Project Euler/40/Solution

Main article: Project Euler/40

Solution

Here is a calculation of the product $d_{1}\times d_{10}\times d_{100}\times d_{1000}\times d_{10000}\times d_{100000}\times d_{1000000}$, where $d_{n}$ is the $n$-th digit of the Champernowne constant $C_{10}=0.123456789101112...$.

We use the established method involving cumulative digit counts ($C_{k}$) for blocks of $k$-digit numbers.

• $C_{1}=9$
• $C_{2}=189$
• $C_{3}=2889$
• $C_{4}=38889$
• $C_{5}=488889$
• $C_{6}=5888889$

Finding the Digits

• Finding $d_{1}$

The first digit ($n=1$) is the first digit of '1'.

$d_{1}=1$

• Finding $d_{10}$

$n=10$. Falls into the 2-digit block ($k=2$), since $C_{1}<10\leq C_{2}$.

Position within block: $m=n-C_{1}=10-9=1$.

Integer index ($0$-based): $num\_index=\lfloor (m-1)/k\rfloor =\lfloor (1-1)/2\rfloor =0$.

Integer $N=10^{k-1}+num\_index=10^{2-1}+0=10$.

Digit position within $N$ ($1$-based): $digit\_pos=(m-1){\pmod {k}}+1=(1-1){\pmod {2}}+1=1$. The 1st digit of 10.

$d_{10}=1$

• Finding $d_{100}$

$n=100$. Falls into the 2-digit block ($k=2$), since $C_{1}<100\leq C_{2}$.

Position within block: $m=100-C_{1}=100-9=91$.

Integer index: $num\_index=\lfloor (91-1)/2\rfloor =\lfloor 90/2\rfloor =45$.

Integer $N=10^{2-1}+45=10+45=55$.

Digit position: $digit\_pos=(91-1){\pmod {2}}+1=90{\pmod {2}}+1=1$. The 1st digit of 55.

$d_{100}=5$

• Finding $d_{1000}$

$n=1000$. Falls into the 3-digit block ($k=3$), since $C_{2}<1000\leq C_{3}$.

Position within block: $m=1000-C_{2}=1000-189=811$.

Integer index: $num\_index=\lfloor (811-1)/3\rfloor =\lfloor 810/3\rfloor =270$.

Integer $N=10^{3-1}+270=100+270=370$.

Digit position: $digit\_pos=(811-1){\pmod {3}}+1=810{\pmod {3}}+1=1$. The 1st digit of 370.

$d_{1000}=3$

• Finding $d_{10000}$

$n=10000$. Falls into the 4-digit block ($k=4$), since $C_{3}<10000\leq C_{4}$.

Position within block: $m=10000-C_{3}=10000-2889=7111$.

Integer index: $num\_index=\lfloor (7111-1)/4\rfloor =\lfloor 7110/4\rfloor =1777$.

Integer $N=10^{4-1}+1777=1000+1777=2777$.

Digit position: $digit\_pos=(7111-1){\pmod {4}}+1=7110{\pmod {4}}+1=2+1=3$. The 3rd digit of 2777.

$d_{10000}=7$

• Finding $d_{100000}$

$n=100000$. Falls into the 5-digit block ($k=5$), since $C_{4}<100000\leq C_{5}$.

Position within block: $m=100000-C_{4}=100000-38889=61111$.

Integer index: $num\_index=\lfloor (61111-1)/5\rfloor =\lfloor 61110/5\rfloor =12222$.

Integer $N=10^{5-1}+12222=10000+12222=22222$.

Digit position: $digit\_pos=(61111-1){\pmod {5}}+1=61110{\pmod {5}}+1=0+1=1$. The 1st digit of 22222.

$d_{100000}=2$

• Finding $d_{1000000}$

$n=1000000$. Falls into the 6-digit block ($k=6$), since $C_{5}<1000000\leq C_{6}$.

Position within block: $m=1000000-C_{5}=1000000-488889=511111$.

Integer index: $num\_index=\lfloor (511111-1)/6\rfloor =\lfloor 511110/6\rfloor =85185$.

Integer $N=10^{6-1}+85185=100000+85185=185185$.

Digit position: $digit\_pos=(511111-1){\pmod {6}}+1=511110{\pmod {6}}+1=0+1=1$. The 1st digit of 185185.

$d_{1000000}=1$

Calculating the Product

We need to compute the product $P$:

$P=d_{1}\times d_{10}\times d_{100}\times d_{1000}\times d_{10000}\times d_{100000}\times d_{1000000}$

$P=1\times 1\times 5\times 3\times 7\times 2\times 1$

$P=1\times 5\times 3\times 7\times 2$

$P=5\times 3\times 7\times 2$

$P=15\times 7\times 2$

$P=105\times 2$

$P=210$

The product of the specified digits is 210.