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# Two non-decreasing sequences of nonnegative integers have different fi

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Joined: 02 Sep 2009
Posts: 59725
Two non-decreasing sequences of nonnegative integers have different fi  [#permalink]

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20 Mar 2019, 05:04
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Difficulty:

55% (hard)

Question Stats:

35% (01:47) correct 65% (02:50) wrong based on 17 sessions

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Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is N. What is the smallest possible value of N ?

(A) 55
(B) 89
(C) 104
(D) 144
(E) 273

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Posts: 8320
Re: Two non-decreasing sequences of nonnegative integers have different fi  [#permalink]

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20 Mar 2019, 07:05
Bunuel wrote:
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is N. What is the smallest possible value of N ?

(A) 55
(B) 89
(C) 104
(D) 144
(E) 273

Let the sequences be
a, b, c, d, e, f, g, h.., and A, B, C, D, E, F, G, H, ...
So, g and G are equal...

As the two sequences are non-negative, we will take the least of one sequence be 0, so let a be 0.
Now, we can write g and G in terms of the first two terms as the later terms are sum related to first two terms..
let us take A, B, C, D, E, F, G, H,..
C=A+B
D=B+C=(A+B)+B=A+2B
E=C+D=(A+B)+(A+B)+B=2(A+B)+B
F=D+E=((A+B)+B)+(2(A+B)+B)=3(A+B)+2B
G=E+F=(2(A+B)+B)+(3(A+B)+2B)=5(A+B)+3B=5A+8B..

Similarly 5a+8b=g

But g=G, so 5a+8b=5A+8B..
We have taken a as 0, so 8b=5A+8B......8(b-B)=5A... here the coefficients 5 and 8 are relative prime.
Thus, A is a multiple of 8 and b-B is a multiple of 8.....
We are looking for least value, so let A be the least possible multiple of 8.. It cannot be 0 as both sequence start with different number.
Thus A=8...(b-B)=5, .. again least values will be when B =8 as it cannot be less than A.. so b=8+5=13

G = 5A+8B=5*8+8*8=40+64=104
let us check g too .. 5a+8b=0*5+8*13=104

C
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Re: Two non-decreasing sequences of nonnegative integers have different fi  [#permalink]

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22 Mar 2019, 09:10
Bunuel wrote:
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is N. What is the smallest possible value of N ?

(A) 55
(B) 89
(C) 104
(D) 144
(E) 273

We can let the first sequence be a, b, c, d, e, f, g, … and the second sequence be r, s, t, u, v, w, x, ….

Notice that in the first sequence, we have:

c = a + b

d = b + c = b + (a + b) = a + 2b

e = c + d = (a + b) + (a + 2b) = 2a + 3b

f = d + e = (a + 2b) + (2a + 3b) = 3a + 5b

g = e + f = (2a + 3b) + (3a + 5b) = 5a + 8b

As we can see from the above, the 7th term of the first sequence, which is g, can be written in terms of the first two terms as 5a + 8b. Using the same logic, we can express the 7th term of the second sequence, which is x, in terms of r and s as 5r + 8s. That is,

5a + 8b = g and 5r + 8s = x

We need to have g = x. So we need to find two sets of distinct nonnegative integers {a, b} and {r, s} such that 5a + 8b = 5r + 8s provided that 0 ≤ a ≤ b and 0 ≤ r ≤ s.

Let’s rewrite the equation as:

5a - 5r = 8s - 8b

5(a - r) = 8(s -b)

Since we are looking for the smallest value of N (the 7th term of each sequence), we need to keep a, b, r and s as small as possible. To that end, let’s let a - r = 8 and s - b = 5, or, equivalently, a = r + 8 and s = b + 5. If we let r = 0, then a = 8. Since b is greater than or equal to a, the smallest value we can obtain for b is also 8. Then, s = 8 + 5 = 13. Let’s work the sequences using the values we obtained above:
a = 8, b = 8, 16, 24, 40, 64, 104

r = 0, s = 13, 13, 26, 39, 65, 104

As we can see, the smallest value of N is 104.

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Re: Two non-decreasing sequences of nonnegative integers have different fi   [#permalink] 22 Mar 2019, 09:10
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