BTGmoderatorDC wrote:1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660




OA E

Source: Official Guide

Since we're adding 24 numbers, we know that:
Six numbers will be in the form 1---
Six numbers will be in the form 2---
Six numbers will be in the form 3---
Six numbers will be in the form 4---

Let's first see what the sum is when we say all 24 numbers are 1000, 2000, 3000 or 4000
The sum = (6)(1000) + (6)(2000) + (6)(3000) + (6)(4000)
= 6(1000 + 2000 + 3000 + 4000)
= 6(10,000)
= 60,000

Since the 24 numbers are actually greater than 1000, 2000, etc, we know that the actual sum must be greater than 60,000

Answer: E

Cheers,
Brent

BTGmoderatorDC wrote:1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660




OA E

Source: Official Guide

Since there are 24 different integers, each of the digits (1, 2, 3 and 4) will appear in each of the place values (thousands, hundreds, tens and ones) exactly 6 times. Therefore, the sum of the 24 integers will be as follows:

6(1000 + 2000 + 3000 + 4000) + 6(100 + 200 + 300 + 400) + 6(10 + 20 + 30 + 40) + 6(1 + 2 + 3 + 4)

6(10,000) + 6(1,000) + 6(100) + 6(10)

60,000 + 6,000 + 600 + 60

66,660

Answer: E