Each of the nine digits 0, 1, 1, 4, 5, 6, 8, 8, and 9 is used once to

Question

Each of the nine digits 0, 1, 1, 4, 5, 6, 8, 8, and 9 is used once to form 3 three-digit integers. What is the greatest possible sum of the 3 integers?

A. 1,752
B. 2,616
C. 2,652
D. 2,775
E. 2,958

PS34550.02

nick1816 · Accepted Answer

Since we need to maximize the sum, first we will maximize the sum of hundreds digits of the 3 numbers. Hundred digits of 3 numbers will be 8, 8 and 9 in any order.

Now we will try to maximize the sum of tens digits of 3 numbers. Tens digits of 3 numbers will be 4, 5 and 6 in any order.

Remaining digits can be put in units place in any order.

Sum of 3 numbers= 100*(8+8+9) + 10*(4+5+6) + (0+1+1) = 2500+150+2 = 2652

ScottTargetTestPrep · Answer

Since we want the greatest possible sum of the 3 integers, we should use 8, 8 and 9 as the hundreds digits of the 3 integers, 4, 5 and 6 as the tens digits, and 0, 1 and 1 as the units digits. Therefore, the greatest possible sum of the 3 integers is:

(800 + 800 + 900) + (40 + 50 + 60) + (0 + 1 + 1) = 2500 + 150 + 2 = 2652

Answer: C

gurmukh · Answer

Three numbers are 
960,851,841
Sum = 2652
Option C is the answer.

Posted from my mobile device

Nipunh1991 · Answer

The way i prefer solving is making blanks and then filling values:

Lets say we have three 3 digit intergers

_ _ _ - 1 integer

_ _ _ - 2nd integer

_ _ _ - 3rd Integer

We need to have the highest combined value, with each digit just repeating once in a 3 digit integer. As we know the maximum number will be influenced most by the number placed in the hundreds place followed by tens and then units.

Hence ill take the highest 3 digits for the 100s digit which can be 9,8 and 8 (in any order), followed by the next highest digits in the 10s digit which can be 6,5 and 4 (in any order) and finally the units digits 1, 1, 0 (in any order), which will result to

9   5   1  - 1st integer
 8   4   1  - 2nd integer
 8   6   0  - 3rd integer

Had the digits in the hundreds place been in any other order among the 3 integers, and 10s and Units, It would still sum up to be the same number.

The sum is 2652. Answer C

Hope this helps a little

2021gmat · Answer

To maximize the sum, pick the 3 largest integers and place them in hundredths place.

9
8
8

Pick the next three to maximize the sum of 10s place
4
5
6

and then place the remaining numbers in ones place.

Do note that the only order that matters in placing the integers in 100s 10s and 1s.

941
851
860
or 
960
841
851

both yield 2652 as the sum.

Posted from my mobile device

itspCv2 · Answer

Given data:In simple terms: each digit can only be used in one integer

The largest sum of xyz+abc+Lmn=(x+a+L)100+(y+b+m)10+(z+c+n)1

To make those brackets max: choose: 100(1st max)+10(2nd max)+1 (3rd max)=100(9+8+8)+10(6+5+4)+1(1+1+0)=2,652

Therefore answer is C.2,652

Hope this helps

bones2020 · Answer

One way to look at this is to look at the 3 highest digits that you have (9,8,8) and place them in the hundreds place since we want to maximize the result of addition and place (0,1,1) in the ones place. You can eliminate D,& E since this would be greater than 900+800+800, and you can eliminate A since 1752 would be to small. Now left with two answer choices, B and C. Since the resultant sum of 0,1,1 in the ones place =2, choose C.

CAMANISHPARMAR · Answer

Maximize hundreds digit.....use one 9 and the two 8's in the hundreds place.. total should be more than 2500...A is out

You would need three 9's in the hundreds place to take the total above 2700; Therefore D and E must be out immediately.

You are down to B vs C. Use 6, 5, and 4 in the tens place. and the two 1's in the units place. Option C is the correct answer.

bumpbot · Answer

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