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A four-digit number is formed using the digits
0, 2, 4, 6, 8 without repeating any one of them.
What is the sum of all such possible numbers?
Answer comes upto 519960
Couldnt find options for same,please help with the process & answer
0, 2, 4, 6, 8 without repeating any one of them.
What is the sum of all such possible numbers?
Answer comes upto 519960
Couldnt find options for same,please help with the process & answer
15 Sep 2024, 01:36
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Hey @keets02
(Please follow the rules for posting when posting questions, here's a helpful link for the same)
There's a way to solve these types of questions that I discovered using some Googling. I'm not sure how much this qualifies to be a GMAT question, but the concept here is interesting and worth learning.
Here's the link of a similar question I stumbled upon which actually lead me to solve this particular question. It's a similar question but with different digits involved, but we'll apply the same concept to this question.
The question states:
First, let's find the total number of such 4 digit numbers possible.
_ _ _ _
We know the first digit can't be 0 for it to be a 4 digit number so first position can be filled in 4 ways (any number out of 2,4,6,8), the next position can be filled in 4 ways (we can consider 0 but since repetition is not allowed, we'll skip the number we choose for the first position), the next in 3 ways and the last position can be filled in 2 ways.
Total number of such 4 digit numbers: 4*4*3*2 = 96
Alright, we have 96 such numbers that we obviously can't manually add. There must be some pattern here, let's try to understand that.
How do you find the sum of any two or more numbers? You start adding the units digits, then the tens digit and so on. So our goal here is to understand how we can find the units digit of the sum, then the tens digit and so on.
Let's start with units digits first.
For this, we'll fix the units digit to be 0: _ _ _ 0
We know there are 4*3*2 = 24 such numbers out of the 96 numbers that end with 0.
Now let's take the units digit to be 2: _ _ _ 2
Now, there are 3*3*2 = 18 such numbers out of the 96 numbers that end with 2. This is again because the first digit can't be 0.
Similarly, you can see that there are 18 such numbers with units digit 4, another 18 with units digit 6 and another 18 numbers with units digit 8.
If we were to sum up the units digit of all such numbers: 24*0 + 18*2 + 18*4 + 18*6 + 18*8
Which gives us 0 + 18*(2 + 4 + 6 + 8) (I took 18 common and now I can see the sum forming 20) => 18*20 => 360
So now we have the sum of the units digit of all the numbers as 360.
We can follow the same pattern to deduce that when you fix the tens digit, you'll again get the sum to be 360. You can try calculating this on your own. The same will happen for the hundreds digit.
So till now we have the sum effectively: 360*10^2 + 360*10^1 + 360*10^0 (Raising powers of 10 for the position of each digit) which gives 39960. But this is only the sum of units, tends and hudredths digits.
To sum up all the digits in the thousand places you'll get a different scenario because you can't fill the first position with 0. So now you only have 4 options - 2,4,6,8.
_ _ _ _
When you fix the first position or the thousand place with 2, you can have 24 such numbers. Similarly, you'll get 24 numbers with 4 in the thousand place, another 24 numbers for 6 and for 8. So to sum up all the thousand digits, you'll get: 24*2 + 24*4 + 24*6 + 24*8 => 24*(2 + 4 + 6 + 8) => 24*20 => 480.
So now the sum of all the digits in the thousandth place will be 480*10^3 => 480,000
So the sum of all such numbers: 480,000 + 39960 => 519960.
(Please follow the rules for posting when posting questions, here's a helpful link for the same)
There's a way to solve these types of questions that I discovered using some Googling. I'm not sure how much this qualifies to be a GMAT question, but the concept here is interesting and worth learning.
Here's the link of a similar question I stumbled upon which actually lead me to solve this particular question. It's a similar question but with different digits involved, but we'll apply the same concept to this question.
The question states:
Quote:
First, let's find the total number of such 4 digit numbers possible.
_ _ _ _
We know the first digit can't be 0 for it to be a 4 digit number so first position can be filled in 4 ways (any number out of 2,4,6,8), the next position can be filled in 4 ways (we can consider 0 but since repetition is not allowed, we'll skip the number we choose for the first position), the next in 3 ways and the last position can be filled in 2 ways.
Total number of such 4 digit numbers: 4*4*3*2 = 96
Alright, we have 96 such numbers that we obviously can't manually add. There must be some pattern here, let's try to understand that.
How do you find the sum of any two or more numbers? You start adding the units digits, then the tens digit and so on. So our goal here is to understand how we can find the units digit of the sum, then the tens digit and so on.
Let's start with units digits first.
For this, we'll fix the units digit to be 0: _ _ _ 0
We know there are 4*3*2 = 24 such numbers out of the 96 numbers that end with 0.
Now let's take the units digit to be 2: _ _ _ 2
Now, there are 3*3*2 = 18 such numbers out of the 96 numbers that end with 2. This is again because the first digit can't be 0.
Similarly, you can see that there are 18 such numbers with units digit 4, another 18 with units digit 6 and another 18 numbers with units digit 8.
If we were to sum up the units digit of all such numbers: 24*0 + 18*2 + 18*4 + 18*6 + 18*8
Which gives us 0 + 18*(2 + 4 + 6 + 8) (I took 18 common and now I can see the sum forming 20) => 18*20 => 360
So now we have the sum of the units digit of all the numbers as 360.
We can follow the same pattern to deduce that when you fix the tens digit, you'll again get the sum to be 360. You can try calculating this on your own. The same will happen for the hundreds digit.
So till now we have the sum effectively: 360*10^2 + 360*10^1 + 360*10^0 (Raising powers of 10 for the position of each digit) which gives 39960. But this is only the sum of units, tends and hudredths digits.
To sum up all the digits in the thousand places you'll get a different scenario because you can't fill the first position with 0. So now you only have 4 options - 2,4,6,8.
_ _ _ _
When you fix the first position or the thousand place with 2, you can have 24 such numbers. Similarly, you'll get 24 numbers with 4 in the thousand place, another 24 numbers for 6 and for 8. So to sum up all the thousand digits, you'll get: 24*2 + 24*4 + 24*6 + 24*8 => 24*(2 + 4 + 6 + 8) => 24*20 => 480.
So now the sum of all the digits in the thousandth place will be 480*10^3 => 480,000
So the sum of all such numbers: 480,000 + 39960 => 519960.
