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

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Math Expert
Joined: 02 Sep 2009
Posts: 48119
If A, B, and C are distinct positive digits and the product of the two  [#permalink]

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14 Dec 2016, 05:52
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Difficulty:

25% (medium)

Question Stats:

82% (01:19) correct 18% (02:12) wrong based on 114 sessions

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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

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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, 10:24
1
2
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 $$B+C=2+3=5 \implies A+B+C=1+2+3=6$$

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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, 01:32
2
1
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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Joined: 07 Dec 2014
Posts: 1069
Re: If A, B, and C are distinct positive digits and the product of the two  [#permalink]

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15 Dec 2016, 13:40
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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Joined: 20 Dec 2017
Posts: 2
Re: If A, B, and C are distinct positive digits and the product of the two  [#permalink]

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20 Dec 2017, 00:37
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
Re: If A, B, and C are distinct positive digits and the product of the two &nbs [#permalink] 20 Dec 2017, 00:37
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