four digit number has first digit is half of third digit, total of second and third digit is eight, forth digit is multiply of first and third digit and total of all four digit is twelve
Answers
In this answer I claim that there is no solution to the above problem. maybe I'm wrong, maybe the question is flawed, or maybe the way I approched the problem is incorrect. But the point of this answer is not to find the answer or prove it doesn't exists. Instead it focuses on giving middle/high school children a path/framework to think and solve math problems in general. Please read the whole answer before reporting / deleting. thanks for reading ...
to simplify the problem, let's think of the digits as
ABCD
so, the following equations can be formed from the given data in problem
A = C/2
B+C = 8
D = A*C
A+B+C+D = 12
but it's still too complex (4 equations, really)
let's try to substitute everything we can using some term, say
I'll choose (but you can choose any one of them)
from the first equation,
similarly,
and,
so, A+B+C+D
( you may notice that the equation above is a quadratic equation in one variable. there are may ways to solve this, but I'll use the mid-term factoring mathod, which may not work everytime though ..... want to learn more about is, consider watching 3blue1brown's lockdown math live streams on youtube, he is an exceptional explainer, believe me you'll love him)
ok, so back at the eq...
which means, either x+2 = 0 or x-1=0
in other words, x = -2 or x=1
two values? which one should we choose .....
remember that when we made the substitution, we assumed C = x ...
that means x must be one of the 10 possible digits (0,1,2,...,9) .... so it can't be -2 because it is not a digit
therefore,
A = x/2 = 1/2 = 0.5
B = 8 - 1 = 7
D = A * 1 = 0.5
wait what? 0.5 is also not a digit ..... does that mean 1 is also a wrong answer?
if you thought that, you're absolutely right! the thing is that there is no possible number that satisfies the conditions given in the question and this has just been proven (by contradiction) because the only possible roots for the equation are both wrong....
yeah! I know, you must be thinking why did you waste our time? I appologize for that ... but if you're still reading this (and study in middle/high school) I can safely assume that you had a great 5 mins, reading a math problem which made you stop and ponder over several steps, trying to understand the real/deeper meaning beind each step .... and doing math is just that, not about getting the correct answer (if it exists).
Nonetheless, I think I gave you a relatively simple framework for solving these type of problems on your own in the future ... and I hope you atleast learned something small and new....
message for the op: if you are the writer of the question, try rechecking the question, maybe you had a slight mistake, or maybe I messed up the equation. if you find that, try to solve the problem on your own using the above steps .... you'll fail the first few times ... but you have to keep trying. and that's how you learn new concepts.
if you think you learned even a small thing, consider saying a thanks ... or if you think I wasted your time, rate it 1 star.....
Step-by-step explanation:
let 1st digit be x
2nd be y
3rd z and 4th is w then go for according to question that they have given