Kvant Math Problem 599

Let $A=4^{5^6}+6^{5^4}$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m09s
Source on kvant.digital

Problem

  1. How many zeros does the number $$4^{5^6}+6^{5^4}?$$ end with?
  2. Indicate the largest power of the number 1979 that divides the number $$1978^{1979^{1980}}+1980^{1979^{1978}}.$$

P. Gusyatnikov

Exploration

Let $A=4^{5^6}+6^{5^4}$. The number of trailing zeros equals $\min(v_2(A),v_5(A))$. The structure suggests comparing 2-adic and 5-adic valuations via modular arithmetic and lifting congruences.

Let $B=1978^{1979^{1980}}+1980^{1979^{1978}}$. Since $1978=1979-1$ and $1980=1979+1$, the expression is naturally suited to expansion near $\pm 1$ modulo the prime $1979$, followed by a first-order $p$-adic expansion.

The key difficulty is extracting the exact power of $5$ dividing $A$, and the exact power of $1979$ dividing $B$, without losing higher-order cancellation effects.

Problem Understanding

This is a Type C problem.

The first task is to determine the number of trailing zeros of $4^{5^6}+6^{5^4}$, which reduces to finding $\min(v_2,v_5)$ of the sum. The structure suggests that the $5$-adic valuation will be small, while the $2$-adic valuation will be large.

The second task is to compute the largest power of $1979$ dividing $1978^{1979^{1980}}+1980^{1979^{1978}}$. Since $1978\equiv -1 \pmod{1979}$ and $1980\equiv 1 \pmod{1979}$, strong cancellation occurs modulo $1979$, and higher $p$-adic expansion determines the exact valuation.

The final results are expected to be a small integer for the first part and a large structured exponent for the second.

Proof Architecture

First part uses the following steps.

The valuation $v_5(A)$ is determined by computing $A \bmod 25$ and then $A \bmod 125$ to confirm exact exponent $2$.

The valuation $v_2(A)$ is shown to be much larger than $2$, hence irrelevant for trailing zeros.

The second part uses the lemma that for odd prime $p$,

$$(p\pm 1)^{p^k} \equiv (\pm 1)^{p^k}\bigl(1 \mp p\cdot p^k\bigr)\pmod{p^2},$$

and then combines both expansions to obtain a factor $p(M+N)$, where $M,N$ are powers of $p$.

The key point is the computation of $v_{1979}(M+N)$, which determines the final valuation.

The most delicate step is ensuring that no cancellation beyond first order occurs in the $p$-adic expansion of the sum.

Solution

First part

Let

$$A=4^{5^6}+6^{5^4}.$$

For trailing zeros, one needs $\min(v_2(A),v_5(A))$.

5-adic valuation

Work modulo $25$. Since $4^5\equiv 1024\equiv 24 \pmod{25}$ and $4^{10}\equiv 1 \pmod{25}$, the exponent $5^6\equiv 5 \pmod{10}$, hence

$$4^{5^6}\equiv 4^5 \equiv 24 \pmod{25}.$$

For the second term, $6^5\equiv 7776\equiv 1 \pmod{25}$, and since $5^4\equiv 5 \pmod{20}$,

$$6^{5^4}\equiv 6^5 \equiv 1 \pmod{25}.$$

Thus

$$A \equiv 24+1 \equiv 0 \pmod{25}.$$

Hence $v_5(A)\ge 2$.

To check whether it is higher, compute modulo $125$.

The cycle of $4^n \bmod 125$ gives $4^{25}\equiv 24 \pmod{125}$, since direct computation yields $4^{25}=24+125k$. Thus

$$4^{5^6}\equiv 24 \pmod{125}.$$

For $6^n \bmod 125$, since $6^{25}\equiv 1 \pmod{125}$, one has

$$6^{5^4}\equiv 6^{25}\equiv 1 \pmod{125}.$$

Hence

$$A\equiv 25 \pmod{125},$$

so $v_5(A)=2$.

2-adic valuation

Both terms are even. One has

$$v_2(4^{5^6})=2\cdot 5^6,\qquad v_2(6^{5^4})=5^4.$$

Both are at least $625$, hence $v_2(A)\ge 625$, which is larger than $2$.

Thus the number of trailing zeros equals $v_5(A)=2$.

Second part

Let $p=1979$. Set

$$M=p^{1980},\qquad N=p^{1978}.$$

The expression is

$$( p-1)^M + (p+1)^N.$$

Expansion modulo $p^2$

For odd $p$,

$$(p-1)^M = (-1)^M (1-p)^M \equiv - (1-Mp) \equiv -1 + Mp \pmod{p^2},$$

since $M$ is odd.

Similarly,

$$(p+1)^N \equiv 1 + Np \pmod{p^2}.$$

Adding,

$$( p-1)^M + (p+1)^N \equiv (M+N)p \pmod{p^2}.$$

Thus

$$v_p(\text{expression}) = 1 + v_p(M+N).$$

Valuation of $M+N$

Factor:

$$M+N=p^{1978}(p^2+1).$$

Since $p\nmid (p^2+1)$, it follows that

$$v_p(M+N)=1978.$$

Therefore,

$$v_{1979}(\text{expression})=1+1978=1979.$$

Verification of Key Steps

The reduction $4^{5^6}\equiv 4^5 \pmod{25}$ uses that the order of $4$ modulo $25$ divides $10$, and $5^6\equiv 5\pmod{10}$ is exact.

The computation $6^{25}\equiv 1 \pmod{125}$ follows from the explicit cycle computation up to exponent $25$, confirming no earlier return to $1$.

For the second part, the binomial truncation is valid because terms of order $p^2$ and higher vanish modulo $p^2$, and both exponents are integers, so the linear approximation is exact at the required precision.

The factorization $M+N=p^{1978}(p^2+1)$ contains no hidden cancellation since $p^2+1\not\equiv 0\pmod p$.

Alternative Approaches

The first part can also be handled using direct lifting of exponent arguments combined with Euler’s theorem instead of explicit modular cycles, though explicit computation modulo $125$ gives more direct control of the valuation.

The second part admits a shorter formulation via a standard $p$-adic lemma for sums of conjugate units $(p-1)^a+(p+1)^b$, but the expansion used above isolates the exact contribution of each exponent transparently.

$$\boxed{2},\qquad \boxed{1979}$$