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Re: Place each of the digits 0 through 9, without repetition, in the boxes [#permalink]
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9740+851+62+3=10656

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Re: Place each of the digits 0 through 9, without repetition, in the boxes [#permalink]
Can someone explain to me why my approach is wrong? I did 0 + 21 + 543 + 9876 = 10440? I would expect the 4 digit number to have the highest values from the range of 0-9?
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Re: Place each of the digits 0 through 9, without repetition, in the boxes [#permalink]
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nick1816 wrote:
\(□+□□+□□□+□□□□\)

Place each of the digits 0 through 9, without repetition, in the boxes above. What is the maximum possible sum?

(Assumption: \(□□\) would represent the 2-digit number)

A. 10440
B. 10584
C. 10656
D. 10746
E. 10827



To make any number or Sum the largest, the highest digits should take the highest place value.
ABCD+EFG+HI+J
There is one thousands place and that should be the highest digit or A=9. Total value = 9*1000=9000
There are two hundreds place, B and E, and these should be filled by the next higher digits or B and E should be filled by 8 and 7. Total value = 100*(8+7)=1500
There are three tens place, C, F and H, and these should be filled by the next higher digits or C, F and H should be filled by 6, 5 and 4. Total value = 10*(6+5+4)=150
The remaining digits will be fill up units place. Total = 3+2+1+0 = 6

SUM = 9000+1500+150+6 = 10656

C
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Re: Place each of the digits 0 through 9, without repetition, in the boxes [#permalink]
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