Kvant Math Problem 26
The numbering pattern is linear.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m24s
Source on kvant.digital
Problem
Suppose that each issue of the Kvant problem book contains five mathematics problems, and the journal is published monthly. Denote by $f(x,y)$ the number of the first problem in the $x$-th issue for the $y$-th year (for example, $f(6, 1970)=26$). Write a general formula for $f(x,y)$, for all $x$, $y$ ($1\le x\le12$, $y\ge 1970$).
Solve the equation $f(x,y)=y$.
Exploration
The numbering pattern is linear. Each issue contains five problems, and there are twelve issues per year, so each year contributes $12 \cdot 5 = 60$ problems. The first issue of 1970 starts with problem $1$, hence the first problem of issue $x$ in year $y$ is obtained by counting how many full issues came before it.
For a fixed year $y$, issue $x$ is preceded by $x-1$ issues of the same year, contributing $5(x-1)$ problems. Before year $y$, there are $y-1970$ full years, contributing $60(y-1970)$ problems. Thus the candidate formula is
$$f(x,y)=60(y-1970)+5(x-1)+1.$$
A quick check:
$$f(1,1970)=1,$$
and
$$f(6,1970)=25+1=26,$$
which matches the statement.
Now solve
$$f(x,y)=y.$$
Substituting the formula gives
$$60(y-1970)+5(x-1)+1=y.$$
Rearranging,
$$59y=60\cdot 1970-5x+4.$$
Compute
$$60\cdot 1970=118200,$$
so
$$59y=118204-5x.$$
Since $1\le x\le12$, the right-hand side ranges from
$$118199$$
down to
$$118144.$$
Dividing by $59$ suggests $y$ is around $2002$. Testing nearby values:
$$59\cdot 2003=118177,$$
hence
$$5x=118204-118177=27,$$
impossible.
Next,
$$59\cdot 2002=118118,$$
giving
$$5x=86,$$
impossible.
A congruence argument is cleaner. From
$$59y=118204-5x,$$
reduce modulo $5$:
$$59y\equiv 118204 \pmod 5.$$
Since
$$59\equiv -1 \pmod 5,\qquad 118204\equiv 4\pmod 5,$$
we get
$$-y\equiv4\pmod5,$$
hence
$$y\equiv1\pmod5.$$
Write $y=5k+1$. Substituting into the equation,
$$59(5k+1)=118204-5x,$$
so
$$295k+59=118204-5x.$$
Then
$$5x=118145-295k=5(23629-59k),$$
thus
$$x=23629-59k.$$
Because $1\le x\le12$, we need
$$1\le23629-59k\le12.$$
Now
$$23629=59\cdot400+29,$$
so
$$x=29-59(k-400).$$
The only values congruent to $29$ modulo $59$ are far outside $[1,12]$. Hence there is no solution.
The delicate point is avoiding arithmetic slips in the linear equation. The congruence argument eliminates all candidates cleanly.
Problem Understanding
We must first derive a formula for the number of the first problem in issue $x$ of year $y$, knowing that each issue contains five problems and that there are twelve issues per year beginning in 1970.
Then we must determine all pairs $(x,y)$ satisfying
$$f(x,y)=y.$$
This is a Type A problem. We must determine all solutions of the equation and prove that no others exist.
The core difficulty is the arithmetic verification that the linear Diophantine equation obtained from the formula has no admissible solutions with
$$1\le x\le12.$$
The expected answer is that no such pair exists, because the equation forces a residue condition incompatible with the allowed range of $x$.
Proof Architecture
First, derive the explicit formula
$$f(x,y)=60(y-1970)+5(x-1)+1,$$
by counting all problems appearing before issue $x$ of year $y$.
Second, substitute this expression into the equation
$$f(x,y)=y$$
to obtain a linear Diophantine equation in $x$ and $y$.
Third, reduce the equation modulo $5$ to deduce a congruence condition on $y$.
Fourth, substitute the resulting parametrization of $y$ back into the equation and solve for $x$.
Fifth, prove that the resulting expression for $x$ cannot lie in the interval $1\le x\le12$.
The most delicate step is the modular computation leading to the contradiction, because a small arithmetic error would produce nonexistent solutions.
Solution
Each year contains $12$ issues, and each issue contains $5$ problems. Hence each year contributes
$$12\cdot5=60$$
problems.
Before year $y$, there are
$$y-1970$$
complete years. Therefore these years contribute
$$60(y-1970)$$
problems.
Within year $y$, before issue $x$, there are
$$x-1$$
issues, contributing
$$5(x-1)$$
problems.
Since numbering begins with problem $1$, the first problem in issue $x$ of year $y$ has number
$$f(x,y)=60(y-1970)+5(x-1)+1.$$
For example,
$$f(6,1970)=60\cdot0+5\cdot5+1=26,$$
which agrees with the statement.
Now solve
$$f(x,y)=y.$$
Substituting the formula for $f(x,y)$ gives
$$60(y-1970)+5(x-1)+1=y.$$
Expanding,
$$60y-118200+5x-5+1=y.$$
Hence
$$59y=118204-5x.$$
Reduce this equation modulo $5$. Since
$$59\equiv-1\pmod5$$
and
$$118204\equiv4\pmod5,$$
we obtain
$$-y\equiv4\pmod5.$$
Thus
$$y\equiv1\pmod5.$$
Write
$$y=5k+1$$
for some integer $k$. Substitute into the equation:
$$59(5k+1)=118204-5x.$$
Therefore
$$295k+59=118204-5x,$$
so
$$5x=118145-295k.$$
Dividing by $5$,
$$x=23629-59k.$$
Since
$$23629=59\cdot400+29,$$
we may write
$$x=29-59(k-400).$$
Hence every possible value of $x$ is congruent to $29$ modulo $59$.
But the interval
$$1\le x\le12$$
contains no integer congruent to $29$ modulo $59$. Therefore no admissible value of $x$ exists.
Thus the equation
$$f(x,y)=y$$
has no solutions.
The general formula is
$$\boxed{f(x,y)=60(y-1970)+5(x-1)+1,}$$
and the solution set of
$$f(x,y)=y$$
is
$$\boxed{\varnothing}.$$
Verification of Key Steps
The first delicate step is the derivation of the formula for $f(x,y)$. A common mistake is to forget the shift by $1$ coming from the fact that the first problem is numbered $1$, not $0$. Checking
$$f(1,1970)=1$$
and
$$f(6,1970)=26$$
confirms the formula.
The second delicate step is the modular reduction. Starting from
$$59y=118204-5x,$$
reducing modulo $5$ gives
$$59y\equiv118204\pmod5.$$
Since
$$59\equiv4\equiv-1\pmod5$$
and
$$118204\equiv4\pmod5,$$
we obtain
$$-y\equiv4\pmod5,$$
hence
$$y\equiv1\pmod5.$$
Replacing $59$ by $-1$ incorrectly as $1$ would reverse the congruence and produce false candidates.
The final delicate step is the exclusion of admissible $x$. From
$$x=23629-59k,$$
every value of $x$ is congruent to
$$23629\equiv29\pmod{59}.$$
The interval $1\le x\le12$ contains only the residues
$$1,2,\dots,12,$$
none of which equals $29$. Hence no solution survives.
Alternative Approaches
Instead of using congruences, one may solve directly for $y$:
$$y=\frac{118204-5x}{59}.$$
For $1\le x\le12$, the numerator ranges from
$$118199$$
to
$$118144.$$
The corresponding values of $y$ lie strictly between
$$2002$$
and
$$2004.$$
Checking
$$59\cdot2002=118118,\qquad 59\cdot2003=118177,\qquad 59\cdot2004=118236$$
shows that none matches $118204-5x$ for an integer $x$ between $1$ and $12$.
The modular method is preferable because it isolates the arithmetic obstruction immediately and avoids checking several nearby integers separately.