Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5. How many such 3-digit numbers are there?
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Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5. How many such 3-digit numbers are there?
This can be solve by probability. I can't get the formula but I could give an intellectual guess. Because 100 is the start of 3-digit numbers, I could get 9 combinations of it. Let me identify them as 101, 112, 123, 134, 145, 156, 167, 178, 189.
Now I will just subtract 1 from the number that I have identified in order to have the next set of numbers in the 200 range. Those are 202, 213, 224, 235, 246, 257, 268, 279.
So that means, 9 combinations for 100 range, then 8 combinations for 200 range, next will be 7 combinations for 300 range, and so on and so forth.
Therefore the solution for me guess would be 9 + 8 + 7 + 6 + 5 + 4 + 3 +2 + 1, which will be 45 combinations.
What price would I get from this? Just kidding!
45 three digit Numbers. Just checked thru a program.
[QUOTE=admin;255]Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5. How many such 3-digit numbers are there?[/QUOTE]
tere are totally 35 numbers for 3 digits
am I right
if the number is xyz
z= x+y
max value of z=9 to satisfy the given condition, x can take any value from 0to 9, but y takes only one value for a given x,i.e 9-x
so there are 9 possibilities for z=9
similarily for z=8,there are 8 possibilities and so on
so required combination is 9+8+7+6+.....+1=9*10/2=45 numbers