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A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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20 Nov 2018, 01:57
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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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Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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20 Nov 2018, 04:46
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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Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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20 Nov 2018, 02:24
Can somebody help me with this problem?



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Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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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 3digit 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 \(a + b = 9.\) Therefore, E is the answer. Answer: E
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Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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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. Answer: E
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Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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16 Apr 2019, 10:18
Dear MathRevolution shouldn't the question stated that : 1 each watch sold for the same price 2 that price is an integer Thanks
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Re: A store sold 72 watches for $a2,34b, where a2,34b is a 5digit integer
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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 5digit integer
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18 Apr 2019, 04:27






