The integer x is positive. What is the remainder when the x is divided by 14 ?
(1) The remainder when 4x is divided by 28 is 12.
(2) The remainder when x is divided by 21 is 3.
correct answer is E
another number properties with "remaining.."
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- logitech
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Statement 1)hwiya320 wrote:The integer x is positive. What is the remainder when the x is divided by 14 ?
(1) The remainder when 4x is divided by 28 is 12.
(2) The remainder when x is divided by 21 is 3.
correct answer is E
4x = 28a+12
x=7a+3 INSUF
Statement 2)
x=21b+3 INSUF
1+2)
X-3 = LCM(21:7)C
X-3=21C
X=21C+3
Insuf
24/14 = 10 as a remainder
43/14 = 1 as a remainder
Hence, E
LGTCH
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Nice approach. I just picked numbers till both matched.X-3 = LCM(21:7)C
X-3=21C
Logitech, do u think we can do this if the remainders were different? I dint think so but let me know ur thoughts
For eg:
Stmt I left a remainder of 4 x=7a+4
Stmt II left a remainder of 3 x=21b+3
- logitech
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Nope, because we treat X-3 as a new number that can be divisible with both 21 and 7.cramya wrote:Nice approach. I just picked numbers till both matched.X-3 = LCM(21:7)C
X-3=21C
Logitech, do u think we can do this if the remainders were different? I dint think so but let me know ur thoughts
For eg:
Stmt I left a remainder of 4 x=7a+4
Stmt II left a remainder of 3 x=21b+3
LGTCH
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I need a more detailed explanation. Can one of you pls elaborate. I dint understand how logitech concluded that 1 and 2 are independently insufficient (algebraically).
thanks a lot
thanks a lot
- logitech
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Statement 1)vittalgmat wrote:I need a more detailed explanation. Can one of you pls elaborate. I dint understand how logitech concluded that 1 and 2 are independently insufficient (algebraically).
thanks a lot
x=7a+3
lets divide X with 14:
x /14 = ( a/2 + 3/14 )
as you see we do not always have an integer with different values of a.
lets say
X = 6b +3
is X divisible by 3 ?
x/3 = 2b+1
so every different value of B we will have an integer so IT IS DIVISIBLE
Last edited by logitech on Mon Dec 01, 2008 1:33 am, edited 2 times in total.
LGTCH
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logitech wrote:Statement 1)vittalgmat wrote:I need a more detailed explanation. Can one of you pls elaborate. I dint understand how logitech concluded that 1 and 2 are independently insufficient (algebraically).
thanks a lot
x=7a+3
lets divide X with 14:
x /14 = ( a/2 + 3/4 ) <--- should be (a/2 + 3/14)
Thanks for the explanation.