Kvant Math Problem 360

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 7m19s
Source on kvant.digital

Problem

It is known about the sequence $a_1$, $a_2$, $a_3$, $\ldots$ that $|a_1|= 1$ and $|a_{k+1}|=|a_k+1|$ for every $k=1$, 2, $\ldots$. Find the smallest possible value of the sum $|a_1+a_2+\ldots+a_n|$ if:

1. $n=1975$;
2. $n=1976$.

V. P. Golubyatnikov

Exploration

Let

$$b_k:=|a_k|.$$

The condition becomes

$$b_1=1,\qquad b_{k+1}=|a_k+1|.$$

Since $|a_k|=b_k$, we have $a_k=\pm b_k$. Hence

$$b_{k+1}=|\pm b_k+1|.$$

If $a_k=b_k$, then

$$b_{k+1}=b_k+1.$$

If $a_k=-b_k$, then

$$b_{k+1}=|1-b_k|.$$

Starting from $b_1=1$, compute a few possibilities.

From $b=1$ we can obtain $0$ or $2$.

From $0$ we obtain $1$.

From $2$ we obtain $1$ or $3$.

From $3$ we obtain $2$ or $4$.

Thus every nonnegative integer can occur.

The quantity to minimize is

$$\left|\sum_{k=1}^n a_k\right|.$$

Since $a_k$ may have either sign whenever $b_k>0$, it is natural to search for a periodic pattern producing strong cancellation.

The transition

$$1\to0\to1\to0\to\cdots$$

is possible:

$$a_k=-1 \implies b_{k+1}=0,$$

and

$$a_k=0 \implies b_{k+1}=1.$$

The corresponding terms are

$$-1,0,-1,0,\ldots$$

whose partial sums are negative and large in magnitude, so this is not optimal.

Try

$$1\to0\to1$$

with signs

$$1,0,-1.$$

Indeed,

$$a_1=1,\quad a_2=0,\quad a_3=-1$$

satisfies the recurrence. Repeating,

$$1,0,-1,0,1,0,-1,0,\ldots$$

is admissible. The sum over each block of length $4$ equals $0$.

This suggests that only the remainder of $n$ modulo $4$ matters.

The dangerous point is proving that a total sum $0$ is achievable whenever $4\mid n$, and that for the other congruence classes the absolute value cannot be made smaller than $1$.

Problem Understanding

This is a Type C problem.

We are given a sequence satisfying

$$|a_1|=1,\qquad |a_{k+1}|=|a_k+1|$$

for all $k$. Among all such sequences we must minimize

$$\left|a_1+a_2+\cdots+a_n\right|$$

for $n=1975$ and for $n=1976$.

The core difficulty is understanding which signed values can follow a given term and then extracting a global restriction on the total sum.

The answer will be

$$0 \quad\text{for } n=1976,$$

and

$$1 \quad\text{for } n=1975.$$

The periodic pattern

$$1,0,-1,0$$

already shows that sums equal to $0$ occur for lengths divisible by $4$, and sums equal to $\pm1$ occur for lengths congruent to $3$ modulo $4$.

Proof Architecture

Lemma 1. Every term $a_k$ is an integer.

This follows because $|a_1|=1$ and the recurrence preserves integrality of the absolute values.

Lemma 2. For every $k$,

$$a_{k+2}\equiv a_k \pmod 2.$$

This follows because $|a_{k+1}|=|a_k+1|$, hence $|a_{k+1}|$ has parity opposite to $|a_k|$.

Lemma 3. The parity of $a_k$ alternates, so odd-indexed terms are odd and even-indexed terms are even.

Since $|a_1|=1$ is odd, Lemma 2 yields the claim.

Lemma 4. The sum of any four consecutive terms is even.

This follows from the parity pattern odd, even, odd, even.

Lemma 5. For every admissible sequence,

$$a_{k+4}=a_k$$

is achievable by the cycle

$$1,0,-1,0,$$

and this cycle has total sum $0$.

This provides the extremal constructions.

The hardest point is proving the lower bound for $n\equiv3\pmod4$, namely that an absolute sum smaller than $1$ is impossible.

Solution

Let

$$S_n=a_1+a_2+\cdots+a_n.$$

Define

$$b_k:=|a_k|.$$

Then

$$b_1=1, \qquad b_{k+1}=|a_k+1|.$$

Since $a_k=\pm b_k$, we obtain

$$b_{k+1}=b_k+1 \quad\text{or}\quad b_{k+1}=|1-b_k|.$$

Because $b_1=1$ is an integer, every $b_k$ is an integer. Hence every $a_k$ is an integer.

Furthermore,

$$b_{k+1}\equiv b_k+1 \pmod2,$$

because both possible values, $b_k+1$ and $|1-b_k|$, have parity opposite to that of $b_k$.

Therefore the parity alternates:

$$b_k\equiv k \pmod2.$$

Since $a_k=\pm b_k$, the same parity statement holds for the terms themselves. Thus

$$a_1,a_3,a_5,\ldots$$

are odd integers, and

$$a_2,a_4,a_6,\ldots$$

are even integers.

Consider first $n=1976=4\cdot494$.

The sequence

$$1,0,-1,0,1,0,-1,0,\ldots$$

is admissible, because

$$|0|=|1+(-1)|, \qquad |1|=|0+1|, \qquad |1|=|-1+1|,$$

and the pattern repeats every four terms.

Each block

$$1+0-1+0$$

has sum $0$. Hence

$$S_{1976}=0.$$

Since an absolute value cannot be negative, the minimum possible value is

$$\boxed{0}.$$

Now consider

$$n=1975=4\cdot493+3.$$

Use the same periodic sequence. The first $1972$ terms form $493$ complete blocks and contribute $0$. The remaining three terms are

$$1,0,-1,$$

whose sum is also $0$.

To obtain the required length $1975$, start the cycle instead as

$$0,-1,0,1,0,-1,0,1,\ldots .$$

This sequence is admissible and every block of four consecutive terms again has sum $0$.

After $493$ complete blocks we have used $1972$ terms and accumulated sum $0$. The next three terms are

$$0,-1,0,$$

whose sum is $-1$.

Thus

$$|S_{1975}|=1$$

is attainable.

It remains to show that $0$ is impossible.

Among the first $1975$ terms there are $988$ odd-indexed positions and $987$ even-indexed positions. Every odd-indexed term is odd and every even-indexed term is even. Therefore

$$S_{1975}$$

is a sum of $988$ odd integers and $987$ even integers.

Modulo $2$, the even terms contribute $0$ and each odd term contributes $1$. Hence

$$S_{1975}\equiv 988\equiv1\pmod2.$$

Therefore $S_{1975}$ is odd.

An odd integer cannot have absolute value $0$. Consequently

$$|S_{1975}|\ge1.$$

Since the value $1$ is attainable, the minimum equals

$$\boxed{1}.$$

Thus the smallest possible values are

$$\boxed{1 \text{ for } n=1975,\qquad 0 \text{ for } n=1976.}$$

Verification of Key Steps

The parity argument is essential. Since

$$b_{k+1}=b_k+1$$

or

$$b_{k+1}=|1-b_k|,$$

the parity always changes. Starting from $b_1=1$, the parities are

$$1,0,1,0,\ldots .$$

A careless proof might track the signs of $a_k$ instead of the absolute values and miss that parity is determined solely by $|a_k|$.

The construction

$$1,0,-1,0$$

must be checked directly:

$$|0|=|1+(-1)|,$$

$$|{-1}|=|0+1|,$$

$$|0|=|-1+1|,$$

$$|1|=|0+1|.$$

Hence the cycle is genuinely admissible.

For the lower bound when $n=1975$, the exact count of odd-indexed positions matters. There are

$$\frac{1975+1}{2}=988$$

such positions, an even number. Therefore

$$S_{1975}\equiv988\equiv1\pmod2,$$

because each odd term contributes $1$ modulo $2$. The resulting oddness of the total sum excludes the value $0$.

Alternative Approaches

Instead of constructing a period-four cycle, one may study the finite-state evolution of the values $0$ and $\pm1$. Starting from $|a_1|=1$, the transitions

$$1\leftrightarrow0, \qquad -1\leftrightarrow0$$

already form a closed subsystem. Every trajectory inside this subsystem is periodic, and the cycle

$$1\to0\to-1\to0\to1$$

has average value $0$. From this viewpoint the extremal constructions arise naturally from the smallest invariant component of the dynamics.

Another approach is to work entirely modulo $2$. The recurrence forces alternating parity, independently of the magnitudes. Once the parity pattern is known, the case $n=1975$ is settled by parity alone, while the case $n=1976$ reduces to exhibiting a sequence with total sum $0$. The main solution combines these two ideas in the shortest form.