M46-24

Question

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

Bunuel · Answer

Official Solution:

What is the sum of all five digit positive integers that can be formed using each of the digits 0, 2, 3, 5, and 7 exactly once?

A. 113,322
B. 4,419,660
C. 4,419,964
D. 4,419,966
E. 4,533,288

Since the numbers in the options are quite large, calculating the entire sum directly would take a long time. Therefore, we should look for a shortcut. Notice that each option has a different unit digit, so focusing on the unit digit is a good strategy.
 
First, let’s find the unit digit of the sum if 0 were allowed as the first digit. Using the digits 0, 2, 3, 5, and 7 exactly once, there are 5! arrangements. In these arrangements, each digit appears equally often in the units place, specifically 5!/5 = 24 times. So, the unit digit of the total sum is the unit digit of:
 
 24 * (0 + 2 + 3 + 5 + 7) = 24 * 17 
 
The unit digit of 24 * 17 is 8. Therefore, the unit digit of this larger sum is 8.
 
Next, let’s find the unit digit of the sum of the numbers where 0 is fixed as the first digit. With 0 as the first digit, the remaining digits (2, 3, 5, 7) form 4! arrangements. Each of these four digits appears 4!/4 = 6 times in the units place. So, the unit digit of the sum of these numbers is the unit digit of:
 
 6 * (2 + 3 + 5 + 7) = 6 * 17 
 
The unit digit of 6 * 17 is 2. Therefore, the unit digit of this excluded sum is 2.
 
Finally, the unit digit of the required sum is:
 
 8 - 2 = 6 .
 
To clarify, we are first calculating the unit digit of the sum assuming 0 is allowed as the first digit, and then subtracting the unit digit of the sum where 0 is fixed as the first digit. This gives us the unit digit of the sum where 0 is not the first digit, which is the unit digit we need for the total sum of all the five-digit numbers.
 
Only one option has 6 as the unit digit, so that option must be correct.

Answer: D

liberoasperiores · Answer

Awesome Question !

Adit_ · Answer

Took a similar approach as the solution during the test. Except I took out the possibilities of 0 at the start at the beginning itself.

Summation of all the digits = 17.
Each digit at the end appears:
3*3*2 = 18 times.
Thus 18*(17) = Last digit = 6.

Answer:  Option D

plum0709 · Answer

This is a great question that’s helpful for learning and I like the solution - it’s helpful.

FlatVase · Answer

Can this be done by taking average of values in each digit and then multiplying by 96?

Bunuel · Answer

Could you please show how you are calculating the average and what exactly you intend to multiply by 96?

FlatVase · Answer

I am just looking for the best solution. I might forget how to solve similar kind of questions in exam.

I am unable to attach the snap from gemini for this question

Bunuel · Answer

Here are alternative solutions that might help: https://gmatclub.com/forum/12-days-of-c ... 52636.html

himssss29 · Answer

I found the solutoin to be tricky, so this was my way to solve-

Since the answer choices all end in different digits, I only need to find the units digit of the total sum — not the full sum.

There are 4×4×3×2×1 = 96 valid five-digit numbers (0 can't be the first digit).

For each choice of first digit, the remaining 4 digits fill the last 4 places, and each digit appears in the units place exactly 4!/4 = 6 times.

First digit = 2:  remaining digits {0,3,5,7}
 Units contribution = 6×(0+3+5+7) = 6×15 = 90

First digit = 3:  remaining digits {0,2,5,7}
 Units contribution = 6×(0+2+5+7) = 6×14 = 84

First digit = 5:  remaining digits {0,2,3,7}
 Units contribution = 6×(0+2+3+7) = 6×12 = 72

First digit = 7:  remaining digits {0,2,3,5}
 Units contribution = 6×(0+2+3+5) = 6×10 = 60

Total units-place sum = 90 + 84 + 72 + 60 = 306

The units digit of the total sum is  6 , which matches only one answer choice:  4,419,966

M46-24

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