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Old 07-27-2007   #1
Brian Y
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If the product of the integers a, b, and c is 1, then what is the SMALLEST possible value of (a+1)(b+1)(c+1)?a) -3 b) -1 c) 0 d) 1 e) 3What is the sum of five consecutive odd integers if the middle integer is 71?a) 213 b) 353 c) 354 d) 355 e) 356?pls show work
 
Old 07-27-2007   #2
zever9000
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1) because a b and c are integers, they must be all 1 or 2 of them must be -1 so (a+1)(b+1)(c+1) = 8 or 0so the answer is c) 02) 67+69+71+73+75=355so d) 355
 
Old 07-27-2007   #3
Maurice H
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replace the variabls with 0 then add each one multiply the three numbers and the product is 1.In short;answer is C(a+1)(b+1)(c+1)?67+69+71+73+75=355 d
 
Old 07-27-2007   #4
forevaz3r0
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ans for 1st one is c) ocoz, since a, b, c are integers, and their product is equal to 1, so it means the possible integers for the 3 unknowns are all 1, or one 1 and two -1 : 1x1x1=1 and 1x-1x-1=1therefore smallest possible for (a+1)(b+1)(c+1) is (1+1)(-1+1)(-1+1) = 0...ans for 2nd one is d) 35571x5=355, as when u take the value before and after 71, for example: 69 and 73... when i minus 2 from 73 i get 71, and den i add the 2 (which i minus away from 73) to 69 , i will still get 71... thus there are 2 numbers before 71, and 2 numbers after 71... thus it will make up of 5 numbers of 71 including itself...
 
Old 07-27-2007   #5
Kathie
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(a+1)(b+1)(c+1)=1(0+1)(0+1)(0+1)=1(1)(1)(1)=1The answer is C 2 67 69 71 73 75------355The answer is D
 
 

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