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M08-35

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Math Expert
Joined: 02 Sep 2009
Posts: 46167

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16 Sep 2014, 00:38
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Difficulty:

35% (medium)

Question Stats:

81% (01:13) correct 19% (01:54) wrong based on 90 sessions

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If operation $$@X$$ is defined as $$@X = X + 2$$ if $$X$$ is even and $$@X = X - 1$$ if $$X$$ is odd, what is $$@(...@(@(@(15)))...)$$ 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

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Math Expert
Joined: 02 Sep 2009
Posts: 46167

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16 Sep 2014, 00:38
1
1
Official Solution:

If operation $$@X$$ is defined as $$@X = X + 2$$ if $$X$$ is even and $$@X = X - 1$$ if $$X$$ is odd, what is $$@(...@(@(@(15)))...)$$ 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

$$@(15) = 14$$. We have to compute $$@(...@(@(@(14)))...)$$ 98 times. Each $$@$$ just adds 2 to the previous result. Therefore, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

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Joined: 06 Mar 2014
Posts: 257
Location: India
GMAT Date: 04-30-2015

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03 Feb 2015, 08:30
Bunuel wrote:
Official Solution:

If operation $$@X$$ is defined as $$@X = X + 2$$ if $$X$$ is even and $$@X = X - 1$$ if $$X$$ is odd, what is $$@(...@(@(@(15)))...)$$ 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

$$@(15) = 14$$. We have to compute $$@(...@(@(@(14)))...)$$ 98 times. Each $$@$$ just adds 2 to the previous result. Therefore, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

if 14 is the first term. common difference (d) = 2 and there are 98 terms then
why Last term is not = 14 + (98-1)(2) ??
why 99 used instead of 98 in the formula : T(last term) = a + (n-1) d
Math Expert
Joined: 02 Sep 2009
Posts: 46167

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03 Feb 2015, 10:10
earnit wrote:
Bunuel wrote:
Official Solution:

If operation $$@X$$ is defined as $$@X = X + 2$$ if $$X$$ is even and $$@X = X - 1$$ if $$X$$ is odd, what is $$@(...@(@(@(15)))...)$$ 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

$$@(15) = 14$$. We have to compute $$@(...@(@(@(14)))...)$$ 98 times. Each $$@$$ just adds 2 to the previous result. Therefore, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

if 14 is the first term. common difference (d) = 2 and there are 98 terms then
why Last term is not = 14 + (98-1)(2) ??
why 99 used instead of 98 in the formula : T(last term) = a + (n-1) d

1. $$@(15) = 14$$
2. $$@(14) = 14 +2=16$$
3. $$@(16) = 14 +2+2=14+2*2=18$$
4. $$@(18) = 14 +2+2+2=14+2*3=20$$
...
99. 14 + 2*98 = 210.
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Joined: 02 Nov 2015
Posts: 21

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14 Aug 2016, 08:40
Bunuel,

Are we sure that this is a 500 level question? Generally, 500 level questions do not involve calculations trick- they are pretty straight forward. I was taking GMAT club test (the 11th set) and found several similar questions to be ranked at 500 level and probably that is the reason the score is not upto the mark.

Just a thought!

Regards
Yash
Manager
Joined: 30 Jun 2015
Posts: 61
Location: Malaysia
Schools: Babson '20
GPA: 3.6

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09 Dec 2016, 18:18
If 14 is the first term. common difference (d) = 2 and there are 98 terms then
why is the Last term not = 14 + (98-1)(2) ?? n=98.. and 14 is the 99th term.

T(last term) = a + (n-1) d

Manager
Joined: 12 Jun 2016
Posts: 221
Location: India
WE: Sales (Telecommunications)

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21 May 2017, 09:35
HI earnit

I too had the same doubt. It took me some time to realise that the first term of the AP is not 14. In fact, the first term is @(14) = 16. After this, everything made sense.

We have $$Tn = a + (n-1)d$$. Where, Tn = Nth term. a = First term, n = number of terms in the sequence, d = common difference of AP
$$Tn = 16+(98-1)*2 = 16+97*2 = 16+ 194 = 210$$.

Lesson Learnt : I now think its better to write the series as Bunuel did here. Just so that the "short cut" of using nth term does not lead me into a trap!

earnit wrote:
Bunuel wrote:
Official Solution:

If operation $$@X$$ is defined as $$@X = X + 2$$ if $$X$$ is even and $$@X = X - 1$$ if $$X$$ is odd, what is $$@(...@(@(@(15)))...)$$ 99 times?

A. 120
B. 180
C. 210
D. 225
E. 250

$$@(15) = 14$$. We have to compute $$@(...@(@(@(14)))...)$$ 98 times. Each $$@$$ just adds 2 to the previous result. Therefore, $$@(...@(@(@(14)))...) 98 \text{ times} = 14 + 2*98 = 210$$.

if 14 is the first term. common difference (d) = 2 and there are 98 terms then
why Last term is not = 14 + (98-1)(2) ??
why 99 used instead of 98 in the formula : T(last term) = a + (n-1) d

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My Best is yet to come!

Intern
Joined: 24 Oct 2016
Posts: 29

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14 Sep 2017, 20:38
I think by A.P. formula T(last term) = a + (n-1)d
the last term should be = 14 + (98-1)(2)=208.
210 is not correct.
Math Expert
Joined: 02 Sep 2009
Posts: 46167

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14 Sep 2017, 22:12
kishor1 wrote:
I think by A.P. formula T(last term) = a + (n-1)d
the last term should be = 14 + (98-1)(2)=208.
210 is not correct.

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Re: M08-35   [#permalink] 14 Sep 2017, 22:12
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