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20 Nov 2018, 01:57
1
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Difficulty:

55% (hard)

Question Stats:

59% (02:05) correct 41% (01:55) wrong based on 61 sessions

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[Math Revolution GMAT math practice question]

A store sold $$72$$ watches for $$a2,34b,$$ where $$a2,34b$$ is a $$5$$-digit integer. What is the value of $$a + b$$?

$$A. 5$$
$$B. 6$$
$$C. 7$$
$$D. 8$$
$$E. 9$$

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20 Nov 2018, 04:46
5
MathRevolution wrote:
[Math Revolution GMAT math practice question]

A store sold $$72$$ watches for $$a2,34b,$$ where $$a2,34b$$ is a $$5$$-digit integer. What is the value of $$a + b$$?

$$A. 5$$
$$B. 6$$
$$C. 7$$
$$D. 8$$
$$E. 9$$

so 72 * price of each watch =a234b
so price of each=a234b/72

for numerator to div by 72,it must be divisible by all factors of 72.

now 72=2^3*3^2=6*9
so in denominator we have 9

now for numerator to be divisible by 9 ,sum of digits must be divisible by 9.(divisibility rule of 9)
so a+2+3+4+b=9+a+b
so if a+b is 9
it becomes 18 which is div by 9

so E
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Joined: 09 Mar 2017
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22 Nov 2018, 02:16
=>

Since $$a234b$$ is a multiple of $$72, a234b$$ is a multiple of both $$8$$ and $$9$$.
The last three digits $$34b$$ of $$a234b$$ form a multiple of $$8$$. So, we must have $$b = 4$$ since $$344$$ is the only 3-digit multiple of $$8$$ beginning with the digits, $$34$$.
Since $$a234b$$ is a multiple of $$9, a + 2 + 3 + 4 + b = a + 2 + 3 + 4 + 4 = a + 13$$ is a multiple of $$9$$. This implies that $$a = 5,$$ and $$a + b = 9.$$

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 5810 Location: United States (CA) Re: A store sold 72 watches for$a2,34b, where a2,34b is a 5-digit integer  [#permalink]

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29 Mar 2019, 09:04
MathRevolution wrote:
[Math Revolution GMAT math practice question]

A store sold $$72$$ watches for $$a2,34b,$$ where $$a2,34b$$ is a $$5$$-digit integer. What is the value of $$a + b$$?

$$A. 5$$
$$B. 6$$
$$C. 7$$
$$D. 8$$
$$E. 9$$

Assuming the price of each watch is the same, we see that a2,34b must be a multiple of 72. That is, it must be a multiple of 8 and 9 since 72 = 8 x 9. We know that for a number to be a multiple of 8, the last 3 digits must be divisible by 8 and for a number to be a multiple of 9, the sum of the digits must be divisible by 9. So here, we need 34b to be divisible by 8 and a + 2 + 3 + 4 + b = a + b + 9 to be divisible by 9.

If 34b is divisible by 8, then b = 4 only. In that case a must be 5 so that a + b + 9 = 18 will be divisible by 9. So a + b = 5 + 4 = 9.

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18 Apr 2019, 04:27
I saw that the question is asking for a+b so I used the rule of multiple of 9

Since a2,34b must be divisible by 9 -> sum of digits = a+b+9 must also be divisible by 9 -> so a+b must be divisible by 9
Only choice E gives this answer
Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5-digit integer [#permalink] 18 Apr 2019, 04:27 Display posts from previous: Sort by # A store sold 72 watches for$a2,34b, where a2,34b is a 5-digit integer

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