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Sum of positive integers x and y

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Source: — Data Sufficiency |
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by cans » Sat Jul 23, 2011 12:54 am
x+y=77
a) x= y+1. get one x and one y and thus xy is unique
Sufficient
b) x and y have same 10 digits. so it must be 3. (if 2, then max sum can be 58 and if 4, then min sum can be 80)
as sum is 77, it means the sum of unit places is 17. and that's possible only with 8 and 9 (9+8=17)
thus no.s are 38,39
unique product.
Sufficient
IMO D
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by clock60 » Sat Jul 23, 2011 1:25 am
one more way to solve st 2, but in essence the same as that of can
x=10a+b
y=10a+m. where a is the same tens digit for x and y and b and m are digits
10a+b+10a+m=77
20a+b+m=77, here a can`t 4 as 20*4=80 that is more than 77, and here a can`t 2 as 20*2=40, and themax value of m+b=18, (40+18=56<77)
so the only value for a is 3
60+b+m=77
b+m=17, the only values for b and m are 9 and 8 so given x and y are 39, and 38 possible to find their product
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by gmatdriller » Sun Jul 24, 2011 12:49 am
Your contributions make sense; thanks "cans" and "clock60"
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