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Call 911: Some Digits are Missing

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Master | Next Rank: 500 Posts
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by dtweah » Tue May 19, 2009 12:33 pm
Given that 29a031 × 342 = 100900b02 where a and b denote two missing
digits, what is the value of a + b ?
(a) 7
(b) 8
(c) 9
(d) 10
(e) 11
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by dumb.doofus » Tue May 19, 2009 1:43 pm
I just used basic multiplication to deduce a = 5 and b = 6

so a + b = 11

any other methods??
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by dtweah » Wed May 20, 2009 5:26 am
dumb.doofus wrote:I just used basic multiplication to deduce a = 5 and b = 6

so a + b = 11

any other methods??
I did the same but here is what they say in the OA

Recall the tests for divisibility by 9 and by 11. Since 342 is divisible by 9, the number 100900b02 is as well. So b has to be 6. But then 100900b02 is divisible by 11. Since 342 is not divisible by 11, we deduce 29a031 must be divisible by 11. So a has to be 5, and a + b = 11.
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