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# On January 1, Jamie did one sit-up. On each of the next 364 days

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Joined: 28 May 2014
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On January 1, Jamie did one sit-up. On each of the next 364 days  [#permalink]

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08 May 2017, 07:19
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Question Stats:

78% (01:19) correct 22% (01:30) wrong based on 171 sessions

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On January 1, Jamie did one sit-up. On each of the next 364 days, she did one more than on the previous day. What was the total number of sit-up she did in 365 days?

(A) 66430
(B) 66795
(C) 66978
(D) 67160
(E) 132860

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Re: On January 1, Jamie did one sit-up. On each of the next 364 days  [#permalink]

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08 May 2017, 07:51
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saswata4s wrote:
On January 1, Jamie did one sit-up. On each of the next 364 days, she did one more than on the previous day. What was the total number of sit-up she did in 365 days?

(A) 66430
(B) 66795
(C) 66978
(D) 67160
(E) 132860

Day 1: 1 sit-up
Day 2: 2 sit-ups
Day 3: 3 sit-ups
Day 4: 4 sit-ups
.
.
.
Day 365: 365 sit-ups

So, we want the sum 1 + 2 + 3 + 4 + . . . 364 + 365

We have a nice formula for the sum of the first n positive integers.
It goes like this: 1 + 2 + 3 + . . . + n = (n)(n + 1)/2

So, 1 + 2 + 3 + 4 + . . . 364 + 365 = (365)(365 + 1)/2
= (365)(366)/2
= (365)(183)

STOP! Before you calculate (365)(183), check the answer choice to see if you can avoid this tedious calculation.
Examine the units digits of (365)(183)

Since 5 x 3 = 15, the units digit of (365)(183) will be 5

Only answer choice B meets this condition, so it must be the correct answer.

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Re: On January 1, Jamie did one sit-up. On each of the next 364 days  [#permalink]

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08 May 2017, 08:08
I should have known about the 5*3. I went ahead with the multiplication instead. Have to remember these properties/shortcuts

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Re: On January 1, Jamie did one sit-up. On each of the next 364 days  [#permalink]

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08 May 2017, 11:11
saswata4s wrote:
On January 1, Jamie did one sit-up. On each of the next 364 days, she did one more than on the previous day. What was the total number of sit-up she did in 365 days?

(A) 66430
(B) 66795
(C) 66978
(D) 67160
(E) 132860

sum of N natural number = N(N+1)/2
So( 365 * 366 )/2
we will get 365 * 183 , no need to do multiplication also ...unit digit have to be 5..only option is B
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On January 1, Jamie did one sit-up. On each of the next 364 days  [#permalink]

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30 Jul 2018, 09:12
1
saswata4s wrote:
On January 1, Jamie did one sit-up. On each of the next 364 days, she did one more than on the previous day. What was the total number of sit-up she did in 365 days?

(A) 66430
(B) 66795
(C) 66978
(D) 67160
(E) 132860

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$

$$a$$ = 1 the first term
$$d$$ = 1 distance
$$n$$ = 364 how many terms to add up

$$\frac{364}{2} (2*1+(364-1)1)$$= 365*182 = 66,430

now i need to add 365, cause one sit-up was not counted. 66,430 +365 = 66,795

pushpitkc is it valid solution ?
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On January 1, Jamie did one sit-up. On each of the next 364 days  [#permalink]

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30 Jul 2018, 09:37
1
1
dave13 wrote:
saswata4s wrote:
On January 1, Jamie did one sit-up. On each of the next 364 days, she did one more than on the previous day. What was the total number of sit-up she did in 365 days?

(A) 66430
(B) 66795
(C) 66978
(D) 67160
(E) 132860

SUM OF N TERMS = $$\frac{n}{2} (2a+(n-1)d)$$

$$a$$ = 1 the first term
$$d$$ = 1 distance
$$n$$ = 364 how many terms to add up

$$\frac{364}{2} (2*1+(364-1)1)$$= 365*182 = 66,430

now i need to add 365, cause one sit-up was not counted. 66,430 +365 = 66,795

pushpitkc is it valid solution ?

Hi dave13

Most of the stuff is correct - the formula is correct, a = d = 1(is correct), but n = 365(it's not 364)

Substituting values, we get $$\frac{365}{2}(2+(365-1)1) = \frac{365*366}{2} = 365*183$$

So, you will get the answer 66795 as it is - without having to add one sit-up which was counted

Hope this helps you!
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On January 1, Jamie did one sit-up. On each of the next 364 days &nbs [#permalink] 30 Jul 2018, 09:37
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# On January 1, Jamie did one sit-up. On each of the next 364 days

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