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find th sum

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by maihuna » Mon May 11, 2009 9:46 am
Find the sum of all numbers between 200 and 400 which are divisible by 7.

8428
8548
8673
8729
9879

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

2550
3050
3240
3300
3600

Find the sum of all two digit numbers which when divided by 4, yields 1 as
remainder.

1076
1150
1210
1250
1340
Charged up again to beat the beast :)
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Source: — Problem Solving |
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by scoobydooby » Mon May 11, 2009 10:18 am
Q1. number of integers between 200 and 400=199
number of integers divisible by 7=>199/7=28

set of integers: 203, 210......399
sum of the integer set: 28*(203+399)/2=301*28=8428
hence, A


Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E


Q3. 2 digit numbers which leave a remainder 1 when divided by 4
of the form 4k+1, k must take values 3, 4, .....,24 as we only need 2 digit numbers

set of numbers: 13+17+21+....+97

Tn=97=13+(n-1)4, solving for n we get n=22

mean of the set: (13+97)/2=110/2=55

sum of the set: 55*22=1210
hence, C
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by hawlia » Mon May 11, 2009 10:28 am
scoobydooby wrote:Q1. number of integers between 200 and 400=199
number of integers divisible by 7=>199/7=28

set of integers: 203, 210......399
sum of the integer set: 28*(203+399)/2=301*28=8428
hence, A


Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E


Q3. 2 digit numbers which leave a remainder 1 when divided by 4
of the form 4k+1, k must take values 3, 4, .....,24 as we only need 2 digit numbers

set of numbers: 13+17+21+....+97

Tn=97=13+(n-1)4, solving for n we get n=22

mean of the set: (13+97)/2=110/2=55

sum of the set: 55*22=1210
hence, C
Acc to me Ans of First Ques is 8729
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by Stuart@KaplanGMAT » Mon May 11, 2009 11:02 am
scoobydooby wrote:Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E
Except that you've double counted 10, 20, 30, ... 100, which are divisible by both 2 AND 5.

So, we have to subtract the multiples of both 2 and 5.

3600 - (10 + 20 + 30 + ... + 100)

= 3600 - 10(55) = 3600 - 550 = 3050... choose (B).

[The shortcut I used to calculate that series, which could have been used on all of these questions (and which Scooby did use), is merely an application of the average forumla.

Sum of terms = (# of terms)(average)

For a set of consecutive numbers, the average is simply:

(1st # + Last #)/2

or in this case:

(10 + 100)/2 = 110/2 = 55]
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by Mr2Bits » Mon May 11, 2009 11:25 am
hawlia wrote:
scoobydooby wrote:Q1. number of integers between 200 and 400=199
number of integers divisible by 7=>199/7=28

set of integers: 203, 210......399
sum of the integer set: 28*(203+399)/2=301*28=8428
hence, A


Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E


Q3. 2 digit numbers which leave a remainder 1 when divided by 4
of the form 4k+1, k must take values 3, 4, .....,24 as we only need 2 digit numbers

set of numbers: 13+17+21+....+97

Tn=97=13+(n-1)4, solving for n we get n=22

mean of the set: (13+97)/2=110/2=55

sum of the set: 55*22=1210
hence, C
Acc to me Ans of First Ques is 8729
Different way to work the first is as follows.

1/2*28*(7(29+57))
= 8428
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by maihuna » Tue May 12, 2009 9:23 am
OA:
8729
3050 Thanks Stuart
1210
Charged up again to beat the beast :)
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