If we add $1996$ to $1997$, we first add the unit digits $6$ and $7$. Obtaining $13$, we write down $3$ and “carry” $1$ to the next column. Thus we make a carry. Continuing, we see that we are to make three carries in total. Does there exist a positive integer $k$ such that adding $1996\cdot k$ to $1997\cdot k$ no carry arises during the whole calculation?
Problem
Source: Baltic Way 1997
Tags: number theory proposed, number theory
Rust
29.01.2011 01:07
$N=1996+1997=3*11^3$. Therefore we can take $k=\frac{10^{242}-1}{3997}$.
diks94
29.01.2011 02:14
@WakeUp if This A trick Question Then the Answer Would Be 0. @Rust how Do We Get \[k=\frac{10^{242} -1}{3997}\]
ocha
29.01.2011 03:55
$\frac{10^{242}-1}{3997} \not \in \mathbb{Z}$ Since $1996k = 2k\cdot 10^3 - 4k$ and $1997k = 2k\cdot 10^3 - 3k$ we can simply pick a three digit number $k$ such that the digits of $2k$ are all less than $5$ and the digits of $3k$ and $4k$ (also three digits numbers) are all greater than $5$. $k=222$ works, so $1996k$ and $1997k$ have all digits less than $5$, so no carries.
$1996\cdot 222 = 443112$, $1997\cdot 222 = 443334$
Rust
29.01.2011 08:21
$10^2=1\mod 11\to 10^{2*11^2}=1\mod 11^3$ and $10^{242}=1\mod 3$. Therefore $k=\frac{10^{242}-1}{3997}\in Z, 3997=1996+1997=3*11^3$. If $k*1997=A$, then $B=k*1996=10^{242}-1-M$. Because $10^{242}-1$ is number with 242 digits 9 $A+B$ give not carryes.
ocha
29.01.2011 08:36
Nice. A small typo: $3\cdot 11^3 = 3993$.