# Math TrickWatch

#1
The eight digits: 6, 5, 5, 4, 4, 3, 2, 1 are used to form two three-digit numbers and one two-digit number. What is the largest possible sum of these numbers?

hmm i can try some to find out,, bt is there an always-works trick?
like writing the biggest number from the 100th or 10th,,?
0
10 years ago
#2
Fill up the places with the highest values first. For example, have 6 and 5 as hundreds, then fill up the tens with the highest remaining digits.
0
10 years ago
#3
652 + 541 + 43 = 1 236
0
#4
(Original post by DanielJBaker)
Fill up the places with the highest values first. For example, have 6 and 5 as hundreds, then fill up the tens with the highest remaining digits.
yeaa thx
im not sure between the two options:
showing examples would be easier for you to see.. so.

652 + 541 + 43 = 1 236 (as Rokker posted) or
653 + 542 + 41 =1236

i know it comes to the same answer ,, bt since it can be not that case everytime,,
which is correct? or should i check both?
0
10 years ago
#5
(Original post by humanbeing)
yeaa thx
im not sure between the two options:
showing examples would be easier for you to see.. so.

652 + 541 + 43 = 1 236 (as Rokker posted) or
653 + 542 + 41 =1236

i know it comes to the same answer ,, bt since it can be not that case everytime,,
which is correct? or should i check both?
You could realise that swapping '2', '1', '3' around won't make any difference.
0
10 years ago
#6
to be honest that is a poor trick!

not really anything special
0
10 years ago
#7
(Original post by humanbeing)
yeaa thx
im not sure between the two options:
showing examples would be easier for you to see.. so.

652 + 541 + 43 = 1 236 (as Rokker posted) or
653 + 542 + 41 =1236

i know it comes to the same answer ,, bt since it can be not that case everytime,,
which is correct? or should i check both?
Any way round would work, as they'll still have the same place value.
0
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