Sum of Two-Digit Numbers

There are 4 possibilities for the tens digit, and 4 for the units digit, which makes for a total of 16 possible 2-digit numbers. This means that when we sum up the tens digits, we'll have 4 10's, 4 20's, and so on, which can be expressed as: 4(10+20+30+40) = 400. For the units digit, we'll have 4 1's, 4 2's, etc., so that'll be: 4(1+2+3+4) = 40. When we sum all of the tens digits and the units digits, we get 440. (Alternately, you can just sum 11+12+13+14+21+22+...+43+44, which is not actually that much longer than solving the problem the intended way.)

As there are 4 given digits 1,2,3,4 therefore, each digit will repeat 4 times in both the tens and ones places in the possible set of numbers.

for example , the two digit numers starting with 1 and contain only 1,2,3,4 are 11,12,13,14 .ending with 1 are -11,21,31,41.

therefore , the sum of all such two digit numbers would be equal to

4( 10(1+2+3+4) ) + 4(1+2+3+4 ) =440

As we only want to consider two-digit numbers, let us denote the units place digit as U and Tens place digit as T.

Then we can observe that for each of the 4 possible values of T, we have 4 corresponding values for U. So, each of {1,2,3,4} appears 4 times at T and 4 times at U.

Now because we have to find the sum, we first sum all the possible digits occurring at U. From the previous explanation, we know that each of {1,2,3,4} appears 4 times at U. So the sum is 4*(1+2+3+4) = 40. Now as in addition, carry the 4. Similarly each of {1,2,3,4} appears 4 times at T. So again the sum is 40. Plus we add the carried 4 and so we get the answer as 440.

we see that 4 4=16 so there are 16 2 digit numbers that have the above condition. they are 11,12,13,14 and 21,21,.... to 43,44. since we know that 1+2+3+4=10 we will multiply 10 by 4(1,2,3,4) that will equal 40. then we see for every 10,20,30 and 40 they are used 4 times so it will become 10 4+ 20 4+ 30 4+40 4 and it will equal 100 4=400 now we just have to sum 400 and 40 which will give us 440.

As the possible digits can be either 1, 2, 3 or 4 and the digits need not be distinct. So if we take summation of all possible 2-digit numbers, we observe each digit is been used 4 times i.e. 11, 21, 31, 41 here 1 gets used 4 times so therefore, summation at UNIT'S PLACE = 4 * (1 + 2 + 3 + 4) = 40 where the 4 gets carried to TEN'S PLACE and now, summation at ten's place = 4 * (1 + 2 + 3 + 4) + 4 [the carry] = 44.

Therefore, summation of all possible 2-digit numbers where digitsd can be 1,2,3 and 4 only is equal to 440

You are finding 2-digit positive numbers and have four possible digits: 1, 2, 3, or 4, which is a basic permutation, n^r or 4^2 or 16 possibilities.

The 16 possible numbers start with either 1, 2, 3, or 4, and end with 1, 2, 3, or 4 so it's easy to find the numbers 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44.

To add quickly, you group them into four equal groups like (11, 14, 41, 44) or (22, 23, 32, 33) which add up to 110 and the final answer should clearly be 4 * 110 = 440.

First, use 1 as the tens digit. You can have 11, 12, 13 and 14 and their sum is 50. If 2 will be the tens digit, you will have 21, 22, 23 and 24 and their sum is 90. The difference between the sums is 40. It can also be seen by difference of two numbers having same ones digit. 21 - 11 = 10, then multiply it by 4 and you will have 40. We can conclude that using 3 as tens digit, the sum of 4 numbers is 130 and using 4 as tens digit, you will have 170. Then add 50, 90, 130 and 170 and you will have 440.

All the numbers have any digit from the set {1,2,3,4} as their tens and/or ones digit. So then, we have 11+12+13+14+21+22+23+24+31+32+33+34+41+42+43+44 =4(1+2+3+4)+4(10+20+30+40)=4(10)+4(100)=440

You can solve this problem just thinking about the possibilities that obey the conditions given on the problem. The possible numbers that are initiated with 1 and that obey the conditions are 11, 12, 13, 14. The possible numbers that are initiated with 2 and that obey the conditions are 21, 22, 23, 24. The possible numbers that are initiated with 3 and that obey the conditions are 31, 32, 33, 34. The possible numbers that are initiated with 4 and that obey the conditions are 41, 42, 43, 44. Then, you must sum all of them, and the sum is equal to 440.

There are two digits in the number which can be 1, 2, 3 or 4.

If we take the first digit as 1, the next digits can be 1, 2, 3 or 4. Hence, there are 4 two digit numbers starting with 1.

It is similar if we take the first digit as 2, 3 or 4. For each of them there are 4 two digit numbers.

If we observe carefully, we would see that each number is repeated 4 times while adding in a column:

11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44,

Hence, the sum of digits of unit’s place = 4 x (1+2+3+4) =40 Hence, the unit’s digit is 0.

The sum of digits of ten’s place = 4 x (1+2+3+4) =40 Hence, the ten’s digit is (40+4) = 44

Hence, the sum is 440

Our Aim :- we have to find 2 digit positive numbers whose digits are either 1,2,3 or 4

so, there are 2 places in a 2 digit number ones and tens place first fix any one place ,let say ones place ane then place the other numbers in tens place

like first take 11 12 13 14 21 22 23 24 31 32 33 34 41 42 43 44 these are the total combinations so add all the numbers ane we get total =440

There are 4^2 different 2-digit positive numbers whose digits are either 1, 2, 3 or 4. The sum of all these 16 numbers is ( 1 4 + 2 4 + 3 4+4 4 + 10 (1 4+2 4+3 4+4*4)). In fact there are 4 numbers which start with1 and 4 which end with 1, and the same for 2, 3 and 4.

tens digit can either be 1,2,3 or 4. ones digit can either be 1,2,3 or 4.

Case 1: Tens digit is 1 there are 4 possible one's digits so the tens digit 1 is used 4 times. (10 4) then add. 1+2+3+4 ( all the possible the ones digit). (40+1+2+3+4) = 50

Case 2: Tens digit is 2 Use the same method from case 1. Tens digit (20 x 4) + ONes digit (1+2+3+4) = 90

Case 3: Tens digit is 3 Tens digit (30x4) + Ones digit (1+2+3+4) = 130

Case 4: Tens digit is 4 Tens digit (40x4) + Ones digit (1+2+3+4) = 170 Add all the cases: 50+ 90+ 130+170 = 440

The numbers are: $11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44,$ for which there are $16$ numbers. To prove that these are all the numbers, since there are $4$ choices for tens digits, and $4$ choices for ones digits, so there are total of $4\times 4=16$ numbers. So the $16$ numbers stated above are all the numbers.

So the sum is: $4\times 10+(1+2+3+4)+4\times 20+(1+2+3+4)+4\times 30+(1+2+3+4)+4\times 40+(1+2+3+4)$ $=4(10+20+30+40)+4(1+2+3+4)$

$=4(100+10)$

$=440.$

So the answer is $\fbox{440}$ . Q.E.D.

Solution 1: There are $4\times 4=16$ such numbers that fulfill the condition. 4 of them have a units digit of 1, 4 of them have a units digit of 2, 4 of them have a units digit of 3, 4 of them have a units digit of 4. Hence, the contribution of the units digit to the sum will be $4 (1+2+3+4)=40$ . Similarily, the contribution of the tens digit to the sum will be $4(10+20+30+40)= 400$ . Hence, the total sum is $400+40=440$ .

Solution 2: We seek the sum $11 + 12 + 13 + 14 + 21 + \ldots + 44$ . There are only 16 terms, and we may sum them up to obtain $440$ .

I'm not sure why people are confused by this. The problem is simple. You are finding the sum of all two-digit numbers which use the digits 1, 2, 3 and 4. Therefore you start with 11 then 12 then 13 then 14 as these are the first four number which use said digits. Continue this using 2 in the tens column to get 21, 22, 23 and 24; the use 3 in the tens column so 31, 32, 33 and 34 and finally 41, 42, 43 and 44. Add these up and you get the answer. Simples :-)