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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If one has 4 digits {1,2,8,9} what are the last two digits [#permalink]
29 May 2005, 14:30

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

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

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?

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.

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.........

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

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)

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.