Mathematics puzzles measure your problem solving skills and help you estimate how well you understand a problem.
One such interesting question is: "I am a 4 digit number between 5000 and 6000. My ones place is half of my tens place. My hundreds place is 3 times my tens place. What number am I?"
How to solve this?
Solving this equation requires simple calculation.
The first digit of the number begins with 5 as the number lies between 5000 and 6000.
Now the work is to find the other digits of the 4 digit number.
Let the digit in the ten's place is x. Going with the estimates given in the question, the digit in the hundreds place is 3x and the one in the ones place is x/2.
Now, since x is the double of the digit that is the ones place, it means it is an even digit like 2, 4, 6 and 8.
Again, the digit in the hundreds place is three times that of the digit in the tens place which means the number is a multiple of 3.
Going with the options, the only number that comes under this criteria is 6.
Hence the right answer is 5621.
Another way to understand this
"So the digits are arranged as follows:
thousands, hundreds, tens, ones.
Thous 5 or 6, hun 3t, tens t, ones (1/2)t0 can’t work for tens column because then the number would have 0 digits in the last 3 spots making it 5000 or 6000 and it must be in between the two. the tens digit must be even because the ones digit is one half that digit so t=2 produces ones as 1 and hundreds as 6
so 5621 is one possibility
lets try t=4 producing ones as 2 and hundreds as 12 which doesn’t work
any larger t will produce an unacceptable hundreds value.
our result is 5621," explains a Quora user.
Another user explains this in a different manner
"Let the four digits be a,b,c, and d . The given conditions are:
6000>(1000a+100b+10c+d)>5000,
so a=5 .
Assuming that each digit is a whole number, and be only a single digit, d must be 1 . If d was anything greater than 1 , b would end up being a two digit number. This means that:
d=1
c=2 since c=2d
b=6 since b=3c
So you are 5621"
