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Find the sum of all the numbers that can be formed by taking all the 26 May 2020, 06:22
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75% (hard)

Question Stats:

59% (02:37) correct 41% (02:32) wrong based on 193 sessions

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Find the sum of all the numbers that can be formed by taking all the digits at a time from 3, 4, 6, 7 and 9 without repetition?
A.7733652
B.7733256
C.7373256
D.7373652
E.7733653
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B
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GMATinsight
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Find the sum of all the numbers that can be formed by taking all the 26 May 2020, 06:34
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MBADream786
Find the sum of all the numbers that can be formed by taking all the digits at a time from 3, 4, 6, 7 and 9 without repetition?
A.7733652
B.7733256
C.7373256
D.7373652
E.7733653
Show more

Total Numbers that can be formed using 3, 4, 6, 7 and 9 without repetition = 5! = 120

Every digit (5 digits) has equal chance to appear at unit place i.e. each digit will appear at unit digit = 120/5 = 24 times

i.e. Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696


i.e. sum of all numbers = 696*10000+696+1000+696*100+696*10+696 = 7733256

Answer: Option B
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Find the sum of all the numbers that can be formed by taking all the 26 May 2020, 06:57
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GMATinsight
MBADream786
Find the sum of all the numbers that can be formed by taking all the digits at a time from 3, 4, 6, 7 and 9 without repetition?
A.7733652
B.7733256
C.7373256
D.7373652
E.7733653

Total Numbers that can be formed using 3, 4, 6, 7 and 9 without repetition = 5! = 120

Every digit (5 digits) has equal chance to appear at unit place i.e. each digit will appear at unit digit = 120/5 = 24 times

i.e. Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696


i.e. sum of all numbers = 696*10000+696+1000+696*100+696*10+696 = 7733256

Answer: Option B
Show more


Can you please explain this part

Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696
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Find the sum of all the numbers that can be formed by taking all the 26 May 2020, 08:37
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[quote="funkyakki"][quote="GMATinsight"][quote="MBADream786"]Find the sum of all the numbers that can be formed by taking all the digits at a time from 3, 4, 6, 7 and 9 without repetition?
A.7733652
B.7733256
C.7373256
D.7373652
E.7733653[/quote]

Total Numbers that can be formed using 3, 4, 6, 7 and 9 without repetition = 5! = 120

Every digit (5 digits) has equal chance to appear at unit place i.e. each digit will appear at unit digit = 120/5 = 24 times

i.e. Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696


i.e. sum of all numbers = 696*10000+696+1000+696*100+696*10+696 = 7733256

Answer: Option B[/quote]


Can you please explain this part

[highlight]Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696[/highlight][/quote]

We have 5 no. With us to form 5 digit no. Without repetion.
Let's say. Unit digit is 9
Then other 4 no. (Tens, hundreds, thousands, ten thousands) can be written as 4! = 24
So there will be 24 no. With 9 as unit digit..
Similar with others.
So, the summation will be
9*24 + 7*24 + 6*24 + 4*24 + 3*24 = 696

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Find the sum of all the numbers that can be formed by taking all the 26 May 2020, 22:18
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funkyakki
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MBADream786
Find the sum of all the numbers that can be formed by taking all the digits at a time from 3, 4, 6, 7 and 9 without repetition?
A.7733652
B.7733256
C.7373256
D.7373652
E.7733653

Total Numbers that can be formed using 3, 4, 6, 7 and 9 without repetition = 5! = 120

Every digit (5 digits) has equal chance to appear at unit place i.e. each digit will appear at unit digit = 120/5 = 24 times

i.e. Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696


i.e. sum of all numbers = 696*10000+696+1000+696*100+696*10+696 = 7733256

Answer: Option B


Can you please explain this part

Sum of unit digits of 120 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696
Show more

We are getting a total fo 120 numbers using the given digits
Every digit has equal chance to appear at unit's place of those 120 numbers

since 5 have 5 digits to use so each digit should appear equal number of times at units place

i.e. the numbers with unit digit 3 will be 120/5 = 24
i.e. the numbers with unit digit 4 will be 120/5 = 24
i.e. the numbers with unit digit 6 will be 120/5 = 24
i.e. the numbers with unit digit 7 will be 120/5 = 24
i.e. the numbers with unit digit 9 will be 120/5 = 24

So sum of the unit digit of all 24 numbers = 24*3+24*4+24*6+24*7+ 24*9 = 24*(3+4+6+7+9) = 696

I hope this help! :)
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Find the sum of all the numbers that can be formed by taking all the 02 Jun 2020, 06:51
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MBADream786
Find the sum of all the numbers that can be formed by taking all the digits at a time from 3, 4, 6, 7 and 9 without repetition?
A.7733652
B.7733256
C.7373256
D.7373652
E.7733653
Show more

an alternate way to determine sum of n digits without repetitions
(n-1)!*(sum of digits)* (11111)
4!*29*11111 = 7733256
OPTION B
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Find the sum of all the numbers that can be formed by taking all the 08 Apr 2025, 18:55
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