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 Post subject: GMAT Sequences
PostPosted: Fri May 03, 2013 9:20 am 
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The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirty-fifth term in the sequence?

A. 2
B. 4
C. 7
D. 9
E. 13

(B) (To understand units and tens, in 75 7 is tens and 5 is units.) We cannot reasonably be expected to write the sequence to the thirty-fifth term. We should therefore expect a repeating pattern within the tens and units digits of the sequence.

Let’s write out the first 8 terms and see what the pattern is:
2, 7, 22, 67, 202, 607, 1822, 5467.
Examining the tens and units digits, we see that the following four-term pattern repeats in those digits:
02, 07, 22, 67, etc.

To find what the tens and units digits of the thirty-fifth term will be, we must first divide the term number (35) by the number of terms in the repeating sequence (4):
35/4 = 8 remainder 3.
This means the four-term sequence fully repeats 8 times, and the remainder tells us how many terms into the repeating sequence the term in question will be.
Three terms into the repeating sequence, the tens and units digits are both 2. The sum of these digits is the answer to the question:
2 + 2 = 4.

The correct answer is choice (B).
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The repeating terms i can understand, but how did you figure out that the tens and units of the number is 2?


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 Post subject: Re: GMAT Sequences
PostPosted: Fri May 03, 2013 9:21 am 
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Quote:
The repeating terms i can understand
We do not have the repeating terms themselves, but we have repeating two last digits.
2, 7, 22, 67, 202, 607, 1822, 5467.

Quote:
but how did you figure out that the tens and units of the number is 2?

We know that the terms of the sequence have a pattern:
...02, ...07, ...22, ...67, ...02, ...07, ...22, ...67, ...02, ...07, ...22, ...67.
So the four endings (02, 07, 22, 67) repeat. Therefore we know that the 4th, 8th, 12th, ... (4n) -th terms end with 67; the 3rd, 7th, 11th, ... (4n + 3)-th terms end with 22, etc.

35 = 32 + 3 = 4 × 8 + 3
So the 32nd term ends with 67; 33rd term ends with 02; 34th term ends with 07 and the 35th term ends with 22.


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The strict mathematical proof of why this pattern will hold:
The last two digits of the resulting number in case of multiplication by 3 and addition of 1 depend solely on the last two digits of the original number.

Suppose R = ab....cxyz, where each letter represents a digit, and there could be any number of any digits in between b and c. We can rewrite R as:
R = ab....cxyz = ab....cx00 + yz

We can see that digits ab....cx do not affect last two digits if R is multiplied by an an integer or if to add a positive number.


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