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9+19+29+39+...+999

by Brent@GMATPrepNow » Sat Jan 23, 2010 3:25 pm
Okay, here's one I just made up.

9+19+29+39+49+ . . . +979+989+999=

A) 49,500
B) 49,800
C) 49,900
D) 50,100
E) 50,400

OA: [spoiler]Rhymes with GLEE :-)[/spoiler]
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by thephoenix » Sat Jan 23, 2010 6:34 pm
IMO E(edited)

my method is a time consuming one
the pattern is

9*100 +(summation 0f 10,20,30,....90)+(summ of 100,110,120.....190)+.....(summ of 900,910,...990)
=450+1450+2450+3450+......9450
=50400(edited)
Last edited by thephoenix on Sat Jan 23, 2010 6:52 pm, edited 2 times in total.
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by papgust » Sat Jan 23, 2010 6:46 pm
Here is my take.

To find the number of terms in the summation series,

999 = 9 + (n-1)*10 [10 is the difference of subsequent terms. 9 is the first term and 999 is the last term]
..
n = 100

Sum of all the terms,

100/2 [2*9 + (100-1)*10]
50 * [1008] = 50,400
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by thephoenix » Sat Jan 23, 2010 6:51 pm
papgust wrote:Here is my take.

To find the number of terms in the summation series,

999 = 9 + (n-1)*10 [10 is the difference of subsequent terms. 9 is the first term and 999 is the last term]
..
n = 100

Sum of all the terms,

100/2 [2*9 + (100-1)*10]
50 * [1008] = 50,400
oopss!! i for got to add 900 in my soln....yes its 50400 ( i edited my post)

and many many thanks to you for framing the eqn
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by sreak1089 » Sun Jan 24, 2010 12:42 am
Two methods to solve it:

1) Recognzie that it is AP (Arithmetic Progression)
Hence, use the formula
nth term = a + (n-1)*d = 9 + (n-1)*10
999 = 10n - 1
10n = 1000 n = 100
Substitute in summation formula:

Sigma (tn) = sigma(10n - 1)
= 10 * (n) * (n+1)/2 - n = (10 * 100 * 101)/2 - 100 = 50500 - 100 = 50400 Hence Eeeeeee :)

2) Recognzie that the it is Equally spaced set:
Hence No. of items = (last - first)/inrement + 1 = (999 - 9)/10 + 1 = 99 + 1 = 100

Since it is equally spaced set, Summation of such a set would be Avg * number of items.
Average of an equally spaced set = (first + last) / 2 = (9 + 999) / 2 = 504
Thus, Sum = 504 * 100 = 50400

Hence E and rhymes with GLEEE :)
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by Stuart@KaplanGMAT » Sun Jan 24, 2010 1:48 am
Brent Hanneson wrote:Okay, here's one I just made up.

9+19+29+39+49+ . . . +979+989+999=

A) 49,500
B) 49,800
C) 49,900
D) 50,100
E) 50,400
Another way we can solve it to simplify the math a bit:

each number in the series is 1 less than a simple number, i.e. 10, 20, 30, 40, ..., 1000.

There are 100 numbers in the series, so we'll have to subtract 100*1 = 100 to get the sum of the series in the question.

We now use the sum of a series of consecutive numbers formula:

sum = ave * # of terms = (10 + 1000)/2 * 100 = 1010 * 50 = 50500

50500 - 100 = 50400... choose (E).

When you have weird numbers in a GMAT question, you can almost always find a way to solve using simpler ones or avoid the math entirely; remember, we're never tested on our ability to do wacky and wild calculations. On this question it's not such a big deal, since there are 100 terms, but a weirder number of terms would have made multiplying by 504 more annoying.
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by sreak1089 » Sun Jan 24, 2010 3:00 am
Awesome !!
Stuart Kovinsky wrote:
Brent Hanneson wrote:Okay, here's one I just made up.

9+19+29+39+49+ . . . +979+989+999=

A) 49,500
B) 49,800
C) 49,900
D) 50,100
E) 50,400
Another way we can solve it to simplify the math a bit:

each number in the series is 1 less than a simple number, i.e. 10, 20, 30, 40, ..., 1000.

There are 100 numbers in the series, so we'll have to subtract 100*1 = 100 to get the sum of the series in the question.

We now use the sum of a series of consecutive numbers formula:

sum = ave * # of terms = (10 + 1000)/2 * 100 = 1010 * 50 = 50500

50500 - 100 = 50400... choose (E).

When you have weird numbers in a GMAT question, you can almost always find a way to solve using simpler ones or avoid the math entirely; remember, we're never tested on our ability to do wacky and wild calculations. On this question it's not such a big deal, since there are 100 terms, but a weirder number of terms would have made multiplying by 504 more annoying.
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by rahul.s » Sun Jan 24, 2010 7:03 am
sreak1089 wrote:Two methods to solve it:

1) Recognzie that it is AP (Arithmetic Progression)
Hence, use the formula
nth term = a + (n-1)*d = 9 + (n-1)*10
999 = 10n - 1
10n = 1000 n = 100
Substitute in summation formula:

Sigma (tn) = sigma(10n - 1)
= 10 * (n) * (n+1)/2 - n = (10 * 100 * 101)/2 - 100 = 50500 - 100 = 50400 Hence Eeeeeee :)

2) Recognzie that the it is Equally spaced set:
Hence No. of items = (last - first)/inrement + 1 = (999 - 9)/10 + 1 = 99 + 1 = 100

Since it is equally spaced set, Summation of such a set would be Avg * number of items.
Average of an equally spaced set = (first + last) / 2 = (9 + 999) / 2 = 504
Thus, Sum = 504 * 100 = 50400

Hence E and rhymes with GLEEE :)
nice approach sreak. quick question though. i didn't quite understand the formula you used for calculating the number of items for evenly spaced numbers. (last - first)/inrement + 1?

didn't quite understand what inrement means. is it increment, i.e. the difference between 2 consecutive numbers?
Last edited by rahul.s on Wed Feb 24, 2010 1:02 am, edited 2 times in total.
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by papgust » Sun Jan 24, 2010 7:25 am
Yeah he meant "increment". It is a shortcut formula to calculate number of terms in a sequence.

If the terms are evenly-spaced say for example, the sequence has the terms that start with 10 and progress with 15, 20, 25,.. till 100. This sequence has each term evenly-spaced by 5 units. To calculate number of terms for these kinds of sequences,

(last term - first term)/increment + 1

Here 5 is the increment.

[(100 - 10)/5] + 1 = 18 + 1 = 19 terms
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by rahul.s » Sun Jan 24, 2010 8:38 am
papgust wrote:Yeah he meant "increment". It is a shortcut formula to calculate number of terms in a sequence.

If the terms are evenly-spaced say for example, the sequence has the terms that start with 10 and progress with 15, 20, 25,.. till 100. This sequence has each term evenly-spaced by 5 units. To calculate number of terms for these kinds of sequences,

(last term - first term)/increment + 1

Here 5 is the increment.

[(100 - 10)/5] + 1 = 18 + 1 = 19 terms
this is definitely worth knowing! thanks
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