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

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Bunuel

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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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Bunuel

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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

GMAT Club Official Explanation:

Since the numbers in the options are quite large, calculating the entire sum directly would take a long time. Therefore, we should look for a shortcut. Notice that each option has a different unit digit, so focusing on the unit digit is a good strategy.

First, let’s find the unit digit of the sum if 0 were allowed as the first digit. Using the digits 0, 2, 3, 5, and 7 exactly once, there are 5! arrangements. In these arrangements, each digit appears equally often in the units place, specifically 5!/5 = 24 times. So, the unit digit of the total sum is the unit digit of:

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

The unit digit of 24 * 17 is 8. Therefore, the unit digit of this larger sum is 8.

Next, let’s find the unit digit of the sum of the numbers where 0 is fixed as the first digit. With 0 as the first digit, the remaining digits (2, 3, 5, 7) form 4! arrangements. Each of these four digits appears 4!/4 = 6 times in the units place. So, the unit digit of the sum of these numbers is the unit digit of:

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

The unit digit of 6 * 17 is 2. Therefore, the unit digit of this excluded sum is 2.

Finally, the unit digit of the required sum is:

8 - 2 = 6.

To clarify, we are first calculating the unit digit of the sum assuming 0 is allowed as the first digit, and then subtracting the unit digit of the sum where 0 is fixed as the first digit. This gives us the unit digit of the sum where 0 is not the first digit, which is the unit digit we need for the total sum of all the five-digit numbers.

Only one option has 6 as the unit digit, so that option must be correct.

Answer: D.

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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

digits given
0,2,3,5,7

total ways to arrange them is 5! ways = 120
sum of digits = 2+3+5+7 ; 17
each digit will appear 4! ways ; 24
17*24*(10000+1000+100+10+1) = 4533288
numbers starting with 0 can be eliminated as it is not possible as ten thousands place with 0 is not possible
numbers 2,3,5,7 can be arranged in (4-1)! ; 6 ways
sum is 17
17*6*(1000+100+10+1) = 113,322
subtract 4522288- 113322
4419964 is correct
OPTION C

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Sum of 5 digit integers = 0+2+3+5+7 = 17
Sum of all combinations of 5 digits numbers = (5-1)! * 17 * 11111 = 24 * 17 * 11111 = ends with 8 - 1)

Now this also has combinations starting with 0, which make it a 4 digit integer
Sum of 4 digits = 17
Sum of all combinations of 4 digit numbers = (3-1)! * 17 * 1111 = 6*17 * 1111 = ends with 2 -2)

Final total = eqn 1 - eqn2 = ends in 6
Answer 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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there will be 96 5 digit numbers that will can be formed using 2,3,5,7,0

each non zero number will get repeated 24 times in each and every place

hence if we sum 4080000+408000+40800+4080+408 (sum of value in place 5,4,3,2,1 digit of all numbers )

answer is 4533288

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So we have 5 slots for each of the 5 numbers which translates into 5! = 120 possible outcomes (Less the number that start with zero)
Each digit appears in each position (5-1)!= 24
Sum of the digit = 0+2+3+5+7=17
Sum of the place values= 10000+1000+100+10+1=111111
To get the total sum of all the permutations = 24*17*11111= 4,533,288
Numbers that start with zero are 4! = 24
Each number appears in (4-1)!= 6 times
Sum of the total digit = 17
Sum of place values in 4!= 1000+100+10+1=1111
To get the total sum of those that start it zero = 6x17*1111= 113332
Total = 4,533,288-113,322= 4,419,966
Ans 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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Total number of 5-digit positive integers = 5! - 4! = 4*4!

Let's first find sum of all 5-digit integers including numbers with 0 as first digit (5! = 120 numbers) = (0+2+3+5+7)*11111*5!/5 = 17*11111*24 = 4,533,288

Sum of 5-digit integers with first number 0 = Sum of 4-digit integers = (2+3+5+7)*1111*4!/4 = 17*1111*6 = 113,322

Sum of all 5-digit positive integers formed using each of the digits 0,2,3,5 and 7 exactly once = 4,533,288 - 113,322 = 4,419,966

IMO B

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I chose D.

My reasoning is as follows. Obviously it doesn't make sense to try to solve this by listing out all of the possible numbers and adding them as that will take way longer than 2 minutes. Let's focus on the units digit instead.

There are 5 possible numbers for the 5 digit number to end in: 0, 7, 5, 3, and 2.

Using permutation formula, let's calculate how many possible 5-digit numbers can be made that end in each of them.

Using the slot method, if we have 5 slots, and the last one is each number, then we have 4 slots which would be 4!. However, since one of the numbers is 0, the first slot can never be 0. So that leaves 3 possible slots for the 0 to be. So for all of the slots which are not the last number or one of the 0s, we can say there are 6*3 = 18 possible numbers for all of the slots that do not end in 0. Since we are only focused on the units digit, we do not need to worry about the number of numbers which end in 0, because that will be 0 anyway.

So now we have 18*7 = units digit 6;
18*5 = units digit 0,
18*3 = units digit 4,
18*2 = units digit 6.

6+6+4= units digit 6, so answer is D.

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glancing at the options, each unit digit is different so y not just calculate the sum of unit digits.

summing the unit digits ending with 0 -> _0
summing the unit digits ending with 2 -> possibilities 3*3*2*1 = 18 -> 18*2 -> _6
summing the unit digits ending with 3 -> 18*3 -> _4
summing the unit digits ending with 5 -> 18*5 -> _0
summing the unit digits ending with 7 -> 18*7 -> _6
summing all unit digits => 0+6+4+0+6 -> _6 -> hence option D

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Since 5 digits, each digit comes 5!/5 =24
for every unit, tenth, 100th ..place , sum =2+3+5+7=17
for 5 digits = 17*11111
for all combinations = 24*17*11111

now we have to remove sum of those combinations in which 0 comes first
for 4 digits = (4!/4)*17*1111

total sum = 17*((24*11111)-(6*1111))= 4,419,966

Ans D

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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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This is a long sum, we can use permutations to solve. Looking at the answer choices, we can also see that they differ by the units digit - finding that number will give us the answer.

Since we have 5 digits, the number of ways we can make a 5 digit number is $5!=120$. Each digit appears in the units place $\frac{120}{5}=24$ times. We can add up all the numbers in the set and multiply it by 24, which gives us $24*17=408$.

We need to now get rid of the five-digit numbers that begin with 0, because that makes them 4-digit, not 5-digit. Using counting principle for each digit slot, we can count this as:
$4*3*2*1=4!=24$

Okay, now we need to find the sum of the units of these 24 numbers. Each number from the set {2,3,5,7} appears $\frac{24}{4}=6$ times, so add up all the numbers in this set and multiply by 6. This gives us:
$17*6=102$

Perfect, we now have the two numbers we need to calculate the units digit. Just subtract them and get:
$408-102=346$

The units digit is 6, so our answer choice must end with 6. Only D fits that bill, thus D is the answer.

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Tot number of permutations: 5! = 120
in those 120 permutations each digit appears an equal amount of time in each position (units, tens, etc): 120 / 5 positions = 24 times in each position.

sum of digits = 0+2+3+5+7=17

SUM OF ALL PERMUTATIONS = 17 * 24 * (10,000 + 1,000 + 100 + 10 + 1) = 17 * 24 * 11,111 = 4,533,288

BUT you must remove all the permutations that have 0 as the first digit (that would make the numbers 4-digits and not 5).
invalid permutations = 4! = 24

24 / 4 positions = 6 times in each position

SUM OF INVALID PERMUTATIONS = 6 * 17 (1,000 + 100 + 10 + 1) = 113,322

all permutations - invalid = 4,419,966
Answer D

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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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0+2+3+5+7=17
total nos including 0 at first position= 5!=120
each no comes at each of the 5 digits for exactly= 120/5=24 times
so total sum of all 5- digit nos including 0 at first place= 24 x17 x( 10^4+ 10^3+10^2+10^2+10^1+1)=24x17x11111
total no of nos with 5 at first place= 4!= 24
each no comes at each digit place for= 24/4= 6 times
so sum of all nos with zero at first place= 6x17x(10^3+10^2+10^2+10^1+1)=6x17x1111
total sum required= 24x17x11111-6x17x1111

lets skip the calculation

check last digit for both
24x17x11111= last digit 8
6x17x1111= last digit= 2
difference= 8-2=6
so our answer must have last digit=6
only possible answer=D

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There will 5! = 120 integers with each digit appearing once and the sum of the digits in the unit position of all these 120 numbers will be 17, repeating (120/5)=24 times as the digit 2 will be in unit position 24 times again digit 5 will be 24 times. Same way the sum of tens unit digits will be 17 24 times. We can write this for all 5 positions of these 120 integers as-

24×17(10^4+10^3+10^2+10^1+10^0)=24+17×11111=4533288

Now when the digit 0 is at the left most those integers are 4 digit integers and should be deducted from the last step. The total number of such integers is 4!=24
Applying the same logic i.e the 4 digits will appear in the same position of these 24 integers 6 times (24÷4=6), so the summation for the 4 digits at each position appears 6 times, that can be written as-
6×17(10^3+10^2+10^1+10^0)=6×17×1111=113,322.

Final Answer 4533288 - 113,322=4419966

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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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Observe that last digit is different, so no need to calculate sum.

To construct 5 digit number out of 0,2,3,5,7 no repitition consider two cases

0 = total cases = 4*3*2*1 = 24

non zero = total cases = 3*3*2*1*4 = 18*4

Now if u add non zero cases u end up with 7 as last digit (2+3+5+7 = 17)

Since we have 18 such cases, it will be 18*7, end up with last digit 6

numbers with last digit zero will not modify this.

Correct Answer: D. 4,419,966

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Since unit digits are all unique, we dont need to find exact sum and just units digit sum is enough.

if 0 is fixed as units digit, number of combinations = 4 * 3 * 2 * 1 = 24
if 2 is fixed as units digit, number of combinations = 3 * 3 * 2 * 1 = 18
if 3 then 18
if 5 then 18
if 7 then 18

(2+3+5+7) * 18 + 0 * 24
17 * 18 = something with units digit 6.

So option D

more concretely, for all positions except for units digit, this holds

for units digit:

if 2 is fixed, number of combinations = 4 * 3 * 2 * 1 = 24
same for 3,5,6 also so

306 * 1 + 306 * 10 + 306 * 100 + 306 * 1000 + (17*24)*10000 =4,419,966

ans: option D

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Digits are 0,2,3,5,7
Total 5 digit numbers possible = 5! = 5x4x3x2x1 = 120
Each of 5 digits appear 120/5, i.e. 24 times at ten thousands place

But, since it is 5 digit number, it cannot start with 0, Hence total possibilities are 120 - 24 = 96
When the numbers are at ten-thousands place, each can appear 24 times, i.e.
(2 x 10000 + 3 x 10000 + 5 x 10000 + 7 x 10000) x 24 = 4080000

At thousands, hundreds, tens and ones places:
We have used 24 places for ten-thousands place and left with 96 more premutations
But of these 96 places, 0 appears 24 times
i.e. when ten-thousands space is fixed for 7, remaining numbers 0,2,3,5 will make 4! = 24 possibilities

excluding 0, remaining permutations = 96 - 24 = 72
remaining 3 numbers can be fixed at 72 /3 = 18 places,
therefore, for each of the positions
(2 + 3 + 5 + 7) x 18 x 1000 = 306000
(2 + 3 + 5 + 7) x 18 x 100 = 30600
(2 + 3 + 5 + 7) x 18 x 10 = 3060
(2 + 3 + 5 + 7) x 18 x 1 = 306

Adding all the values
4080000 + 306000 + 30600 + 3060 + 306 = 4419966

Hence, Answer Option D

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Number of 5 digit numbers formed with 0 2 3 5 7 can be found by 4*4*3*2*1 = 96

If we take all combinations of these 5 digits, we can write these numbers in 5! = 120 ways.
Each digit appears 5!/5 = 24 times in each position

Sum of digits = 0+2+3+5+7 = 17

Sum of all numbers using these five digits = 24*17(1+10+100+1000+10000) = 24*17*11111 = 4,533,288

We have to subtract numbers starting with zero from this.

Number of combinations with 0 in the ten thousands place = 4!
Each digit appears 4!/4 = 6 times in each position

Sum of all numbers starting with 0 = 6*17(1+10+100+1000) = 6*17*1111 = 113,322

Subtracting this 4533288 - 113322 = 4,419,966

Hence the sum of all five digit positive integers that can be formed with 0 2 3 5 7 is 4,419,966

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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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?

Total ways of arranging 5 numbers = 5!= 5*4*3*2*1= 120
Sum of all digits = 0+2+3+5+7= 17
In all arrangements each digit appears = 4! (24) times
Place value sum= 10000+1000+100+10+1= 11111

Total sum (of all number formed including 0 at beginning) =17*24*11111= 4533288

The five digit number made of 0,2,3,5,7 can't have 0 at the beginning, or it will be considered a 4 digit number. So we will subtract number that begin with 0.
When 0 is in beginning, then the remaining 4 digits can be arranged in = 4! ways = 24
In arrangements beginning with 0, all digits appears= 3! = 6
Place value sum= 1000+100+10+1= 1111
Sum of numbers that begin with 0= 17*6*1111= 113,322

Sum of arrangements of 5 numbers - Sum of arrangements of numbers that begin with 0 = 4,533,288-113,322= 4,419,966

truedelulu

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So we have the formula in this case:
Sum of integers formed using 5 digits without repetition = (5-1)!*(0+2+3+5+7)*11111 = 4,533,288 (1)
However, the numbers with 0 in the ten-thousandth place (i.e. 02357) are four digit numbers, hence we have to subtract sum of 4 digit numbers from (1)
Sum of 4-digit numbers = (4-1)!*(2+3+5+7)*1111 = 113,322
=> Final sum = 4,533,288 - 113,322 = 4,419,966

Answer: D.

For quick calculation:
Obviously the final sum would be a very big number so eliminate A
(5-1)!*17*11111=24*17*1111 => Unit digit = 8
(4-1)!*17*1111=6*17*1111 => Unit digit = 2
=> 8-2 = 6 (only D satisfies)

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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