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Manager
Joined: 14 Apr 2010
Posts: 219

Kudos [?]: 222 [0], given: 1

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03 May 2010, 00:13
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1. If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990. What is the least possible value of n?

2. if each term is the sum a1+a2+....+an is either 7 or 77 and the sum is equal to 350, which of the following could equal n?
38,39,40,41,42

3. Ben needs to form a committee of 3 from a group of 8 engineers to study designs. if two of the engineers are too inexperienced to serve together on the committee. how many different committees can ben form?

Kudos [?]: 222 [0], given: 1

Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 143 [1], given: 6

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03 May 2010, 11:29
1
KUDOS
]1. If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990. What is the least possible value of n?

Sol - Product of n numbers = $$1*2*3.... n = n ! = 990 * K = 9* 10 *11 * K$$

The least possible value of n will be 11, as 11 is a prime number which is there in the product of the n numbers.

Kudos [?]: 143 [1], given: 6

Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 143 [0], given: 6

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03 May 2010, 11:30
2. if each term is the sum a1+a2+....+an is either 7 or 77 and the sum is equal to 350, which of the following could equal n?
38,39,40,41,42

Sol- I guess, there is some typo error in the question.

for the sum to be 350, the number of terms in the sequence has to be a multiple of 10 ( because 7 * 10 = 70, which also means if we add 7 ten times then only we will get a multiple of 10 in the sum. Its hard for me to explain this ne but if you use some number plugging you will get the idea what I am trying to explain here ..)

So the only options which satisfies this logic is 40. as per the options given.

Kudos [?]: 143 [0], given: 6

Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 143 [0], given: 6

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03 May 2010, 11:30
3. Ben needs to form a committee of 3 from a group of 8 engineers to study designs. if two of the engineers are too inexperienced to serve together on the committee. how many different committees can ben form?

Sol - Number of committees in which 2 inexperienced engineers wont be together = Total number committees - Number of committees in which they are together.

Total number of C = $$8C3 = \frac{8*7*6}{3*2}$$ = 56
Number of C in which they are together = $$1*1* 6 = 6$$

Required number = 56-6 = 50

Please confirm the OA for all the questions...!

Kudos [?]: 143 [0], given: 6

Senior Manager
Joined: 12 Jan 2010
Posts: 255

Kudos [?]: 87 [2], given: 28

Schools: DukeTuck,Kelogg,Darden

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03 May 2010, 11:34
2
KUDOS
Ben needs to form a committee of 3 from a group of 8 engineers to study designs. if two of the engineers are too inexperienced to serve together on the committee. how many different committees can ben form?

2 inexperienced engineers
6 engineers

2 conditions
Condition 1 : All 3 from 6 engineers = C(6,3) = 20 ways

Condition 2
Need 1 from 2 inexperienced engineers = C(2,1) = 2
2 from 6 engineers = C(6,2) = 15
Ways = 15*2 = 30

No of different committees = 30+20 = 50.
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Kudos [?]: 87 [2], given: 28

Manager
Joined: 05 Mar 2010
Posts: 205

Kudos [?]: 37 [1], given: 8

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05 May 2010, 05:25
1
KUDOS
question 2

sum of all is 350 and each term is either 7 or 77
If all the numbers are 7 then n is 50 (not in option)
If one number is 77 then n =40 (350-77 =273) divide it by 7 you get 39. So 39+1 =40

IMO 40
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Kudos [?]: 37 [1], given: 8

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