multiples of 4 but not 6

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wishkaro: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 18, 2009 3:13 am; Thanked: 3 times

by wishkaro » Tue May 05, 2009 9:04 am

Always get wrong answer in these kind of question ...

What is the sum of all the terms between 50 and 450 which are only multiples of 4 but not 6?
A 20968
B 28968
C 20768
D 20568
E 20468
OA A

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Source: Beat The GMAT — Problem Solving | Tue May 05, 2009 9:04 am

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: multiples of 4 but not 6

by Vemuri » Tue May 05, 2009 9:55 am

I am also struggling mate

Let me put down my thoughts:

Sum of all terms between A & B can be obtained by using the formula = (# of numbers/2) * (first term + last term)

first term is : 52 (mutliple of 4 but not 6)
last term is : 448 (mutliple of 4 but not 6)

448-52 = 396 numbers are there in between these numbers. Dividing this number by 4 will give us the number of numbers that are divisible by 4, i.e 99 numbers. Now, these 99 numbers have numbers that are multiples of 6 as well. Every 3rd number is a multiple of 6. So, in effect only 66 numbers are multiples of 4 & not 6. Using the formula:

66/2 * (52+448) ==> 33*500 ==> 16500

Obviously, I am doing something wrong :roll:

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wishkaro: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 18, 2009 3:13 am; Thanked: 3 times

by wishkaro » Tue May 05, 2009 6:38 pm

Vemuri .. thanks for reply ... I found the answer but struggling to understand it .. anybody can explain

4 * 13 + ….. + 4*112 – 24*(3 + …. 18)
= 4*(13 + ….112) – 24 *( 3 + …18) = 4(100)/2*(13 + 112) – 24 *(16/2)*(18+3) = 200*(125) -24*8*21 = 25000 – 4032 = 20968

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jitendragneogi: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Aug 09, 2008 6:03 am; GMAT Score:490

by jitendragneogi » Wed May 06, 2009 2:31 pm

I think we need to answer the question this way:

Reqd SUM = SUM of all the multiples of 4s - SUM of all the numbers in the first category those are also multiple of 6s.

Calculation for the multiples of 4s:
Available numbers start from 50 ends at 450. So, the numbers those are multiple of 4s starts from 52 and ends at 448.

Last terrm in AP = {First term in AP + (No of terms - 1) * Difference}
=> 448 = 52 + (n-1) * 4
=> n = 100

SUM of all the multiple of 4s = { No of terms * (First number + last number) } / 2

SUM = {100 * (51+448) } / 2 = 25000

Similarly we can find out the SUM of all the digits which are a multiple of 24 in the range of 50 to 448. ( 24 = 4 * 6)

SUM = { (72 + 432) * 16 } / 2 = 4032

So, the reqd SUM = 25000 - 4032 = 20968

IMO A

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DeepakR: Master | Next Rank: 500 Posts; Posts: 158; Joined: Tue Sep 30, 2008 8:47 amFollowed by: 1 member; Thanked: 23 times; GMAT Score:660

by DeepakR » Wed May 06, 2009 6:58 pm

[quote="wishkaro"]Vemuri .. thanks for reply ... I found the answer but struggling to understand it .. anybody can explain

4 * 13 + ….. + 4*112 – 24*(3 + …. 18)
= 4*(13 + ….112) – 24 *( 3 + …18) = 4(100)/2*(13 + 112) – 24 *(16/2)*(18+3) = 200*(125) -24*8*21 = 25000 – 4032 = 20968[/quote]

I guess the solution seems to have ignored numbers like 60,84 etc..which are multiples of both 4 and 6 by taking only multiples of 24

I got something like this:

Sum=(52+56+60+...448) - (60+72+84+...444)
= 25000 - 8316 = 16684.

Can someone please explain what i am missing here ?..

-Deepak

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kvamsy: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Tue Mar 25, 2008 9:00 pm

by kvamsy » Thu May 07, 2009 12:41 am

I agree with Deepak , Solution as below

4*(13+...+112) - (LCM of 4,6= 12)*(5+6+....+37)

Which gives 16684. I too got the same , but I don't know about the answers mentioned above.

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Pranay: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Tue May 05, 2009 1:58 am; Thanked: 1 times

by Pranay » Thu May 07, 2009 1:20 am

jitendragneogi wrote:I think we need to answer the question this way:

Reqd SUM = SUM of all the multiples of 4s - SUM of all the numbers in the first category those are also multiple of 6s.

Calculation for the multiples of 4s:
Available numbers start from 50 ends at 450. So, the numbers those are multiple of 4s starts from 52 and ends at 448.

Last terrm in AP = {First term in AP + (No of terms - 1) * Difference}
=> 448 = 52 + (n-1) * 4
=> n = 100

SUM of all the multiple of 4s = { No of terms * (First number + last number) } / 2

SUM = {100 * (51+448) } / 2 = 25000

Similarly we can find out the SUM of all the digits which are a multiple of 24 in the range of 50 to 448. ( 24 = 4 * 6)

SUM = { (72 + 432) * 16 } / 2 = 4032

So, the reqd SUM = 25000 - 4032 = 20968

IMO A

Hi Jitendra,

I agree with the concept of sum of multiples of 4 -sum of multiples of 6

but,

in the calculation you have done it this way,

Answer = multiples of 4 - multiples of 24

which leaves numbers that are multiples of 4 and divisible by 6 but not divisible by 24 ex., 60.

Please expain.

Thanks in advance

Pranay

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