Question

You have to use the numbers 1,2,3,4 and 5. Add those numbers and the sum should be equal to 50. You can combine the numbers to form tens (i.e 2 and 4 is 24) but you can only use the number once.Q: 12345=50
You can only use the number once and only addition is the operation accepted.

Answer 1

Answer with explanation:

⇒Keep in mind

Sum of evens is Even.

Sum of two odd is even.

Sum of the digits from 1 to 5 , when used once =1+2+3+4+5=15

Now, Sum of the digits also can't be 50 when operation Addition is used three times between the numbers or Digits.

45+1+2+3=51

54+....=can't be equal to 50.

43+5+2+1=51

42+5+3+1=51

41+5+2+3=53

Similarly, when number begin from 3, that is 35 +4+......=can;t be equal to 50.

Now, we have to check when operation addition is used twice between digits or numbers can the sum be 50.

That, is sum of two even and one odd or Sum of two odd and one even should be equal to 50.

→As unit place can be filled in two ways that is 2, and 4 and tens place can be filled by remaining of four numbers.

So, total number of even numbers using the digits 1,2,3,4,and 5 =4 ×2=8

These numbers are, 12,32,42,52,14,24,34,and 54.

→And total number of numbers which are odd using the digits 1,2,3,4,and 5 are=4 ×3=12, which are, 21,31,41,51,13,23,43,53,15,25,35,45.

When you take any of these two,two digit numbers between these 20 numbers  and combine with another distinct remaining digit ,it can't be equal to 50.

For example:

→ 12 +34+5≠50

→14+35+2≠50

→21+24+5≠50

Answer 2

I think we must use two numbers which have two digits and one of them must be one digited number.If we try it we can say our numbers are xy,ab and c.

xy + ab + c = 50

10x+y+10a+b+c = 50

10(x+a) + y+b+c = 50

Think about y+b+c. Maximum it can be 5+4+3 = 12 but it can't because there is a number which is multiple of 10.So it can be 10 maximum and x+a must be 4.

We can only use 1+3. 40 + y+b+c = 50 , y+b+c = 10. And other numbers are 2+4+5 = 11.  10 doesn't equal to 11 and it says there isn't any solution for this problem :)