Number system puzzle problem 8: Leftmost digit of a natural number at least 7 digits long is 3. When it is dropped, number becomes one-fifth...Read on.
CAT Number system puzzle problem 8
The leftmost digit of a natural number at least 7 digit long is 3. When the leftmost digit is dropped, the number formed by the rest of the digits becomes one-fifth of the original number. What is the product of the non-zero digits of the original number?
Time to solve: 90 seconds.
Hint: Solve using the very basic concepts intelligently.
Solution to the Number system puzzle problem 8: Key pattern identification using place value concept, asking most important question and analyzing for answer
Forgetting for a moment the uncertainty of the number of digits, let us assume the desired number $n$ digits long to be,
$N=(3p)$, where $p$ is the number formed by the digits left after dropping leftmost digit 3,
Or, $N=3\times{10^n}+p$..............(1)
The number $p$ is, $\frac{1}{5}$th of $N$, that is,
$N=5p$...................................(2)
From equation (1) and equation (2),
$5p=3\times{10^n}+p$,
Or, $4p=3\times{10^n}$.
What it means is, multiplying $p$ by 4, you have to get 3 followed by all zeroes, $n$ in number.
How can it happen?
Better still, how can it happen the easiest way?
The easiest and the only way must be,
$4\times{75}\times{10^{n-2}}=300\times{10^{n-2}}$, where $p=75\times{10^{n-2}}$.
$\Rightarrow 4\times{75}\times{10^{n-2}}=3\times{10^n}=N-p$, the place value of the most significant digit 3.
So, the most significant two digits of the number $p$ formed by dropping leftmost digit of $N$, must be $75$, the number of digits following will all have to be $0$.
So, $N=375\times{10^{n-2}}$.
The desired product,
$3\times{7}\times{5}=105$.
There is no need to worry about the uncertain length of the desired number.
If you identify this not so difficult mechanism, there is no need to go into equations at all to get the answer.
Crux of the mechanism, the key pattern is,
For the original number $N$ to be 5 times the number formed by dropping the most significant digit 3, the leftmost two digits of the shortened number are to be 75 with rest all zeroes, so that 4 times 75 equal to 300 becomes the leftmost three digits on $N$ followed by all zeroes. Adding the shortened number results in $N$ to be 5 times the shortened number.
This is shown below visually,
End note
At first the problem may seem to be daunting and very difficult.
But whatever be the problem, if you start solving by putting what you know on paper and analyze it, it is just a matter of time before you would discover the key to the solution.
To go through more such selected hard CAT Quant problems with step by step easy solutions, click here.