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.papgust wrote: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?
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.
Can you help me understand how you get 64 for each digit?
