is the correct choice which gives C=4
so A+B+C=7+6+4=17
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If A, B and C represent different digits in the multiplication, AAB * 29 Mar 2026, 00:46
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Got the correct answer but it took me 4:55 :?
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