Inverting 4-digits numbers.

Number Theory Level 3

Let $n$ be a four digits number with non-zero units digit. Let $f(n)$ be the number that we get from inverting the order of the digits of $n$ , for example : $f(3259)=9523$ .

Find $n$ such that $f(n)=4n+3$ .

Note : remember that $n$ is a four digits number with non-zero units digit.

The answer is 1997.

1 solution

Kaushal Agrawal
Jan 13, 2015

since 4n+3 is also a 4 digit number, leading digit must be 1 or 2. It can't be 2 since in that case n is either 2xy8 or 2xy9 and clearly the unit's digit of RHS (4n+3) becomes 5 and 9 respectively instead of 2 (inverting the digits of n to get 8yx2 or 9yx2). Hence n is of the form 1xyz.

Similar argument tells us that last digit of n must be 2 or 7. But 1xy2 is also not possible as RHS > 4000 and LHS < 4000. Hence n is 1xy7.

We observe that x>6 since otherwise, RHS < 7000 whereas LHS >7000. Check for the 3 resultant equations obtained by taking n=17y7 n=18y7 n=19y7

You will get y as a digit only for the last case where it comes out to be 9. Hence n=1997.

Inverting 4-digits numbers.

The answer is 1997.

1 solution

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