Maximize this!
If A , B , C , and D are distinct digits, find the maximum possible value of the following sum:
+ A C B D
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4 solutions
My method:
The following number can be expressed as 1 0 ( a + c ) + b + d .
For the number to be maximum , 1 0 ( a + c ) must be maximum and since a,c are distict 1 digit numbers , a + c ≤ 1 7 where ( a , c ) : ( 9 , 8 ) , ( 8 , 9 ) .
For b + d to be maximum , ( b , d ) : ( 7 , 6 ) , ( 6 , 7 ) .
Thus the maximum sum 1 0 ( a + c ) + b + d = 9 7 + 8 6 = 9 6 + 8 7 = 1 8 3 .
it can be also 173. because 82+91=173, here i choose A=8, C=9, B=2 and D=1.
largest 4 digits = 9 ,8 ,7,6
they should be arranged in order to get the maximum value
a and c should be the largest so they can be 9 and 8
for b and d we are left with 7 and 6
so a+c = 17 and b+d = 13
after that we get ...ab + cd = 10 ( a + c) + (b+d) = 183.
Yup, that's the idea :)
But 98+76 equals 174....? Right idea, right answer, wrong digits.
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Thanks for pointing that out! His original solution stated "a and c should be the largest so they can be 9 and 8 for b and d we are left with 7 and 6 " which is correct, but then wrongly stated that "so ab =98 and cd= 76 ". I've updated this solution.
To get the greatest value we will have to take the maximum value for the letters A and C ,which is 9 & 8 and then for B and D we take the next consecutive greater values that are 7 & 6.Thus the numbers formed are 97 +86 = 183.
You put 9 and 8 in the tens column and 7 and 6 in the ones column.
97 + 86 = 96 + 87 = 1 8 3
97 + 86 = 183
A & C should be the largest (as they represents 10's place) so i choose 9 & 8
Next B &D should also be as large as possible therefore 7 & 6. Any other combo won't give a bigger value.