Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 18:13

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If one has 4 digits {1,2,8,9} what are the last two digits

Author Message
Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
Followers: 1

Kudos [?]: 37 [0], given: 0

If one has 4 digits {1,2,8,9} what are the last two digits [#permalink]

### Show Tags

29 May 2005, 15:30
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?
SVP
Joined: 05 Apr 2005
Posts: 1711
Followers: 5

Kudos [?]: 79 [0], given: 0

### Show Tags

29 May 2005, 16:27
Quote:
{1,2,8,9}

i got the followings:

8+2 (or 9+1) =10
9+2 (or 8+1+2) = 11
9+1+2 = 12
9+8 = 17
9+1+8= 18
9+2+8= 19
9+8+2+1= 20
Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

29 May 2005, 17:08
The answer is a two digit number

sum = 1289 + 1298 + 2189 + ... etc
Senior Manager
Joined: 17 Apr 2005
Posts: 373
Location: India
Followers: 1

Kudos [?]: 27 [0], given: 0

### Show Tags

30 May 2005, 07:04
sparky wrote:
If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?

80.

As we would have a total of 256 different no. with a equal representation of each of the 4 digits in all the places, we would have the sum of all units digits as (1+2+8+9)*64 = 1280. So the units digits is 0.
The tens digit is 1280 + 128 ( carry ) = 8 with 140 carried to the hundreth place.Hence 80.

HMTG.
Senior Manager
Joined: 21 Mar 2004
Posts: 444
Location: Cary,NC
Followers: 3

Kudos [?]: 72 [0], given: 0

### Show Tags

30 May 2005, 15:53
sparky wrote:
If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?

Total no of 4 digits nos = 4! = 24.

Each digit comes an equal no of times in units and tens place.

sum of units = (1+2+8+9) * 4 = 80
sum of tens= (1+2+8+9) * 4 = 80

therefore sum of all units and tens = 880

last 2 digits = 80
_________________

ash
________________________
I'm crossing the bridge.........

Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

01 Jun 2005, 10:37
sparky wrote:
If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?

My solution

Here we go,

Sum of units digits is 3!*(1+2+8+9) = 6*20 = 120 => put 0 carry over 12
Sum of tens digits is 3!*(1+2+8+9) + 12 = 120 + 12 = 132 => put 2 carry over 13

tens and units digits are, therefore, 20

The entire sum is 133,320, if I am not mistaken
Senior Manager
Joined: 17 Apr 2005
Posts: 373
Location: India
Followers: 1

Kudos [?]: 27 [0], given: 0

### Show Tags

01 Jun 2005, 21:40
sparky wrote:
sparky wrote:
If one has 4 digits {1,2,8,9} what are the last two digits (tens and units) of a sum of all possible four digit numbers that can be constructed from those 4 digits?

My solution

Here we go,

Sum of units digits is 3!*(1+2+8+9) = 6*20 = 120 => put 0 carry over 12
Sum of tens digits is 3!*(1+2+8+9) + 12 = 120 + 12 = 132 => put 2 carry over 13

tens and units digits are, therefore, 20

The entire sum is 133,320, if I am not mistaken

sparky, I think that you are assuming that the numbers cannot be repeated.

there can be a total of 4^4 numbers, yours just give 4*3!(24)

HMTG.
Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

01 Jun 2005, 21:52
hehe, when I wrote the problem I meant no repeats, and I said in the problem itself that 'all possible four digit numbers that can be constructed from those 4 digits'

if you did it for 4^4, it's cool as long as you understand the concept it all that matters, the rest is details.
Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

03 Jun 2005, 13:19
That's what I got. I don't have an answer from a book for this one.
Senior Manager
Joined: 17 May 2005
Posts: 272
Location: Auckland, New Zealand
Followers: 1

Kudos [?]: 13 [0], given: 0

### Show Tags

03 Jun 2005, 13:20
sparky wrote:
That's what I got. I don't have an answer from a book for this one.

well thats what i got too...so hopefully its right! hehe
03 Jun 2005, 13:20
Display posts from previous: Sort by