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# 3/ If each term in the sum a1 + a2 + a3 +....+an is either 7

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SVP
Joined: 05 Jul 2006
Posts: 1512
Followers: 5

Kudos [?]: 240 [0], given: 39

3/ If each term in the sum a1 + a2 + a3 +....+an is either 7 [#permalink]

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15 Oct 2006, 09:53
3/ If each term in the sum a1 + a2 + a3 +....+an is either 7 or 77 and the sum equals 350, what could be equal to n?

350 = 7*50

7+77+7+77+7+77+7+77+7+7 = 7(12*4+1+1) = 350

N COULD BE 10
Manager
Joined: 02 Oct 2006
Posts: 54
Location: Paris
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Kudos [?]: 0 [0], given: 0

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15 Oct 2006, 10:15

I don't get it...
Manager
Joined: 01 Oct 2006
Posts: 242
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Kudos [?]: 7 [0], given: 0

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15 Oct 2006, 11:41
Assume there are n1 terms of 7, and n2 terms of 77.

7n1 + 77n2= 350
n1+11n2=50

If n2=1,n1=39. Thus n1+n2=40
If n2=2, n1=28 Thus n1+n2=30
If n2=3, n1= 17. Thus n1+n2 = 20
If n2=4, n1=6. Thus n1+n2=10

Are there any options given for the answer?
Manager
Joined: 02 Oct 2006
Posts: 54
Location: Paris
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Kudos [?]: 0 [0], given: 0

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16 Oct 2006, 01:47
Yes, it must be so...

I think the options were...

38, 40, 42, 45, 46

Something like that
Senior Manager
Joined: 05 Oct 2006
Posts: 267
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Kudos [?]: 13 [0], given: 0

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16 Oct 2006, 02:06
n could be 10,20,30,40

n,no of 7s,no of 77s = 10,30,1
20,28,2
30,17,3
40,6,4
Manager
Joined: 02 Oct 2009
Posts: 197
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Kudos [?]: 18 [0], given: 4

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10 Jan 2010, 18:53
Nice work, got me stumped...
Re: zAMZIM POST 3   [#permalink] 10 Jan 2010, 18:53
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