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Multiples

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by akshatgupta87 » Mon Apr 11, 2011 9:47 am
Q.) How many multiples of 4 are there between 16 and 260?

for this question do we include start and end numbers?
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by pankajks2010 » Mon Apr 11, 2011 10:03 am
as the question is to find out multiples of r between 16 and 260, we would not consider 16 and 260..
Thus, the possible multiples are 20, 24, 28.....256
As this is an AP we can find out the number of terms using the formula: Tn=A + (n-1)d
Answer is 60
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by pankajks2010 » Mon Apr 11, 2011 10:06 am
pankajks2010 wrote:as the question is to find out multiples of r between 16 and 260, we would not consider 16 and 260..
Thus, the possible multiples are 20, 24, 28.....256
As this is an AP we can find out the number of terms using the formula: Tn=A + (n-1)d
Answer is 60
an even simpler method is to just find out the numbers between 4 & 65, which would again be 60. Sorry for including unnecessary progressions and formulas and calculations...
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by srcc25anu » Mon Apr 11, 2011 10:07 am
for such questions the formula is [(last number - first number) / 4 ] +1 (for both inclusive)
= (260-16)/4 and add 1 to the solution
we get 62 multiples of 4 between 16 and 260 (inclusive)

for first and last number exclusive, the formula is (last number - first number) / 4 -1 (for in between first and last numbers)
= 260-16 / 4 and subtract 1 from the solution we get 60 multiples of 4 between 16 and 260 (both exclusive)
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by akshatgupta87 » Mon Apr 11, 2011 10:15 am
so inclusive is not mentioned here..
therefore 60
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by srcc25anu » Mon Apr 11, 2011 10:24 am
Thats right.
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by Brent@GMATPrepNow » Tue Apr 12, 2011 6:47 am
If you're not good at remembering formulas, then another option is to look for a pattern.
So, we want to know how many terms are in the sequence 20, 24, 28, . . .. 252, 256

Notice that:
20 = 4(5)
24 = 4(6)
28 = 4(7)
32 = 4(8)
.
.
.
252 = 4(63)
256 = 4(64)

So, the number of terms in the sequence 20, 24, 28, . . .. 252, 256 is equal to the number of integers from 5 to 64 inclusive.

Well we have a formula (an easy formula) for finding this. We know that the number of integers from x to y inclusive is equal to y-x+1

So, the number of integers from 5 to 64 inclusive is equal to 64-5+1 = 60
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by Ian Stewart » Tue Apr 12, 2011 3:05 pm
Brent Hanneson wrote:If you're not good at remembering formulas, then another option is to look for a pattern.
So, we want to know how many terms are in the sequence 20, 24, 28, . . .. 252, 256
(...)
So, the number of terms in the sequence 20, 24, 28, . . .. 252, 256 is equal to the number of integers from 5 to 64 inclusive.

Well we have a formula (an easy formula) for finding this.
Or, if you wanted to continue without using a formula, the number of integers between 5 and 64 inclusive is equal to the number of integers between 1 and 60 inclusive (just subtract 4 from the start and end values), which is clearly 60.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by vzzai » Wed Apr 13, 2011 4:22 am
Hi, Would this formulae help?
Last - First/Increment - 1 (Numbers inclusive)
Last - First/Increment + 1 (In between)
Thank you,
Vj
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