# Max and Minnie each add up sets of three-digit positive integers. Each of them adds three different three-digit integers whose nine digits are all different. Max creates the largest possible sum. Minnie creates the smallest possible sum. The difference between Max’s sum and Minnie’s sum is (A) 594 (B) 1782 (C) 1845 (D) 1521 (E) 2592

2
by Deleted account
Did you get the answer (C)
:)
i am going wrong somewhere, let me try again (and sorry for late reply my internet connection is distrbing me as hell)
Yep! (C)
:)

2016-04-03T21:25:56+05:30
Solutions

Consider three three-digit numbers with digits RST, UVW and XYZ. The integer with digits RST equals 100R + 10S + T, the integer with digits UV W equals 100U + 10V + W, and the integer with digits XY Z equals 100X + 10Y + Z. Therefore,

RST + UV W + XY Z = 100R + 10S + T + 100U + 10V + W + 100X + 10Y + Z

= 100(R + U + X) + 10(S + V + Y ) + (T + W + Z)

We note that each of R, S, T, U, V, W, X, Y, Z can be any digit from 0 to 9, except that R, U and X cannot be 0.

Max wants to make 100(R + U + X) + 10(S + V + Y ) + (T + W + Z) as large as possible. He does this by placing the largest digits (9, 8 and 7) as hundreds digits, the next largest digits (6, 5 and 4) as tens digits, and the next largest digits (3, 2 and 1) as units digits.

We note that no digits can be repeated, and that the placement of the digits assigned to any of the place values among the three different three-digit numbers is irrelevant as it does not affect the actual sum.

Max’s sum is thus 100(9 + 8 + 7) + 10(6 + 5 + 4) + (3 + 2 + 1) = 2400 + 150 + 6 = 2556. Minnie wants to make 100(R + U + X) + 10(S + V + Y ) + (T + W + Z) as small as possible. She does this by placing the smallest allowable digits (1, 2 and 3) as hundreds digits, the next smallest remaining digits (0, 4 and 5) as tens digits, and the next smallest digits (6, 7 and 8) as units digits. Minnie’s sum is thus 100(1 + 2 + 3) + 10(0 + 4 + 5) + (6 + 7 + 8) = 600 + 90 + 21 = 711. The difference between their sums is 2556 − 711 = 1845.

2016-04-03T22:09:09+05:30
Max and minnie take Three three-digit numbers.

# Max creates largest sum
# Minnie creates smallest sum

Solution:

Let the three numbers be ABC, DEF, GHI

Sum : ABC + DEF + GHI

Now, For Max,

The sum should be maximum number possible.
For this, we need to have the largest Hundreds digit, then second largest tens digit and remaining next units digit so that we can make a big number.

So, first we will put the largest numbers  in the hundreds place i.e. in place of A, D and  G.
A = 9
D = 8
G = 7

(Note : Repitition of same digits is not allowed)

Later, next big numbers at tens place.
B = 6
E = 5
H = 4

Remaining, at units place.
C = 3
F = 2
I = 1

So,Max's Sum = ABC + DEF + GHI
= 963 + 852 + 741
= 2556

Now, For Minnie,

The sum should be minimum number possible.
For this, we need to have the smallest Hundreds digit, then second smallest tens digit and next units digit so that we can make a small number.

Note : Here, we have to make smallest number so we can use zero. Zero i smallest number but since we can't use it at hundreds place, we will use it all tens place in the first number.

So, first we will put the smallest numbers  in the hundreds place i.e. in place of A, D and  G.
A = 1
D = 2
G = 3

Later, next small numbers at tens place (including zero)
B = 0
E = 4
H = 5

Remaining, at units place.
C = 6
F = 7
I = 8

So, Minnie's sum = ABC + DEF + GHI
= 106 + 247 + 358
= 711

Difference of their sum : 2556 - 711 = 1845

So, the correct answer is (C) 1845.