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In the first week of the Year, Nancy saved $1. [#permalink]

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11 Nov 2012, 18:35

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In the first week of the Year, Nancy saved $1. In each of the next 51 weeks, she saved $1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

In the first week of the Year, Nancy saved $1. In each of the next 51 weeks, she saved $1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. $1,326 B. $1,352 C. $1,378 D. $2,652 E. $2,756

Does anyone have a better way to do this?

The total amount of money will be 1 + 2 + 3 + 4 + ... + 52 (in the 52nd week, she will save $52)

Sum of first n consecutive positive integers = n*(n+1)/2 Sum = 52*53/2 = 26*53 The product will end with 8 since 6*3 = 18 so answer must be (C)
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Re: In the first week of the Year, Nancy saved $1. [#permalink]

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19 May 2013, 09:00

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For an even list of consecutive numbers where a = the first number and z = the last number and (in this case) a total of 52 consecutive numbers-- (a+z)(.5*number of consecutive integers)---> (1+52)*26=1378
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Re: In the first week of the Year, Nancy saved $1. [#permalink]

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03 Jun 2014, 06:14

How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

The sum of first n consecutive positive integers is given by the formula n(n+1)/2. For GMAT, it is a good idea to know this formula. It could simplify many calculations. Test prep companies discuss all such useful formulas in their curriculum. The Official Guides only give practice questions.

You should also know the more generic formula of sum of an Arithmetic Progression. From that you can easily derive this formula.

Sum of an Arithmetic Progression will be n*Average where n is the number of terms in the AP and Average will be the average value of the terms. The average value can be found as (First term + Last term)/2

If first term is a, last term is a + (n-1)*d where d is the common difference.

Average = (a + a + (n-1)*d)/2

Sum = n*(a + a + (n-1)*d)/2 = n/2(2a + (n-1)*d)

In case of consecutive integers starting from 1, a = 1 and d = 1

Re: In the first week of the Year, Nancy saved $1. [#permalink]

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03 Jun 2014, 22:47

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momentofzen wrote:

How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

If you don't recall the formula, there is one method:

Re: In the first week of the Year, Nancy saved $1. [#permalink]

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04 Jun 2014, 11:08

PareshGmat wrote:

momentofzen wrote:

How did you know about that formula (n(n+1))/2 ? I've read through the GMAT Official Guide Math Review but don't recall learning that. It seems if you know that formula the question is very easy, but if you don't the problem can be a big time drain. Do you recommend other guides I should be reading to study up on tips and formulas for problems like these?

If you don't recall the formula, there is one method:

Sum of 3rd & 3rd last digit = 3 + 50 = 53 . . . . This is repeated 26 times

So, 26 * 53 = 1378

Answer = C

Yes! Definitely better/easier/faster to apply logic than to memorize another formula. This way, you can use the logic even if you don't start at one. Or if you are counting by 2's or 3's. For example, what if Nancy saved $51 the first week of the year, and increased her savings by $1 each week after that.

Start at $51. Know that there are 52 weeks in a year, so the last week she will save 102. So the sequence will be 51, 52, 53... 100, 101, 102. See how adding the first and last is the same as adding the second and the second-to-last, and the same as the third and third-to-last? All like pairs are $153. If there are 52 items, then there are 26 pairs. $153 x 26 = $3978

Counting by 3's: First week is $51. Last week will be (52 - 1) x 3 = 153 more, or 51+153=204. So: 51, 54, 57... 198, 201, 204. First + last: 51 + 204 = $255. Times 26 pairs: $255 x 26 = $6630

Counting by 3's, odd number of weeks: Let's say Nancy only saved for 51 weeks, because she bought holiday presents for everyone in the last week. First week is $51. Last week will be (51 - 1) x 3 = 150 more, or 51+150=201. So: 51, 54, 57... 195, 198, 201. First + last: $51 + $201 = $252. Now, there are 25 pairs + 1. The one is in the very middle. So $252 x 25 = $6300. Still have to add that last number that wasn't part of a pair. What was it? Nancy saved for 51 weeks, so that's 25 weeks on one side, 25 on the other, the middle is the 26th week. 26th week is (26 - 1) x $3 = $75 more than week 1, or 51 + 75 = 126. $6300 + $126 = $6426

Yes! Definitely better/easier/faster to apply logic than to memorize another formula. This way, you can use the logic even if you don't start at one. Or if you are counting by 2's or 3's. For example, what if Nancy saved $51 the first week of the year, and increased her savings by $1 each week after that.

It's certainly better to apply logic than just learn up formulas for specific situation because you may find you have no formula for a given situation in the question. Also, you need to know exactly when the formula is applicable for example here you must know that the formula is applicable for first n positive integers only.

That said, n(n+1)/2 is a very basic and useful formula. You should know it for GMAT, not because you may not be able to do a question without it but because you may spend unnecessary amount of time on a question when other people will be able to run away on it with the formula.
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In the first week of the Year, Nancy saved $1. In each of the next 51 weeks, she saved $1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. $1,326 B. $1,352 C. $1,378 D. $2,652 E. $2,756

Let's first set up the pattern of Nancy's savings. The first week she saved $1, the second week she saved $2, the third week she saved $3, and so forth. Therefore, the total amount of money she will have saved at the end of 52 weeks will be: $1 + $2 + $3 + $4 + … + $52. The pattern is obvious, but the arithmetic looks daunting because we need to add 52 consecutive integers. To shorten this task, we can use the formula: sum = average x quantity.

We know that Nancy saved money over the course of 52 weeks, so our quantity is 52.

To determine the average, we add together the first amount saved and the last amount saved and then divide by 2. Remember, this technique only works when we have an evenly spaced set.

The first quantity is $1 and the last is $52. Thus, we know:

average = (1 + 52)/2 = 53/2

Now we can determine the sum.

sum = average x quantity

sum = (53/2) x 52

sum = 53 x 26 = 1,378

Answer is C.

Note: If we did not want to actually multiply out 26 x 53, we could have focused on units digits in the answer choices. We know that 26 x 53 will produce a units digit of 8 (because 6 x 3 = 18), and the only answer choice that has a units digit of 8 is answer choice C.
_________________

Scott Woodbury-Stewart Founder and CEO

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In the first week of the Year, Nancy saved $1. In each of the next 51 weeks, she saved $1 more than she had saved in the previous week. What was the total amount that Nancy saved during the 52 weeks?

A. $1,326 B. $1,352 C. $1,378 D. $2,652 E. $2,756

Here's another approach (that doesn't require formulas).

We want to add 1+2+3+4+...+51+52 So, let's add them in pairs, starting from the outside and working in. 1+2+3+4+...+51+52 = (1+52) + (2+51) + (3+50) + . . . = 53 + 53 + 53 + ....

How many 53's are there in our new sum? Well, there are 52 numbers in the sum 1+2+3+..+52, so there must be 26 pairs, which means there are 26 values in our new sum of 53 + 53 + 53 + ....

So, what is (26)(53)? Fortunately, if we examine the answer choices, we see that we don't even need to calculate (26)(53)

Why not? Notice that when we multiply (26)(53), the units digit in the product will be 8 (since 6 times 3 equals 18).

Since only 1 answer choice ends in 8, the correct answer must be

I often get confused between n(n-1)/2 and n(n+1)/2 for consecutive integers.

When do we use n(n-1)/2 could you please explain a little. Thanks in advance.
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-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- When nothing seem to help, I would go and look at a Stonecutter hammering away at his rock perhaps a hundred time without as much as a crack showing in it. Yet at the hundred and first blow it would split in two. And I knew it was not that blow that did it, But all that had gone Before.