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another number properties with "remaining.."

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another number properties with "remaining.."

by hwiya320 » Sun Nov 30, 2008 5:29 pm
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
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Re: another number properties with "remaining.."

by logitech » Sun Nov 30, 2008 5:42 pm
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
Statement 1)

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
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by cramya » Sun Nov 30, 2008 5:50 pm
X-3 = LCM(21:7)C

X-3=21C
Nice approach. I just picked numbers till both matched.

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
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by logitech » Sun Nov 30, 2008 6:01 pm
cramya wrote:
X-3 = LCM(21:7)C

X-3=21C
Nice approach. I just picked numbers till both matched.

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
Nope, because we treat X-3 as a new number that can be divisible with both 21 and 7.
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by vittalgmat » Mon Dec 01, 2008 12:58 am
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
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by logitech » Mon Dec 01, 2008 1:12 am
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
Statement 1)

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.
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by vittalgmat » Mon Dec 01, 2008 1:26 am
logitech wrote:
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
Statement 1)

x=7a+3

lets divide X with 14:

x /14 = ( a/2 + 3/4 ) <--- should be (a/2 + 3/14)



Thanks for the explanation.

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