Number sequence prob

[papgust]

What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

OA is 711040. Can someone give a good clear approach?

[Stuart@KaplanGMAT]

What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

Can someone give a good clear approach?

In the future, please post the answer choices - the solution below will show all the steps, but it's extremely unlikely that you'd have to go beyond the first or second one if we had choices to eliminate.

We need to start by counting the number of possible 4-digit numbers we can make. Each digit has 4 possible values, so there are 4*4*4*4 = 256 different numbers.

Here's the key to solving this question: each digit appears with equal frequency in each spot in the number. So, 64 of our numbers end in 1, 64 end in 2, 64 end in 3 and 64 end in 4. Similarly, 64 have a tens digit of 1, 64 have a tens digit of 2, 64 have a tens digit of 3 and 64 have a tens digit of 4. The same goes for the hundreds and thousands digit.

Let's start by summing the units digits:

64(1) + 64(2) + 64(3) + 64(4) = 64(1+2+3+4) = 64(10) = 640

So, the units digit will be "0" and we "carry the 64".

At this point, we eliminate any answers that don't end in "0".

On to the tens digit: again, the numbers sum to 640, but we need to add the 64 carry-over, so the tens digit sums to 704. Put a "4" in the tens digit and carry 70 over to the hundreds column.

At this point, we eliminate any answers that don't end in "40".

Hundreds digit: sum of 640, 70 carry-over, so the hundreds digit sums to 710. Put a "0" in the hundreds digit and carry 71 over to the thousands digit.

Eliminate any answers that don't end in "040".

Last part: sum of 640, 71 carry-over, so the remaining numbers are 711.. final answer: 711040.

[Stuart@KaplanGMAT]

As with most questions, there are multiple approaches. Here's a more mathematical one (that's quicker if you don't mind long multiplication):

We know the smallest number we can make is 1111 and the largest number we can make is 4444.

We also know that our numbers will be evenly distributed in the middle (i.e. 1112 is balanced by 4443; 1113 is balanced by 4442). So, we can solve using the average formula.

Finally, we know that there are 4*4*4*4 = 256 numbers in our set.

Average = sum of terms/# of terms

sum of terms = average * # of terms

sum of terms = (1111+4444)/2 * 256 = 5555/2 * 256 = 5555*128 = 711040

Again, if we had answer choices, chances are very good that we wouldn't have to do all the multiplication to narrow it down to just one answer. Depending on how spread out the answer choices were, we could probably also round off the numbers and estimate (e.g. we could round 5555 down to 5500 and 128 up to 130, or if the answers were very spread out, round 5555 down to 5000 and 128 up to 150).

[uttam.albela]

What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

Can someone give a good clear approach?

In the future, please post the answer choices - the solution below will show all the steps, but it's extremely unlikely that you'd have to go beyond the first or second one if we had choices to eliminate.

We need to start by counting the number of possible 4-digit numbers we can make. Each digit has 4 possible values, so there are 4*4*4*4 = 256 different numbers.

Here's the key to solving this question: each digit appears with equal frequency in each spot in the number. So, 64 of our numbers end in 1, 64 end in 2, 64 end in 3 and 64 end in 4. Similarly, 64 have a tens digit of 1, 64 have a tens digit of 2, 64 have a tens digit of 3 and 64 have a tens digit of 4. The same goes for the hundreds and thousands digit.

Let's start by summing the units digits:

64(1) + 64(2) + 64(3) + 64(4) = 64(1+2+3+4) = 64(10) = 640

So, the units digit will be "0" and we "carry the 64".

At this point, we eliminate any answers that don't end in "0".

On to the tens digit: again, the numbers sum to 640, but we need to add the 64 carry-over, so the tens digit sums to 704. Put a "4" in the tens digit and carry 70 over to the hundreds column.

At this point, we eliminate any answers that don't end in "40".

Hundreds digit: sum of 640, 70 carry-over, so the hundreds digit sums to 710. Put a "0" in the hundreds digit and carry 71 over to the thousands digit.

Eliminate any answers that don't end in "040".

Last part: sum of 640, 71 carry-over, so the remaining numbers are 711.. final answer: 711040.

Another easy approach extending the logic mentioned above:

We have 64 numbers with each of unit digit 1,2,3,4
Sum of values of all these unit digits = 64(4+3+2+1)=640
We have 64 numbers with each 1,2,3,4 at tens digit
Sum of values of all these at tens digits = 64(40+30+20+10)=6400
Sum of all digits for hundred place=64000
Sum of all digits at thousand place =640000

Sum all these to get the answer.

Simple

[papgust]

Thanks Stuart! I like your second approach than the first one.

Stuart/uttam.albela,

Extremely sorry for being so ignorant! Can you help me understand how you get 64 for each digit?

[uttam.albela]

Thanks Stuart! I like your second approach than the first one.

Stuart/uttam.albela,

Extremely sorry for being so ignorant! Can you help me understand how you get 64 for each digit?

hi papgust,

We have 4 places for digits.
(thousand place) (hundred place) (tens place) (unit place)
Now how many numbers can you place on say unit place if only 1,2,3,4 are allowed? 4 ways.
Similarly 4 ways for 10's place, 100's place, and 1000's place.

So total numbers = 4*4*4*4=256

Out of these 256, how many will be ending with 1?
how many will be ending with 2?
how many will be ending with 3?
how many will be ending with 4?
As we are living in free world, we are not discriminating between 1 or 2 or 3 or 4. So 256/4 =64 numbers will end with 1.
64 will end with 2.
64 will end with 3.
64 will end with 4.
Equality prevailed once again in this world.

Similarly we can think of 10s place. Out of 256 numbers how many have 1 at 10s place? How many have 2 at 10s place and so on? Even at this place, no discrimination at all. Which leads to 64 numbers with 1 or 2 or 3 or 4 at 10's place.