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What is the sum of all 3-digit positive integers such that all the dig

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happypuppy
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What is the sum of all 3-digit positive integers such that all the dig [#permalink] New post   24 Jan 2022, 20:18
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39% (02:12) correct 61% (02:37) wrong based on 38 sessions
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What is the sum of all 3-digit positive integers such that all the digits of each of the number is even?

(A) 55,500
(B) 1300
(C) 44,400
(D) 247,275
(E) 54,400
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E
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What is the sum of all 3-digit positive integers such that all the dig [#permalink] New post   24 Jan 2022, 20:33
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Here, my approach was to find the number of occurrence of each digit at combination:


Hundreds Digits : 2,4,6,8
Tens Digit: 0, 2, 4, 6, 8
Unit Sum: 0, 2, 4, 6, 8


Hundreds Digit Sum for all numbers: (2+4+6+8)*(25) = 500
Tens Digit Sum for all numbers: (0+2+4+6+8)*(20) = 400
Unit Sum Sum for all numbers: (0+2+4+6+8)*(20) = 400

= 500 * 100 + 400 * 10 + 400

Answer (E)
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Re: What is the sum of all 3-digit positive integers such that all the dig [#permalink] New post   25 Jan 2022, 18:08
Roy867 wrote:
Here, my approach was to find the number of occurrence of each digit at combination:


Hundreds Digits : 2,4,6,8
Tens Digit: 0, 2, 4, 6, 8
Unit Sum: 0, 2, 4, 6, 8


Hundreds Digit Sum for all numbers: (2+4+6+8)*(25) = 500
Tens Digit Sum for all numbers: (0+2+4+6+8)*(20) = 400
Unit Sum Sum for all numbers: (0+2+4+6+8)*(20) = 400

= 500 * 100 + 400 * 10 + 400

Answer (E)


Why are you multiplying the sums initially with 25, 20, 20 respectively? Thank you in advance.
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Re: What is the sum of all 3-digit positive integers such that all the dig [#permalink] New post   26 Jan 2022, 03:00
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hundred place = 2 4 6 8 4 number
ten place = 0 2 4 6 8 5 number
one place = 0 2 4 6 8 5 number
total number = 4*5*5 = 100
smallest number = 200
largest number = 888
average = (200+888)/2 = 544
sum = 544*100 = 54400
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Re: What is the sum of all 3-digit positive integers such that all the dig [#permalink] New post   18 Apr 2024, 22:28
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Charli08 wrote:
Roy867 wrote:
Here, my approach was to find the number of occurrence of each digit at combination:


Hundreds Digits : 2,4,6,8
Tens Digit: 0, 2, 4, 6, 8
Unit Sum: 0, 2, 4, 6, 8


Hundreds Digit Sum for all numbers: (2+4+6+8)*(25) = 500
Tens Digit Sum for all numbers: (0+2+4+6+8)*(20) = 400
Unit Sum Sum for all numbers: (0+2+4+6+8)*(20) = 400

= 500 * 100 + 400 * 10 + 400

Answer (E)

Why are you multiplying the sums initially with 25, 20, 20 respectively? Thank you in advance.

­Total possible such numbers could be 4*5*5 = 100
So at unit's and ten's place, each of 0, 2, 4, 6, 8 can come (100/5) 20 times. At hundred's place, each of 2, 4, 6, 8 can come (100/4) 25 times.
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What is the sum of all 3-digit positive integers such that all the dig [#permalink] New post   20 Apr 2024, 09:16
Numbers will include the 200s, 400s, 600s and 800s. Within each 00s tens digit can be 0, 2, 4, 6 and 8, and this is the same for ones digit

So there are 5*5=25 possible numbers for each 00s, so 100 total numbers

The avg. value for the set is largest number (888) plus smallest number (200) = 1088/2 = 544

100 total numbers * avg value of 544 = 54400­
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What is the sum of all 3-digit positive integers such that all the dig [#permalink]
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