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Math Expert V
Joined: 02 Sep 2009
Posts: 58367
What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A) 444440
B) 610000
C) 666640
D) 711040
E) 880000

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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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7
Bunuel wrote:
What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A) 444440
B) 610000
C) 666640
D) 711040
E) 880000

Total Numbers that can be formed by using digits 1, 2, 3 and 4 = 4x4x4x4 = 256

Now sum of all these numbers can be calculated by looking at an observation that

Any digit can come at any of the 4 places i.e. each digit will appear at unit digit 256/4 = 64 times

i.e. 1 will be used as Unit digit in 64 numbers
i.e. 2 will be used as Unit digit in 64 numbers
i.e. 3 will be used as Unit digit in 64 numbers
i.e. 4 will be used as Unit digit in 64 numbers

i.e. Sum of Unit digits of all the number = 64(1+2+3+4) = 640

i.e. Number becomes _ _ _ 0 AND 64 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at ten's digit 256/4 = 64 times

i.e. 1 will be used as Tens digit in 64 numbers
i.e. 2 will be used as Tens digit in 64 numbers
i.e. 3 will be used as Tens digit in 64 numbers
i.e. 4 will be used as Tens digit in 64 numbers

i.e. Sum of Tens digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+64=704)

i.e. Number becomes _ _ 4 0 AND 70 GOES AS CARRY FORWARD
Any digit can come at any of the 4 places i.e. each digit will appear at Hundreds digit 256/4 = 64 times

i.e. 1 will be used as Hundreds digit in 64 numbers
i.e. 2 will be used as Hundreds digit in 64 numbers
i.e. 3 will be used as Hundreds digit in 64 numbers
i.e. 4 will be used as Hundreds digit in 64 numbers

i.e. Sum of Hundreds digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+70=710)

i.e. Number becomes _ 0 4 0 AND 71 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at Thousands digit 256/4 = 64 times

i.e. 1 will be used as Thousands digit in 64 numbers
i.e. 2 will be used as Thousands digit in 64 numbers
i.e. 3 will be used as Thousands digit in 64 numbers
i.e. 4 will be used as Thousands digit in 64 numbers

i.e. Sum of Thousands digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+71=711)

i.e. Number becomes 7 1 1 0 4 0

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Math Expert V
Joined: 02 Aug 2009
Posts: 7959
What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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3
1
sharathnair14 wrote:
InstantMBA wrote:
n = 4 * 4 * 4 * 4 = 256

<x> = (4444 + 1111) / 2 = 2777.5

Sum = number of integers x average value

n * <x> = 256 * 2777.5 = 711040

This solution seems fishy!
Isn't the averge equal to the first and the last terms equal to the average of the entire set only when the elements are equidistant? This isn't even a consecutive set for that matter.
Help me Out!

Yes, you are correct...We cannot do this straightforward..
However, here he is lucky that there is a pattern that suits this calculation..
There are always two numbers that will add up to 5555, and this is why..
Largest and smallest are 4444+1111=5555....Next largest and smallest are 4443+1112=5555....1234+4321=5555 and so on
Now since each set of two numbers equals 5555, and there are 256 numbers, so 256/2 sets, SUM becomes $$5555*\frac{256}{2} = 711040$$

Another quick way will be ..
There are total 4*4*4*4=256 numbers possible, so each place ( for examples at ones or at hundreds), there are 256 places and each of the 4 digits( 1, 2,3, 4) will be equally distributed..
Thus at each place each digit will come 256/4=64 times....

One way...
64 sets of (1,2,3,4), so 64*(1+2+3+4)=640. Sum of place values in each number - 1000+100+10+1=1111... So 640*1111

OR
Second way could be
1) 1 at ones place will give 64*1...at tens place = 64*1*10, at hundreds place = 64*1*100 and at thousands place = 64*1*1000
Total 64*1*1+64*1*10+64*1*100+64*1*1000=64*1(1+10+100+1000)=64*1*1111
2) similarly for digit 2,---- 64*2*1111
3) Digit 3 ---64*3*1111 and
4) Digit 4, it will be ---64*4*1111

Sum = $$64*1111*(1+2+3+4)=640*1111=711040$$
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What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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22
7
n = 4 * 4 * 4 * 4 = 256

<x> = (4444 + 1111) / 2 = 2777.5

Sum = number of integers x average value

n * <x> = 256 * 2777.5 = 711040

General Discussion
Senior Manager  Joined: 15 Oct 2015
Posts: 299
Concentration: Finance, Strategy
GPA: 3.93
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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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1
GMATinsight wrote:
Bunuel wrote:
What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A) 444440
B) 610000
C) 666640
D) 711040
E) 880000

Total Numbers that can be formed by using digits 1, 2, 3 and 4 = 4x4x4x4 = 256

Now sum of all these numbers can be calculated by looking at an observation that

Any digit can come at any of the 4 places i.e. each digit will appear at unit digit 256/4 = 64 times

i.e. 1 will be used as Unit digit in 64 numbers
i.e. 2 will be used as Unit digit in 64 numbers
i.e. 3 will be used as Unit digit in 64 numbers
i.e. 4 will be used as Unit digit in 64 numbers

i.e. Sum of Unit digits of all the number = 64(1+2+3+4) = 640

i.e. Number becomes _ _ _ 0 AND 64 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at ten's digit 256/4 = 64 times

i.e. 1 will be used as Tens digit in 64 numbers
i.e. 2 will be used as Tens digit in 64 numbers
i.e. 3 will be used as Tens digit in 64 numbers
i.e. 4 will be used as Tens digit in 64 numbers

i.e. Sum of Tens digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+64=704)

i.e. Number becomes _ _ 4 0 AND 70 GOES AS CARRY FORWARD
Any digit can come at any of the 4 places i.e. each digit will appear at Hundreds digit 256/4 = 64 times

i.e. 1 will be used as Hundreds digit in 64 numbers
i.e. 2 will be used as Hundreds digit in 64 numbers
i.e. 3 will be used as Hundreds digit in 64 numbers
i.e. 4 will be used as Hundreds digit in 64 numbers

i.e. Sum of Hundreds digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+70=710)

i.e. Number becomes _ 0 4 0 AND 71 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at Thousands digit 256/4 = 64 times

i.e. 1 will be used as Thousands digit in 64 numbers
i.e. 2 will be used as Thousands digit in 64 numbers
i.e. 3 will be used as Thousands digit in 64 numbers
i.e. 4 will be used as Thousands digit in 64 numbers

i.e. Sum of Thousands digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+71=711)

i.e. Number becomes 7 1 1 0 4 0

Are you kidding me!!!!!!!!  Shocking solution process.
Who's gonna understand that for d day?
Aren't you gonna give us something of value for GMAT?
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2974
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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2
Nez wrote:
GMATinsight wrote:
Bunuel wrote:
What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A) 444440
B) 610000
C) 666640
D) 711040
E) 880000

Total Numbers that can be formed by using digits 1, 2, 3 and 4 = 4x4x4x4 = 256

Now sum of all these numbers can be calculated by looking at an observation that

Any digit can come at any of the 4 places i.e. each digit will appear at unit digit 256/4 = 64 times

i.e. 1 will be used as Unit digit in 64 numbers
i.e. 2 will be used as Unit digit in 64 numbers
i.e. 3 will be used as Unit digit in 64 numbers
i.e. 4 will be used as Unit digit in 64 numbers

i.e. Sum of Unit digits of all the number = 64(1+2+3+4) = 640

i.e. Number becomes _ _ _ 0 AND 64 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at ten's digit 256/4 = 64 times

i.e. 1 will be used as Tens digit in 64 numbers
i.e. 2 will be used as Tens digit in 64 numbers
i.e. 3 will be used as Tens digit in 64 numbers
i.e. 4 will be used as Tens digit in 64 numbers

i.e. Sum of Tens digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+64=704)

i.e. Number becomes _ _ 4 0 AND 70 GOES AS CARRY FORWARD
Any digit can come at any of the 4 places i.e. each digit will appear at Hundreds digit 256/4 = 64 times

i.e. 1 will be used as Hundreds digit in 64 numbers
i.e. 2 will be used as Hundreds digit in 64 numbers
i.e. 3 will be used as Hundreds digit in 64 numbers
i.e. 4 will be used as Hundreds digit in 64 numbers

i.e. Sum of Hundreds digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+70=710)

i.e. Number becomes _ 0 4 0 AND 71 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at Thousands digit 256/4 = 64 times

i.e. 1 will be used as Thousands digit in 64 numbers
i.e. 2 will be used as Thousands digit in 64 numbers
i.e. 3 will be used as Thousands digit in 64 numbers
i.e. 4 will be used as Thousands digit in 64 numbers

i.e. Sum of Thousands digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+71=711)

i.e. Number becomes 7 1 1 0 4 0

Are you kidding me!!!!!!!!  Shocking solution process.
Who's gonna understand that for d day?
Aren't you gonna give us something of value for GMAT?

Logic and basic approaches only help on D-Day... I have much more experience with D-Days... So a suggestion is that don't get tricked with "Tricky ways"' to solve maths on D-Day Fort he question above, All I can say is that it's simple addition written in detail.
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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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Quote:
Logic and basic approaches only help on D-Day... I have much more experience with D-Days... So a suggestion is that don't get tricked with "Tricky ways"' to solve maths on D-Day Fort he question above, All I can say is that it's simple addition written in detail.

Hi GMATinsight,
Could "Logic and Basic Approach" help me on D day, if I didn't hone in on the LABA skill before the D day?
If I solve like high school math teacher alone, how will that help me on gmat's 2 minutes test?
Thanks for providing another approach to the solution.
Just learning, and i'm a big fluke.
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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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Nez wrote:
Quote:
Logic and basic approaches only help on D-Day... I have much more experience with D-Days... So a suggestion is that don't get tricked with "Tricky ways"' to solve maths on D-Day Fort he question above, All I can say is that it's simple addition written in detail.

Hi GMATinsight,
Could "Logic and Basic Approach" help me on D day, if I didn't hone in on the LABA skill before the D day?
If I solve like high school math teacher alone, how will that help me on gmat's 2 minutes test?
Thanks for providing another approach to the solution.
Just learning, and i'm a big fluke.

Yes, Basic approaches (Like High schools maths) are good enough but the idea is "To Be Lazy" and reduce as many steps as possible. "Think More" and "Write Less" and soon you will see that your speed has improved substantially and your mental calculations have improved significantly. Make sure that you don't miss on basic aptitude of

1) Using Options in most of the PS questions and eliminate wrong options to get right Option
2) Do the reverse engineering and validate the questions with Options to find the correct option at the earliest. E.g. in the question above we could have stopped right after calculating Thousands digit of the sum as it is applicable only with one option (correct one) and after calculation of tens digit of sum you could have eliminated 2 options already.
3) Data Sufficiency needs to be practised and the basic need to be understood that you don't have to reach to final value of unknown variable but you have to just stop immediately when you have figured that statement is SUFFICIENT or NOT

I hope that helps!
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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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2
10
Total no. of possibilities: 4^4

Sum of digits = 10

As each digit has an equal chance of occurring and since there are 4 digits, Each digit will appear 4^4 / 4 = 4^3 = 64

Sum of units digits = 10 * 1 * 64 = 640
tens digit = 10 * 10 * 64 = 6400
hundreds digit = 10 * 100 * 64 = 64,000
thousands = 10 * 1000 * 64 = 640,000

total = 640000 + 64000 + 6400 + 640 = 711,040
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What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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since repetition is allowed, as many as 4^4 = 256 numbers can be formed with the digits 1, 2, 3, 4

so, a total of 256/4 = 64 1000s, 64 2000s, 64 3000s, 64 4000s will sum to (64000 + 128000 + 192000 + 256000) = 640,000

now, lets drop 1 zero, every time we move on to the next decimal places and sum the numbers up, which implies

640,000 + 64000 + 6400 + 640 = 711040 = D the answer

thanks
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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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InstantMBA wrote:
n = 4 * 4 * 4 * 4 = 256

<x> = (4444 + 1111) / 2 = 2777.5

Sum = number of integers x average value

n * <x> = 256 * 2777.5 = 711040

This solution seems fishy!
Isn't the averge equal to the first and the last terms equal to the average of the entire set only when the elements are equidistant? This isn't even a consecutive set for that matter.
Help me Out!
Senior Manager  G
Joined: 09 Aug 2017
Posts: 491
Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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Any digit can come at any of the 4 places i.e. each digit will appear at unit digit 256/4 = 64 times
Please elaborate why did you divide 256 by 4?

GMATinsight wrote:
Bunuel wrote:
What is the sum of all four digit integers formed using the digits 1, 2, 3 and 4 (repetition is allowed)

A) 444440
B) 610000
C) 666640
D) 711040
E) 880000

Total Numbers that can be formed by using digits 1, 2, 3 and 4 = 4x4x4x4 = 256

Now sum of all these numbers can be calculated by looking at an observation that

Any digit can come at any of the 4 places i.e. each digit will appear at unit digit 256/4 = 64 times

i.e. 1 will be used as Unit digit in 64 numbers
i.e. 2 will be used as Unit digit in 64 numbers
i.e. 3 will be used as Unit digit in 64 numbers
i.e. 4 will be used as Unit digit in 64 numbers

i.e. Sum of Unit digits of all the number = 64(1+2+3+4) = 640

i.e. Number becomes _ _ _ 0 AND 64 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at ten's digit 256/4 = 64 times

i.e. 1 will be used as Tens digit in 64 numbers
i.e. 2 will be used as Tens digit in 64 numbers
i.e. 3 will be used as Tens digit in 64 numbers
i.e. 4 will be used as Tens digit in 64 numbers

i.e. Sum of Tens digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+64=704)

i.e. Number becomes _ _ 4 0 AND 70 GOES AS CARRY FORWARD
Any digit can come at any of the 4 places i.e. each digit will appear at Hundreds digit 256/4 = 64 times

i.e. 1 will be used as Hundreds digit in 64 numbers
i.e. 2 will be used as Hundreds digit in 64 numbers
i.e. 3 will be used as Hundreds digit in 64 numbers
i.e. 4 will be used as Hundreds digit in 64 numbers

i.e. Sum of Hundreds digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+70=710)

i.e. Number becomes _ 0 4 0 AND 71 GOES AS CARRY FORWARD

Any digit can come at any of the 4 places i.e. each digit will appear at Thousands digit 256/4 = 64 times

i.e. 1 will be used as Thousands digit in 64 numbers
i.e. 2 will be used as Thousands digit in 64 numbers
i.e. 3 will be used as Thousands digit in 64 numbers
i.e. 4 will be used as Thousands digit in 64 numbers

i.e. Sum of Thousands digits of all the number = 64(1+2+3+4) = 640
with carry forward 64, the sum becomes (640+71=711)

i.e. Number becomes 7 1 1 0 4 0

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Posts: 9701
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Re: What is the sum of all four digit integers formed using the digits 1,  [#permalink]

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sharathnair14 wrote:
InstantMBA wrote:
n = 4 * 4 * 4 * 4 = 256

<x> = (4444 + 1111) / 2 = 2777.5

Sum = number of integers x average value

n * <x> = 256 * 2777.5 = 711040

This solution seems fishy!
Isn't the averge equal to the first and the last terms equal to the average of the entire set only when the elements are equidistant? This isn't even a consecutive set for that matter.
Help me Out!

InstantMBA is not wrong at all. The solution is perfect, provided he/she understood why he/she was doing it this way.

What is the average of 1, 2, 5, 8, 9? (elements are not consecutive or equidistant)
It will be 5. Why? Because deviation of 2 from 5 is the same as deviation of 8 from 5. Similarly, deviation of 1 from 5 is the same as that of 9 from 5. The corresponding numbers from the beginning and from the end are equidistant from the centre 5.

Now look at our numbers in increasing order:
1111
1112
1113
1114
1121
1122
...
(2777.5)
...
4433
4434
4441
4442
4443
4444

Since we have 4 clean non zero consecutive digits to make the numbers, the numbers will be equidistant from the centre 2777.5. The avg of each pair of two numbers (one from each end e.g. 1111 and 4444, 1112 and 4443 etc) will be the same 2777.5
That is why you can use this method.

Now think what happens if the digits given are 1, 2, 5 and 7.
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