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joemama142000
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the 22 Feb 2006, 16:03
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the following is in the sequence?

105,786,966, 1025



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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the 22 Feb 2006, 18:10
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the following is in the sequence?

105,786,966, 1025
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I can't find a shortcut
so applying brute force, substract 67,66,64 from 105,786,966,1025 if the number can be divided by eight that is your answer

you know it has to be even so that helps you to eliminate quickly some options

if you try 786-66 = 720 that is a multiple of 8 so there is your answer

it took me about 30 seconds to solve it but i think was a bit of luck
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the 22 Feb 2006, 18:15
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Only 786...

subtract 64, 66 and 67 (One by one) from each number and then divide the new number by 8... if the quotient is an integer then its in the sequence...
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the 23 Feb 2006, 22:25
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ps_dahiya
Only 786...

subtract 64, 66 and 67 (One by one) from each number and then divide the new number by 8... if the quotient is an integer then its in the sequence...
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How did you know to do that? :?
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the 23 Feb 2006, 23:06
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jlui4477
ps_dahiya
Only 786...

subtract 64, 66 and 67 (One by one) from each number and then divide the new number by 8... if the quotient is an integer then its in the sequence...

How did you know to do that? :?
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a1 = 64
a2 = 66
a3 = 67
a4 = 8 + a1 = 72
a5 = 8 + a2 = 74
a6 = 8 + a3 = 75
a7 = 8 + a4 = 80
a8 = 8 + a5 = 82
a9 = 8 + a6 = 83

Now there are three sequences

1. a1, a4, a7 i.e 64, 72, 80....... First term is 64 and difference bw terms is 8
2. a2, a5, a6 i.e 66, 74, 82....... First term is 66 and difference bw terms is 8
3. a3, a6, a9 i.e 67, 75, 83....... First term is 67 and difference bw terms is 8

Nth term of the sequence is = a + (n-1)d

Now lets test the given numbers
105: If there then can only be in seq 3 because odd number will be in seq 3 only. So 105 = 67 + (n-1) * 8 solve this and n in not an integer. So this can not be in any sequence

786: This can only be in seq 1 and 2 because its even number. So 786 = 64 + (n-1) * 8. Solving this we get n as non-integer. So this is not in seq1. Now test for seq 2. 786 = 66 + (n-1) * 8 . Bingo, we get n = 91. So this will be in sequence

966: Can only be in seq1 and seq2. Test for seq1 and seq2 and we get n as non-integer in both seq

1025: Can only be in seq3. Test and we get n as non-integer.
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A sequence, a1=64, a2 =66, a3=67, an=8 + an-3, which of the 24 Feb 2006, 23:40
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Oh, ok! Thanks! When I saw an=8+an- 3. I took it as the number an represented minus 3 and not n-3. :oops:



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