Problem Solving

Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

OA: B

Can you show your workings to this question and also, why is C not correct ?

Let's start with the digit 1 and begin with the 1000's place. The remaining 3 digits can be placed 3 x 2 ways = 6 ways. So,
the digit 1 will appear in the 1000's places 6 times.

Similarly, the digit 1 will appear in the other three places 6 times for the same reason.

So, 6x1*1000 + 6*1*100 +6x1*10 +6x1x1 represents the sum of all the digit 1's. Organizing this a bit better:

6X1x(1000+100+10+1) =1111x6x1

Each of the remaining three digits can be placed the same way, so for the digits 2,5 ,6:

1111x6x2 = 1111x12

1111x6x5=1111x30

1111x6x6=1111x36

Organizing these to add up: 1111x(6+12+30+36)= 84x1111 = 93324 B

ardz24 wrote:Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

From the four digits, the total number of possible arrangements = 4! = 24.
Thus, the number of integers that can be formed from the four digits = 24.

Given a set of values that is symmetrical about the median:
Sum = (number of values)(median of the values).

The set of 4-digit integers that can be formed from the digits 1, 2, 5 and 6 is symmetrical about the median:
...2516, 2561, 2615, 2651, 5126, 5162, 5216, 5261...
In the set above, the integers in green constitute the two middle numbers.
Working from the center out, we get the following differences:
2651-2615 = 36 and 5162-5126 = 36.
2615-2561 = 54 and 5216-5162 = 54.
2561-2516 = 45 and 5261-5216 = 45.
Since the differences to the left of center are the same as those to the right of center, the set is symmetrical about the median.

Thus, the sum can be calculated as follows:
Median = (average of the two middle numbers) = (2651+5126)/2 = 7777/2.
Sum = (number of values)(median of the values) = (24)(7777/2) = (12)(7777) = integer with a units digit of 4.

The correct answer is B.

ardz24 wrote:Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

Alternate approach:

There are 4 positions: thousands place, hundreds place, tens place, units place.
There are 4 choices for each position: 1, 2, 5 or 6.
As noted in my post above, the total number of possible 4-digit integers = 4! = 24.

Each digit will appear in each position 26/4 = 6 times.
Thus, in each position, there will be six 1's, six 2's, six 5's, and six 6's.
Sum of the digits for each position = 6*(1+2+5+6) = 84.

Sum for the thousands place = 84*1000 = 84000.
Sum for the hundreds place = 84*100 = 8400.
Sum for the tens place = 84*10 = 840.
Sum for the units place = 84*1 = 84.

Sum of all the integers = 84000 + 8400 + 840 + 84 = integer with a units digit of 4.

The correct answer is B.

The easiest way is to cheat with the units digit. If you've got 24 numbers that look like so:

six of the form _ _ _ 1
six of the form _ _ _ 2
six of the form _ _ _ 5
six of the form _ _ _ 6

then when you add them up, your units digit is 6*1 + 6*2 + 6*5 + 6*6 => _ 4 (well, 84, but we only want the units digit).

That means the sum must end in 4, and only B does.

Unnecessary but fun follow up: once we've got that sum for the units digit, by the way, we can use it for all the other columns like so:

Units column = 84
Tens column = 84
Hundreds column = 84
Thousands column = 84

So the sum of those columns would be 84 + 840 + 8400 + 84000, or 93324.

Another quick and dirty approach that could game us to the right answer is estimation.

We know that we've got 6 numbers are around 1000, 6 that are around 2000, 6 that are around 5000, and 6 that are around 6000.

That means our sum should be more or less 6 * (1000 + 2000 + 5000 + 6000), or more or less 84000. Since we rounded everything down (all our 1000ish numbers are > 1000), the sum must be greater than 84000, but not *too* much greater.

Only B is in the neighborhood, so we're done.

In the spirit of my other answers, here's one more.

Suppose I take 1256 and cycle it: 1256, 2561, 5612, 6125. That sum is 15554.

The cycle is determined by the first arrangement of numbers (1256). By fixing one of those and holding the others constant, I can find the number of cycles. Holding out of the four digits gives me 3! ways to arrange the others, so I've got 6 cycles.

Each cycle will have the same sum, so my total sum = 6 * 15554, or 93324.

Thanks Matt for showing such a wonderful approach!!!