BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Counting Problems

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Wed Jun 06, 2007 10:19 am

Counting Problems

by light_speed » Sat Mar 01, 2008 7:57 am
if n=8^11 -8, what is the units digit of n?
A 0
B 1
C 4
D 6
E 8

Ans: C

In the sequence 1,2,4,8,16,32...each term after the first is twice the previous term. What is the sum of the 16th,17th and 18th term in the sequence?

A. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)

Ans:E

If 10^50 -74 is written as an integer is base 10 notation, what is the sum of the digits in that integer

A. 424
B 433
C 440
D 449
E 467

Ans: C
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Sun Jul 01, 2007 5:56 pm
Thanked: 3 times

by smushkas » Sat Mar 01, 2008 8:57 am
Question 1.
You should keep track of only units digits, or only multiply them.
8^2 = 64
8^3 = 512____________==> 4 * 8 = 32
8^4 = 4'096___________==> 2 * 8 = 16
8^5 = 32'768__________==> 6 * 8 = 48
8^6 = 262'144_________==> 8 * 8 = 64 and so on..
As you can see there is a pattern of repeating numbers [ 4, 2, 6, 8 ], so 10th ==> [ 8 * 8 ] = 64, 11th ==> [ 4 * 8 ] = 32. We really do not care about the tenth digit (it could be any integer from 0 to 9), we must figure out unit's digit, which is 2.This is enough for us to determine the unit's digit of the sum, 32 - 8 = 24 and units digit is 4.

Question 3.

10^50 = 1 with 50 zeroes, right? If we subtract 74 from this number we'll have 6, 2, and 48 numbers of 9.
Then we can calculate the sum ==> [ 48 * 9] + 6 + 2 = 432 + 8 = 440

Hope these might help!
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Sun Jul 01, 2007 5:56 pm
Thanked: 3 times

by smushkas » Sat Mar 01, 2008 11:27 am
Question 2.

To find a nth term in the sequence of numbers where "each term after the first is twice the previous term", we use a formula n = 2^(n-1). We need to find values of terms [ 2^16] , [ 2^17 ] , [ 2^18 ]. According to our formula values of 16th = 2^(16 - 1), 17th = 2^(17 - 1), and 18th = 2^(18 - 1).
This can be written like ===> [ 2^15 + 2^16 + 2^17 ] = [ 2^15 + 2 * 2^15 + 2 * 2 * 2^15 ] = [ 2^15 + 2 * 2^15 + 2^2 * 2^15 ]
Then we take out from the bracket term 2^15 ===> 2^15 * [ 1 + 2 + 4 ] = 2^15 * [ 7 ] = 7 * [2^15]

Hope it is clear!
Top
Legendary Member
Posts: 789
Joined: Sun May 06, 2007 1:25 am
Location: Southern California, USA
Thanked: 15 times
Followed by:6 members

my way.. counting

by resilient » Sun Mar 30, 2008 12:22 pm
Smushkas has a nice approach but I am curious why you are multiplying the units digit with by 8 when you are supposed to subtract.

There is an annoying flaw here where brackets were supposed to be used in the question.

my way:
8^1 = 8 8-8 =0
8^2 = 64 64-8=56
8^3 = 512 512-8= 504
8^4 = 4'096 4096-8= 4088 (now know the sequence)
8^5 = 32'768 32768-8= 32760
8^6 = 262'144 262,138 (sequence confirmed)

suquence is 0,6,4,8 0,6,4,8,0,6,4,8

eleventh number in sequence is 4


BUt I am curious why the other way!
Appetite for 700 and I scraped my plate!
Top
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Sat Aug 16, 2008 7:10 pm
Thanked: 1 times

by ohwell » Wed Nov 05, 2008 7:44 pm
Hi everyone, I know this was dealt with a long time ago, but I thought I could add something that makes it even easier, IMO. I really don't want to calculate powers of 8, so, I simplified it to powers of 2. After all, 8 = 2^3.

So, I ultimately arrive at 8 * (2^30 - 1). I then follow the same approach as you:

2^1 - 1 = 1
2^2 -1 =3
2^3 - 1 = 7
2^4 -1 = 15

This is the cycle. So, I am getting to a 3 as the units digit for 2^30 but still need to multiply by the 8, and 8 * 3 = 24, with a new unit digit of 4.

Cheers
Top
Legendary Member
Posts: 1159
Joined: Wed Apr 16, 2008 10:35 pm
Thanked: 56 times

by raunekk » Wed Nov 05, 2008 10:03 pm
nice approach @ ohwell..
Top
Legendary Member
Posts: 829
Joined: Mon Jul 07, 2008 10:09 pm
Location: INDIA
Thanked: 84 times
Followed by:3 members

Re: Counting Problems

by sudhir3127 » Wed Nov 05, 2008 10:38 pm
light_speed wrote:if n=8^11 -8, what is the units digit of n?
A 0
B 1
C 4
D 6
E 8

Ans: C

In the sequence 1,2,4,8,16,32...each term after the first is twice the previous term. What is the sum of the 16th,17th and 18th term in the sequence?

A. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)

Ans:E

If 10^50 -74 is written as an integer is base 10 notation, what is the sum of the digits in that integer

A. 424
B 433
C 440
D 449
E 467

Ans: C
1st question is a cyclicity problem
we know its 4 in case of 8 ..

so we can deduce the equation to
8^3 - 8
512- 8 = hence the units digit is 4.

thus C .

2: its in Gometric mean ..

16th term = a*r^n-1
a = 1
r = 2
n 16
16th term is
2^15

similarly

17th term is 2^16
18th term is 2^17

2^15 + 2^16 + 2^17
2^15 ( 1+2+4)

hence 7*2^15.

3rd question is also cyclicity

we need to understand a simple logic here

10^3 - 74 = 926 (one 9)
10 ^ 4 - 74 = 9926 (two 9s)
10 ^ 5 - 74 = 99926 (three 9s)

thus if we follow this pattern till 50 we get..

10^50 - 74 = 48 (9s) followed by 26

Hence it is 48*9 + 2 + 6 = 440

hope it makes sense .. do let me know if u have any doubts,,,

And please make sure u post only one question per post..
Top
Post new topic Post Reply
• Page 1 of 1