Problem Solving

hey guys,

i am getting confused over this silly thing. when a question asks: What is the sum of all the even integers between 99 and 301 ?

To first count the number of integers between 99 and 301, isn't the equation last-first+1?
so then wouldn't the answer be 203?

in the book it says there are 202 terms. so 101 even numbers.

hmm....can you explain please

thx.

I think since its not mentioned whether they are to be considered inclusive, thats why we have got 202 numbers, out of which 101 are even and 101 odd..

for ex from 1 to 100 there are 100 integers bth inclusive 50 odd and 50 even
and from 1 to 101 there are still 50 evens but 51 odds and therefore 101 int both inclusive
so from 99 to 301 we have 301-99 = 202+1=203 int but we will have one less even so in total we have 101 even and 102 odds

sanalnnair wrote:I think since its not mentioned whether they are to be considered inclusive, thats why we have got 202 numbers, out of which 101 are even and 101 odd..

thats what i thought too... but

look at this example: even numbers between 1 and 5 not inclusive. (there are 2). but 4-2 =2/2 =1 +1 = 2 correct!!

but now even numbers between 1 and 5 inclusive = 5-1=4+1/2 = a fraction incorrect!!!!

i thought the following were the rules: inclusive = last - first + 1. Unless the count is not inclusive like from 1-3, how many even #'s. 3-1-1=1.

do you see where i am getting confused!!!

Gurpinder wrote:

sanalnnair wrote:I think since its not mentioned whether they are to be considered inclusive, thats why we have got 202 numbers, out of which 101 are even and 101 odd..

thats what i thought too... but

look at this example: even numbers between 1 and 5 not inclusive. (there are 2). but 4-2 =2/2 =1 +1 = 2 correct!!

but now even numbers between 1 and 5 inclusive = 5-1=4+1/2 = a fraction incorrect!!!!

i thought the following were the rules: inclusive = last - first + 1. Unless the count is not inclusive like from 1-3, how many even #'s. 3-1-1=1.

do you see where i am getting confused!!!

the rule is true for integers
but for even and odd integers we need to include last even or last odd integers
for ex b/n 99 to 301 for #of even int we need to find out from 100 t0 301 or from 99 to 300 and same for odd int...
then the eqn will work

thephoenix wrote:for ex from 1 to 100 there are 100 integers bth inclusive 50 odd and 50 even
and from 1 to 101 there are still 50 evens but 51 odds and therefore 101 int both inclusive
so from 99 to 301 we have 301-99 = 202+1=203 int but we will have one less even so in total we have 101 even and 102 odds

thx for the reply.

so there is no eq. eh? i gotta do it your way?

#s between (inclusive) 1 - 10 = last-first+1 = 10-1=9+1=10.
#s between (not inclusive) 1 - 10 = last-first+1 = 9-2=7+1 = 8

is the step in red ok?

if i were to take even numbers from both values then even numbers in the 10 = 5 and even numbers in the 8 would = 4?

right?

Gurpinder wrote: thx for the reply.

so there is no eq. eh? i gotta do it your way?

#s between (inclusive) 1 - 10 = last-first+1 = 10-1=9+1=10.
#s between (not inclusive) 1 - 10 = last-first+1 = 9-2=7+1 = 8

is the step in red ok?

if i were to take even numbers from both values then even numbers in the 10 = 5 and even numbers in the 8 would = 4?

right?

yeah the red one is correct but u will land up in trouble if u miss these lines

but for even and odd integers we need to include last even or last odd integers

in your ex if u have to find out # of even int b/n 1 to 11 inc. u may end up in same confusion if u do 11-1=(10+1)/2
now thats incorrect as we asked even we have to either take the starting no. as even or the last no. as even which is either 11-2 or 10-1 abnd then add 1 and divide by 2
i hope u got it now

thephoenix wrote:

Gurpinder wrote: thx for the reply.

so there is no eq. eh? i gotta do it your way?

#s between (inclusive) 1 - 10 = last-first+1 = 10-1=9+1=10.
#s between (not inclusive) 1 - 10 = last-first+1 = 9-2=7+1 = 8

is the step in red ok?

if i were to take even numbers from both values then even numbers in the 10 = 5 and even numbers in the 8 would = 4?

right?

yeah the red one is correct but u will land up in trouble if u miss these lines

but for even and odd integers we need to include last even or last odd integers

in your ex if u have to find out # of even int b/n 1 to 11 inc. u may end up in same confusion if u do 11-1=(10+1)/2
now thats incorrect as we asked even we have to either take the starting no. as even or the last no. as even which is either 11-2 or 10-1 abnd then add 1 and divide by 2
i hope u got it now

perfect!

thank you!

Gurpinder wrote:hey guys,

i am getting confused over this silly thing. when a question asks: What is the sum of all the even integers between 99 and 301 ?

To first count the number of integers between 99 and 301, isn't the equation last-first+1?
so then wouldn't the answer be 203?

in the book it says there are 202 terms. so 101 even numbers.

hmm....can you explain please

thx.

The simplest way to count integers, in my opinion, is: [(LAST-FIRST)/Increment]+1 (if inclusive)
and [(LAST-FIRST)/Increment]-1 (if non-inclusive).
Note that for even or odd integers we need to include the last even or last odd integer, depending on the question. eg: How many even integers are there between 16 and 567? Solution: since 567 is odd, the greatest even integer between 16 and 567 is 566, so: [(566-16)/2]+1=(550/2)+1=226