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by maihuna » Fri Oct 30, 2009 11:53 am
How many multiples of 4 are there between 16 and 260?

128
64
62
61
60
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by satish.nagdev » Sat Oct 31, 2009 7:54 am
IMO its 62

260/4 = has 66 numbers with 4 as factor plus we'll leave 4, 8, 12, 16 = 4 numbers hence 62 in total
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by crackgmat007 » Sat Oct 31, 2009 11:19 am
satish.nagdev wrote:IMO its 62

260/4 = has 66 numbers with 4 as factor plus we'll leave 4, 8, 12, 16 = 4 numbers hence 62 in total
260/4 = 65 numbers, this includes 260 as well. Hence 64. Applying same logic, we get 60.

Edited to avoid confusion
Last edited by crackgmat007 on Sat Oct 31, 2009 1:50 pm, edited 1 time in total.
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by byby030384 » Sat Oct 31, 2009 12:39 pm
How many multiples of 4 are there between 16 and 260?

Two Keyis not to include 16 and 260.

Divide 4 by 20 and 256 and you will have 5 and 64.

Next 64 - 5 + 1 = 60.
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by mehravikas » Sat Oct 31, 2009 7:04 pm
I think it should be 62
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by satish.nagdev » Sat Oct 31, 2009 8:28 pm
crackgmat007 wrote:
satish.nagdev wrote:IMO its 62

260/4 = has 66 numbers with 4 as factor plus we'll leave 4, 8, 12, 16 = 4 numbers hence 62 in total
260/4 = 65 numbers, this includes 260 as well. Hence 64. Applying same logic, we get 60.

Edited to avoid confusion
oh yea thanks man for correcting .... :)
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by Abdulla » Sat Oct 31, 2009 10:49 pm
IMO it's 62

= (Last term - First term) / Increment +1
= { ( 260-16 ) / 4 } +1
= ( 244 / 4 ) +1
= 61+1
= 62 ..#
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by sanjana » Sat Oct 31, 2009 11:07 pm
Abdulla wrote:IMO it's 62

= (Last term - First term) / Increment +1
= { ( 260-16 ) / 4 } +1
= ( 244 / 4 ) +1
= 61+1
= 62 ..#
It says between,so we should not include 16 and 260.
If the question said inclusive then the answer would be 62
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by Abdulla » Sat Oct 31, 2009 11:09 pm
sanjana wrote:
Abdulla wrote:IMO it's 62

= (Last term - First term) / Increment +1
= { ( 260-16 ) / 4 } +1
= ( 244 / 4 ) +1
= 61+1
= 62 ..#
It says between,so we should not include 16 and 260.
If the question said inclusive then the answer would be 62
Oh yeah !! thanks sanjana :D
Abdulla
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