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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Thu Oct 10, 2013 7:08 am

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

OA [spoiler]{B}[/spoiler]

R A H U L

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Source: Beat The GMAT — Problem Solving | Thu Oct 10, 2013 7:08 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 10, 2013 7:22 am

theCodeToGMAT wrote: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

OA [spoiler]{B}[/spoiler]

Let's examine a few terms and look for a pattern.
02, 07, 22, 67, 202, 607, 1822, 5467, . . .

So, the pattern goes 02, 07, 22, 67, etc.

This pattern repeats every 4 terms.
So, the 4th term ends in 67
The 8th term ends in 67
The 12th term ends in 67
.
.
.
The 32nd term ends in 67
From here, the 33rd term ends in 02
The 34th term ends in 07
The 35th term ends in [spoiler]22[/spoiler]

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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sanjoy18: Senior | Next Rank: 100 Posts; Posts: 81; Joined: Tue Jun 11, 2013 10:24 pmFollowed by: 1 member; Thanked: 7 times

by sanjoy18 » Thu Oct 10, 2013 12:15 pm

Lets first term T1, second term T2 ..so on
I found the below pattern
T1= 2 = (5*3^0 -1)/2
T2 = 7 = (5*3^1-1)/2
T3=22= (5*3^2 -1)/2
........................
T35= (5*3^34-1)/2

Now our job is to find the last two digit of 3^34

Last two digit can be found very by binomial expansion
3^34= (3^2)17
=(9)^17
=(10-1)^17
last two digit would be 17c1 *10 -1 = 69

hence last two digit of T35= (69*5-1)/2= 44/2=22

hence answer 2+2=4
hence B

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