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adityaprateek15
Joined: 26 May 2023
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25 Dec 2025, 06:04
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If all the possible n-digit numbers using n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! x {sum of all the digits) x {111.......} n times.
(5-1)!*(0+2+3+5+7)*11111 = 4533288
But, this also includes those numbers that start with 0 and if the number starts with 0, it becomes a 4-digit integer. Subtract the possible numbers with first digit as 0.
Remaining digit = (2,3,5,7)
(4-1)!*17*1111 = 113322
final answer = 4533288 - 113322 = 4419966 (D)
(5-1)!*(0+2+3+5+7)*11111 = 4533288
But, this also includes those numbers that start with 0 and if the number starts with 0, it becomes a 4-digit integer. Subtract the possible numbers with first digit as 0.
Remaining digit = (2,3,5,7)
(4-1)!*17*1111 = 113322
final answer = 4533288 - 113322 = 4419966 (D)
Bunuel
msignatius
25 Dec 2025, 06:56
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Okay this one was quite tough!
First thing first, let's see what hints are offered by the stem.
We need all possible 5 digit numbers that can be formed using the digits 0, 2, 3, 5 and 7 once each.
First thing: 0 can't be the first digit. So we have 2, 3, 5 and 7.
Secondly, I know for a fact it cannot be A, as 113,322 is like less than two 5 digit numbers that begin with the 7. Also, the other options are in the 4 millions, remarkably close by.
Now, to find the actual answer, let's add each digit-place individually.
The first digit, ten-thousands, will include 2, 3, 5 or 7, as stated above.
Now, to understand how many times each of these digits will be added, we can include the 0 in the equation for the time-being, for the ten-thousands digit even.
For instance, if we take 7, we do all possible combinations of 3, 5, 2 and 0 for the remaining four digits. That's 24 times.
Hence, 7*24 + 3*24 + 2*24 and 5*24 + 0*24, or really just 17*24 = 408 is our sum in the ten-thousands space.
We, until now have 4,080,000.
But lets try to remove all the values from this that are actually 4 digit numbers.
We have 4 numbers we can take, since we are using 0 in the ten-thousands space, in the thousands space - 2, 3, 5, 7.
We, again, can arrange these numbers 24 ways (4!), or 6 ways for each.
Hence, we will have 2*6 + 3*6 + 5*6 + 7*6 = 102 thousands, or
102,000.
For hundreds digit, we will have 2*6 + 3*6 + 5*6 + 7*6 = 102 added once again. (Ways to arrange 3 numbers = 6 ways, but 6 ways for each of the 4 numbers usable in the thousands digits - as we cannot use 0, of course. This again gives us 24).
But we now add the 102 starting from the next digit.
102,000 + 10,200 = 112,200.
Now, we will see this is exactly how it goes for the tens and ones spaces too -
Add 1020 and 102 to the mix.
112,200 + 1,020 + 102 = 113,322.
See something familiar in that number? It's the first option. This is just what we will minus from the final sum. I am keeping this in for now to simplify calculation, and I will deduct it in the end.
‐-----------
Now, we repeat this for further digit places.
For the thousands place, we can pick either one of 2, 3, 5, 7 or 0. With these 5 numbers, the total number of combinations of 4 numbers permissible is 4! or 24. However, we are leaving 1 number out in each case, hence this will repeat 5 times to give a total combinations of 120.
Among these, each number will have 120 / 5 = 24 occurences in the thousands space.
That's a total of
24*3 + 24*5 + 24*0 + 24*2 + 24*7 (the numbers) = 408, yet again what we observed above.
However, this time we don't need to subtract the instances containing 0s.
And this time, we add this starting from the second digit, to get,
4,080,000 + 408,000 = 4,488,000.
Now, for the hundreds digit, we can combine each of the 5 digits, and the total number of permissible combinations is 6 (3!). This, however, is multiplied by the 5 possible selections in the thousands digit, and 4 possible in the ten-thousands, to get, yet again - 6*5*4 = 120 instances.
Divide hy 5 again, and you get 24 instances in each case.
Add another 408 here, but this time from the next digit.
4,488,000 + 40,800.
And, when we do this for the tens and ones digits respectively, we will always find the 408 to add up.
4,488,000 + 40,800 + 4,080 + 408 = 4,533,288.
This is option E, haha.
But now we know to minus option A, or 113,322 from this from our computation above right?
We do:
4,533,288 - 113,322 = 4,419,966
That is the answer - D.
=======
First thing first, let's see what hints are offered by the stem.
We need all possible 5 digit numbers that can be formed using the digits 0, 2, 3, 5 and 7 once each.
First thing: 0 can't be the first digit. So we have 2, 3, 5 and 7.
Secondly, I know for a fact it cannot be A, as 113,322 is like less than two 5 digit numbers that begin with the 7. Also, the other options are in the 4 millions, remarkably close by.
Now, to find the actual answer, let's add each digit-place individually.
The first digit, ten-thousands, will include 2, 3, 5 or 7, as stated above.
Now, to understand how many times each of these digits will be added, we can include the 0 in the equation for the time-being, for the ten-thousands digit even.
For instance, if we take 7, we do all possible combinations of 3, 5, 2 and 0 for the remaining four digits. That's 24 times.
Hence, 7*24 + 3*24 + 2*24 and 5*24 + 0*24, or really just 17*24 = 408 is our sum in the ten-thousands space.
We, until now have 4,080,000.
But lets try to remove all the values from this that are actually 4 digit numbers.
We have 4 numbers we can take, since we are using 0 in the ten-thousands space, in the thousands space - 2, 3, 5, 7.
We, again, can arrange these numbers 24 ways (4!), or 6 ways for each.
Hence, we will have 2*6 + 3*6 + 5*6 + 7*6 = 102 thousands, or
102,000.
For hundreds digit, we will have 2*6 + 3*6 + 5*6 + 7*6 = 102 added once again. (Ways to arrange 3 numbers = 6 ways, but 6 ways for each of the 4 numbers usable in the thousands digits - as we cannot use 0, of course. This again gives us 24).
But we now add the 102 starting from the next digit.
102,000 + 10,200 = 112,200.
Now, we will see this is exactly how it goes for the tens and ones spaces too -
Add 1020 and 102 to the mix.
112,200 + 1,020 + 102 = 113,322.
See something familiar in that number? It's the first option. This is just what we will minus from the final sum. I am keeping this in for now to simplify calculation, and I will deduct it in the end.
‐-----------
Now, we repeat this for further digit places.
For the thousands place, we can pick either one of 2, 3, 5, 7 or 0. With these 5 numbers, the total number of combinations of 4 numbers permissible is 4! or 24. However, we are leaving 1 number out in each case, hence this will repeat 5 times to give a total combinations of 120.
Among these, each number will have 120 / 5 = 24 occurences in the thousands space.
That's a total of
24*3 + 24*5 + 24*0 + 24*2 + 24*7 (the numbers) = 408, yet again what we observed above.
However, this time we don't need to subtract the instances containing 0s.
And this time, we add this starting from the second digit, to get,
4,080,000 + 408,000 = 4,488,000.
Now, for the hundreds digit, we can combine each of the 5 digits, and the total number of permissible combinations is 6 (3!). This, however, is multiplied by the 5 possible selections in the thousands digit, and 4 possible in the ten-thousands, to get, yet again - 6*5*4 = 120 instances.
Divide hy 5 again, and you get 24 instances in each case.
Add another 408 here, but this time from the next digit.
4,488,000 + 40,800.
And, when we do this for the tens and ones digits respectively, we will always find the 408 to add up.
4,488,000 + 40,800 + 4,080 + 408 = 4,533,288.
This is option E, haha.
But now we know to minus option A, or 113,322 from this from our computation above right?
We do:
4,533,288 - 113,322 = 4,419,966
That is the answer - D.
=======
Bunuel
25 Dec 2025, 07:07
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Bookmarks
Let's exploit the fact that the possible results have all different unit digits ==> this means it's enough for us to find the last digit and we crack the problem!
Starting with the 2 in the end, we have the following result:
* ,* , *, *, 2 ==> Hence 4! combinations possible, minus the numbers starting with 0 and ending with 2, hence 3!
so with the 2 in the end we have a total of 4!-3! = 24-6 = 18 possiblities
Exactly the same applies for 3, 5 and 7. ( 0 we do not care )
Hence we get:
18 times ending with 2
18 ending with 3
18 with 5
18 with 7
so the unit digits are:
18*2 = ..6 ; 18*3= ..4; 18*5 =..0 and 18*7=..6
Hence the unit digit is 6
IMO D!
Starting with the 2 in the end, we have the following result:
* ,* , *, *, 2 ==> Hence 4! combinations possible, minus the numbers starting with 0 and ending with 2, hence 3!
so with the 2 in the end we have a total of 4!-3! = 24-6 = 18 possiblities
Exactly the same applies for 3, 5 and 7. ( 0 we do not care )
Hence we get:
18 times ending with 2
18 ending with 3
18 with 5
18 with 7
so the unit digits are:
18*2 = ..6 ; 18*3= ..4; 18*5 =..0 and 18*7=..6
Hence the unit digit is 6
IMO D!
