Detailed mathematical approach

Problem Summary

Problem 154 asks for the number of coefficients in the expansion of \((x+y+z)^{200000}\) that are divisible by \(10^{12}\). Every coefficient on the layer \(a+b+c=200000\) of Pascal's pyramid has the form

$\binom{200000}{a,b,c}=\frac{200000!}{a!b!c!},\qquad a,b,c\ge 0,\quad a+b+c=200000.$

Because \(10^{12}=2^{12}5^{12}\), a coefficient qualifies exactly when it contains at least twelve factors of 2 and at least twelve factors of 5. The implementations never expand the polynomial; they reduce each candidate to two arithmetic tests on precomputed factorial valuations.

Mathematical Approach

The problem is really about the prime exponents inside multinomial coefficients. Once those exponents are written in a reusable form, the huge trinomial layer becomes a structured counting problem.

Prime valuations are the real divisibility test

For any prime \(p\), let \(v_p(m)\) denote the exponent of \(p\) in the prime factorization of \(m\). Then for a trinomial coefficient we have

$v_p\!\left(\binom{n}{a,b,c}\right)=v_p(n!)-v_p(a!)-v_p(b!)-v_p(c!),\qquad a+b+c=n.$

With \(n=200000\), divisibility by \(10^{12}\) is therefore equivalent to the pair of inequalities

$v_2\!\left(\binom{n}{a,b,c}\right)\ge 12,\qquad v_5\!\left(\binom{n}{a,b,c}\right)\ge 12.$

No other prime matters. The entire search reduces to deciding, for every admissible triple \((a,b,c)\), whether the 2-adic and 5-adic valuations are both large enough.

Factorial valuations are built by a prefix recurrence

The implementations precompute two tables:

$V_2(m)=v_2(m!),\qquad V_5(m)=v_5(m!),\qquad 0\le m\le n.$

The key recurrence is

$V_p(0)=0,\qquad V_p(m)=V_p(m-1)+v_p(m).$

Here \(v_p(m)\) is obtained by repeatedly dividing \(m\) by \(p\) until divisibility stops. This recurrence is exactly what the code computes when it walks from \(1\) up to \(200000\). Mathematically it matches Legendre's formula,

$V_p(m)=\sum_{r\ge 1}\left\lfloor\frac{m}{p^r}\right\rfloor,$

but the cumulative table is more convenient for millions of repeated queries.

Each coefficient becomes two constant-time inequalities

Once the tables are ready, define the fixed thresholds

$T_2=V_2(n)-12,\qquad T_5=V_5(n)-12.$

Then a triple \((a,b,c)\) contributes to the answer exactly when

$V_2(a)+V_2(b)+V_2(c)\le T_2,$

$V_5(a)+V_5(b)+V_5(c)\le T_5.$

This is the central invariant of the solution. The enormous value of \(\binom{200000}{a,b,c}\) is never formed explicitly; the implementation only compares three array lookups against two fixed bounds.

Symmetry collapses the search to one ordered wedge

The trinomial coefficient is symmetric in \(a\), \(b\), and \(c\), so the implementations enumerate only ordered triples

$a\le b\le c,\qquad a+b+c=n.$

After choosing \(a\) and \(b\), the third value is forced as \(c=n-a-b\). The ordering condition gives the sharp bounds

$0\le a\le \left\lfloor\frac{n}{3}\right\rfloor,\qquad a\le b\le \left\lfloor\frac{n-a}{2}\right\rfloor.$

Every accepted ordered triple stands for several coefficients obtained by permuting its coordinates. Its multiplicity is

$ \operatorname{mult}(a,b,c)= \begin{cases} 1,& a=b=c,\\ 3,& a=b\ne c\ \text{or}\ a\ne b=c,\\ 6,& a,b,c\ \text{all distinct}. \end{cases} $

So the full triangular layer of Pascal's pyramid is replaced by one sixth-sized ordered region together with a small orbit-size correction.

Worked Example: testing one ordered triple

The real computation uses \(n=200000\), but a smaller example makes the arithmetic transparent. Take \(n=20\) and the ordered triple \((a,b,c)=(4,6,10)\).

For the factor 5,

$V_5(20)=4,\qquad V_5(4)=0,\qquad V_5(6)=1,\qquad V_5(10)=2,$

so

$v_5\!\left(\binom{20}{4,6,10}\right)=4-(0+1+2)=1.$

For the factor 2,

$V_2(20)=18,\qquad V_2(4)=3,\qquad V_2(6)=4,\qquad V_2(10)=8,$

hence

$v_2\!\left(\binom{20}{4,6,10}\right)=18-(3+4+8)=3.$

This coefficient has only one factor of 5 and three factors of 2, so it fails the \(10^{12}\) test immediately. Because \(4\), \(6\), and \(10\) are all distinct, a successful triple of this shape would contribute 6 coefficients to the final count.

How the Code Works

Precomputing the factorial valuations

The C++, Python, and Java implementations allocate two arrays of length \(n+1\). For each \(m=1,2,\dots,n\), they count how many times \(m\) is divisible by 2 and by 5, then add those counts to the previous prefix totals. After this pass, every query \(v_2(m!)\) or \(v_5(m!)\) becomes an \(O(1)\) array lookup.

Enumerating the ordered region

They then loop over \(a\) from \(0\) to \(\lfloor n/3\rfloor\), over \(b\) from \(a\) to \(\lfloor(n-a)/2\rfloor\), and set \(c=n-a-b\). For each ordered triple, the implementation evaluates the two valuation inequalities above. When both hold, it adds 1, 3, or 6 according to whether the triple has three equal entries, exactly two equal entries, or three distinct entries.

Serial and parallel accumulation

The mathematical test is the same in all three languages. The C++ and Python implementations split the outer \(a\)-range into chunks so different workers accumulate disjoint partial totals and combine them at the end. The Java implementation performs the same ordered scan serially.

Complexity Analysis

Building the factorial valuation tables costs \(O(n\log n)\) in the straightforward repeated-division model used by the implementations, and it uses \(O(n)\) memory for the two prefix arrays.

The ordered enumeration visits

$\sum_{a=0}^{\lfloor n/3\rfloor}\left(\left\lfloor\frac{n-a}{2}\right\rfloor-a+1\right)=\Theta(n^2)$

candidate pairs \((a,b)\), so the dominant running time is \(O(n^2)\). Restricting to \(a\le b\le c\) saves roughly a factor of six compared with scanning every permutation separately, but the asymptotic order remains quadratic. Memory stays linear because only the two valuation tables and a few counters are needed.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=154
2. Multinomial theorem: Wikipedia - Multinomial theorem
3. Legendre's formula: Wikipedia - Legendre's formula
4. \(p\)-adic valuation: Wikipedia - p-adic valuation
5. Factorial: Wikipedia - Factorial