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12 Days of Christmas GMAT Competition 2025 - Day 8: What is the sum of

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Dereno

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Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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Let’s first solve this problem, by taking the 5 digits (0,2,3,5,7)

The five digits can be arranged in 5! Ways = 120.

How many times a digit repeats at each position = Number of Ways / Total Digits

= 120/5

= 24

Number of ways = (11111) * Times a digit repeats * (sum of digits)

= (11111) * 24 * (0+2+3+5+7)

= 4533288.

Since, Zero (0) cannot come at the Ten Thousands place. We need to remove all numbers which has zero at the initial place.

Excluding zero, we have 4 digits : (2,3,5,7)

The four digits can be arranged in 4! = 24 ways.

Number of times a digit repeating at each positions = 24/(Number of digits)

= 24/4

= 6

Number of ways = (1111) * 6 * (2+3+5+7)

= 1111 * 6 * 17

= 113322.

Total number of ways

= 4533288 - 113322

= 4419966.

Option D

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Given that 0,2,3,5 and 7 are each used once to form 5 digit numbers. Now let us look exclusively at the last digit. Numbers with 0 in the last digit will appear 24 times (4!), numbers with 2,3,5 and 7 will appear 18 times each (3*3*2*1)

Therefore, the sum of unit digits alone will be 0*24+2*18+3*18+5*18+7*18 =306

We now know that the sum will end with a unit digit 6 and the 30 will be carried forward to the next. The sum for the digit to the left of the unit digit will also have the same sum (306). Adding 30 to 306, we have 336. So the sum will have the digit prior to the unit digit also to be 6. At this point, we can continue along the same lines or look at the options. Only one of the options end with a 6.

Therefore, Option D

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Strategize: Total sum of all digits (including the first number subtract with 0)
- Total sum of 5 digits without repetiton can be: 4!*(0+2+3+5+7)*(1000+1000+100+10+1)= 24*17*11111 (1)
- Subtract with numbers start with 0 - put 0 first as the first digit, the remaining appears 3 times: 3!*17*1111 (2)
The calculation is so complicated, to be honest, I observe the answer and see the option A is 113,322, which is quite equal to (2), answer E is the maximum calculation for (1) and I run the quick check to validate my assumption. The final answer would be (1) - (2) = 4,419,966. D

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Given 5 different digits; 0,2,3,5,7
The number of ways to arrange 5 distinct items is 5= 5x4x3x2x1=120
Exclude numbers starting with 0 so we have 4 digits = 4x3x2x1=24
valid umbers= 120-24=96
Sum of the digits= 2+3+5+7=17
number of digits = highest power +1
5 digit umber- highest power is 10^4 so 10^4 + 10^3 + 10^2 + 10+1=11111
Final sum= 24 x 17 x 11111= 4,533,288
Final answer is E

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25 Dec 2025, 00:35

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The sum would be sum of 10^4 * 17 + 10^3 *17 + 100*17 + 10*17 + 1*17 hence unit digit would be 8 from options answer E.

Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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Gmat860sanskar

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25 Dec 2025, 00:50

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Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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Tricky question Indeed

We have 5 digits - 0, 2, 3, 5, and &

Total permutation = 5! =120 ; if 120 / 5 = 24 (So for every position, a number will appear for 24 times)

However 0 can't be the starting number,

so total permutation where 0 will start is 4! =24

so in this 24 invalid permutation each number (2,3,5 and 7) will appear for 24/4 = 6 times

so total - invalid = 24 - 6 = 18

so per position 18 times a number will appear;

Let's add the total 2 + 3 + 5 + 7 = 17

let's see what number will appear in unit digit = 17 * 18 = 306

So our total addition will end with 6; Our answer is D

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25 Dec 2025, 01:02

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Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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The units place of all the numbers are different, so finding the unit digit is enough to answer this question.

_ _ _ _ _

The unit digit can be filled in 5 ways

Let's assume that the unit digit is 7

The first place can be filled in 3 ways, the second place can be filled in 3 ways, the hundreds place can be filled in 2 ways and the tens place can be filled in 1 way.

Total = 3 * 3 * 2 = 18 ways.

This repeats for all the digits. hence, Units place of the sum = 18 * (0 + 2 + 3 +5 7) = 18 * 17 = Ends with 6.

Option D

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So we need to create 5 digit numbers using each of the digits 0,2,3,5,7. Let's place 2 at the end , then there will be 3 choices to place any of the digit except 0 in first place , similarly 3 choices in second place , 2 choices in 3rd place and 1 choice in fourth place. So there will be 3*3*2*1 = 18 choices for 2 at unit place. That sums to 36. Now continue keeping 3 ,5 ,7 , it will lead to 54,90,126 that sums overall 306. So unit digit of Answer will be 6. From POE we get to answer D. But still we will move forward. Now 30 will be carried over. So Let's continue the same pattern with ten's place. Similarly it will sum up to same 306 with addition of 30 to 336. So ten's place will be 6. Now with 100's place Overall sum will be 306+33 =339 so 100's place will be 9. Thousand's place will be 306+33 = 339 , which will be same as 9. Now the pattern for keeping the digit at first place will be different. If we keep 2 at first place in 5 digit , there will be 4*3*2*1 =24 choices . Similarly will be for 3 ,5 ,7. So total sum will be 24*(2+3+5+7) = 408 and now add 33 more then final sum will be 441. So final answer will be 4419966.

So answer is D.

Bunuelur

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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gchandana

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25 Dec 2025, 01:12

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23570
23750
32750
.
.
.
.

We need to find the sum of all the arrangements possible here.
Note, we want five-digit integers, and we have 0, so we should be careful not to consider cases where 0 is in the ten thousands place, like 02357 or 03572, etc.

We can approach this in this way.
Let's fix 0 in the units place(_ _ _ _ 0), then the remaining four numbers(2, 3, 5, 7) can be arranged in 24 ways (4!). Which means, 0 appears 24 times in the units place.
Then let's fix 2 in units place(_ _ _ _ 2), even though this appears 24 times, we cannot consider those cases where 0 is in the ten thousands place.
How many such cases?
Fix 0 in the ten-thousands place(0_ _ _ 2), the remaining 3 digits can be arranged in 6 ways(3!), and we have to exclude these 6 cases. Then the total is 24-6 = 18 cases. So 2 appears 18 times.
The same applies to 3, 5, and 7 as well.

So the total sum in the units place will be 0*24 + 18(2 + 3 + 5 + 7) = 306.
Notice here that 306 is the sum for every other place as well(tens, hundreds, thousands, and ten thousands).
30 from units place gets carried to tens place, sum = ______6
So in tens place, it is 306+30 = 336, 33 gets carried to hundreds place, sum = _____66
In hundreds place, it is 306+33 = 339, 33 gets carried, ____966.

We can stop here, as only one option with 966 in the end. Option D.

Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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adityamntr

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25 Dec 2025, 01:54

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Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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lets assume even 0 as starting diigit will be alid 5 digit num.

if so sum of all number will be
all digits can come on certain position 4! times by fixing one position and taking pemuatiaons of all remaining
so total = ( 7 + 5 + 3 + 2) (4!) ( 10000 + 1000 + 100 + 10 +1)
= 17 * 24* 11111

no removing the 4digit num from tabove, we can resue same formula adjusted for 4 digits using 2 3 5 7
= 17* (3!) ( 1111)
=17*6*1111

17*(266664 - 6666)
unit digit will be
7*8= 6
answer is D 4419966

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25 Dec 2025, 03:25

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Total permutation = 5!=120
Numbers starting with 0= 4!=24
Total Valid numbers=120-24=96
As, only 2,3,5,7 can be the first digit,
and each appears $\frac{96}{4=24times}$.

And the SUM of values = 2+3+5+7=17
Place value sum=10000+1000+100+10+1=11111

Total sum of all 5 digits positive integers=24*17*11111
=> 45,33,288/-
Hence option E is the correct answer

Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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Mardee

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We are given 5 integers which are 0,2,3,5,7 from which 5 digit values are considered. We are to find sum of all such combinations

Total possiblies = 5! - 4! = 96 (All possible combinations - Combinations with 0 as first digit)
Now, to find entire sum
Sum of each digit = 0+2+3+5+7 = 17
Sum of place values = 10000 + 1000 + 100 + 10 + 1 = 11111
No of times each non zero digit appears in each position = 96/4 = 24

Final value = 24 * 17 * 11111 = 4533288

E. 4533288

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AS teh first terms can only be a non - zero digit => we can have them in 4! way i.e 24.
All other non-zero digits will come 18 times each (as they would not come 6 times when they are the first digits)

Sum of the the thousands => 4!*(2+3+5+7) *10000 => 17*24*10000 => 4080000
Sum of other terms => 18*17*[1000+100+10+1] => 18*17*1111 => 339966
Total = 4080000+339966 => 4419966 => D

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Five digit means first digit cannot be 0.

Permutation of 5 numbers (including those with 0 in first place): 5!=120

Each digit(0,2,3,5,7) appears equally often in each place: 120/5 = 24 times per position

0+2+3+5+7=17

24*17*11111=4533288

Remove those with 0 in first place: 4!=24
Each digit(2,3,5,7) appears equally often in each place: 24/4 = 6 times

2+3+5+7=17

6*17*1111=113322

4533288-113322=4419966

IMO D

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Total numbers = 5!-4! = 120-24 = 96

For all valid numbers, digits 2,3,5,7 appear equally often.
Total digit positions = 96*5 = 480
and they are equally shared by 4 digits = 480/4 = 120
Sum of digits = 17
Sum of place values = 10000+1000+100+10+1 = 11111

Total sum of numbers = 120*17*11111 = 4419966

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Without constraints
5!=120

Using simetry, all the digits appear the same number of times
120/5=24

7+5+3+2=17 and each digit occurs 24 times
24*17=408

408*1 + 408*10 + 408*100 + 408*1000 + 408*10000 = 408*11111

As 0 can not be the ten thousands digit we will subtract the sum of the unallowed numbers
4!=24

Using simetry, all the digits appear the same number of times
24/4=6

7+5+3+2=17 and each digit occurs 6 times
6*17=102

102*1 + 102*10 + 102*100 + 102*1000 = 102*1111

408*11111 - 102*1111 = 408*10000 + 306*1111 = 4419966

Answer D

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Count all valid five-digit numbers formed using the digits 0,2,3,5,7 exactly once, so numbers cannot start with 0.

Sum of numbers with all permutations of the digits (including 0): 5!=120
The distribution of digits is uniform across each position: 120/5=24
24*0+24*2+24*3+24*5+24*7=408 each digit.

We must subtract the sum of the numbers that start with 0 from the total sum of all permutations of the digits, since those are not valid five-digit numbers.
Sum of numbers with all permutations of the digits (with 0 as the first digit): 4!=24
The distribution of digits is uniform across each position: 6/4=6
6*2+6*3+6*5+6*7=102 each digit.

In the 10000's: 408*10000
In the 1000's: (408-102)*1000 = 306*1000
In the 100's: (408-102)*100 = 306*100
In the 10's: (408-102)*10 = 306*10
In the 1's: (408-102)*1 = 306*1

408*10000+306*1111=4419966

The answer is D

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Count all distinct five-digit numbers where the digits come from the set {0, 2, 3, 5, 7}, each digit is used exactly once, and the first digit cannot be 0.

Total permutations without restrictions
120 (5!)

Every digit 0,2,3,5,7 occurs with equal frequency at each place value, so every digit occurs
24 (120/5)

24*(0+2+3+5+7)=24*17=408

408*(1+10+100+1000+10000)=408*11111=4533288

We have to subtract the sum of numbers not allowed.

Permutation of 4 numbers (excluding 0 as it is in the first place)
24 (4!)

Every digit 2,3,5,7 occurs with equal frequency at each place value, so every digit occurs
6 (24/4)

6*(2+3+5+7)=6*17=102

102*(1+10+100+1000)=113322

Total: 4533288-113322=4419966

The correct answer is D

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25 Dec 2025, 05:43

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Forming five-digit positive integers using 0, 2, 3, 5, 7 exactly once, so numbers cannot start with 0.

Summing of all permutations (including those starting with 0):
Total permutations of 5 digits = 5 ! = 120.
Each digit appears equally often in each position: 4 ! = 24 times per position
Sum of digits = 0 + 2 + 3 + 5 + 7 = 17
Sum of place values = 10000 + 1000 + 100 + 10 + 1 = 11111
Total sum = 24 × 17 × 11111 = 4 , 533 , 288
Total sum=24×17×11111 = 4533288 (This includes numbers that start with 0, which are to be discarded)

Now lets find the numbers that start with 0 and subtract.

Fixing 0 in the first place : remaining digits: 2, 3, 5, 7
Number of such permutations = 4 ! = 24
Each of the 4 digits appears 3! = 6 times in each position
Sum of digits = 17
Sum of place values (last 4 digits) = 1000 + 100 + 10 + 1 = 1111
Sum to subtract = 6 × 17 × 1111 = 113322
Sum to subtract=6×17×1111=113322

Step 3: Final Answer 4533288 − 113322 = 4419966
Ans should be D

Bunuel

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

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25 Dec 2025, 05:56

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5 digits permutation - 120, fixing 0 in first place, 4! =24, hence 96, 24 appereance per digit, sum of digits is 17, hence 17*24*10000=4080000, add 1111 and the solution becomes 453,288