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If A, B, and C are distinct positive digits and the product of the two

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If A, B, and C are distinct positive digits and the product of the two [#permalink]

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14 Dec 2016, 04:52
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If A, B, and C are distinct positive digits and the product of the two-digit integers AB and AC is 156, what is the sum of the digits A, B, and C?

A. 4
B. 5
C. 6
D. 10
E. 25
[Reveal] Spoiler: OA

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Re: If A, B, and C are distinct positive digits and the product of the two [#permalink]

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14 Dec 2016, 09:24
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Bunuel wrote:
If A, B, and C are distinct positive digits and the product of the two-digit integers AB and AC is 156, what is the sum of the digits A, B, and C?

A. 4
B. 5
C. 6
D. 10
E. 25

$$\overline{AB} \times \overline{AC}=156$$

We have $$(10A+B)(10A+C)=100A+10AB+10AC+BC=156$$

Since $$100A\leq 156 \implies A \leq 1 \implies A=1$$

Hence $$\overline{1B} \times \overline{1C}=156=12 \times 13$$

So $$B+C=2+3=5 \implies A+B+C=1+2+3=6$$

The answer is C
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Re: If A, B, and C are distinct positive digits and the product of the two [#permalink]

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15 Dec 2016, 00:32
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Option C)

Given: AB * AC = 156

: A + B + C = ?

$$156 = 2^2 * 3^1 * 13^1 = 4^1 * 3^1 * 13^1 = 12^1 * 13^1$$

Two numbers AB & AC have same 10's digit: As per above factorization it could only be 1, hence A = 1 and B & C = 2 & 3 or 3 & 2

We need not to get exact values of B & C, as in either case $$A + B + C = 1 + 2 + 3 = 1 + 3 + 2 = 6$$
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Re: If A, B, and C are distinct positive digits and the product of the two [#permalink]

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15 Dec 2016, 12:40
Bunuel wrote:
If A, B, and C are distinct positive digits and the product of the two-digit integers AB and AC is 156, what is the sum of the digits A, B, and C?

A. 4
B. 5
C. 6
D. 10
E. 25

the only two 2 digit factors of 156 are 12 and 13
1+2+3=6
C

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Re: If A, B, and C are distinct positive digits and the product of the two [#permalink]

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19 Dec 2017, 23:37
I would advise to do prime factorization of 156 = 2*78=2*2*39=2*2*3*13=12*13 so you get the two numbers and they are in the same format as mentioned i.e AB*AC . Thus you get respective values of A,B and C. Thus you get your answer quiet easily.
A+B+C=1+2+3=6

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Re: If A, B, and C are distinct positive digits and the product of the two   [#permalink] 19 Dec 2017, 23:37
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If A, B, and C are distinct positive digits and the product of the two

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