Canada Perfect Number

A Canada perfect number or Canada perfect number (from the English Canada perfect number ) is a natural number whose sum of the nontrivial divisors is equal to the sum of the squares of its digits in the decimal system. ${\displaystyle n\in \mathbb {N} }$

In other words:

A composite number is called a Canada-perfect number if and only if : ${\displaystyle n\in \mathbb {N} }$

${\displaystyle \sum _{1\leq i\leq c(n)}\ell _{i}^{2}=\sum _{{d|n} \atop {1<d<n}}d}$, where the digits are in the decimal representation of the number . ${\displaystyle \ell _{1},\ell _{2},\ldots \ell _{c(n)}}$ ${\displaystyle n}$

On the occasion of Canada's 125th birthday , these numbers were defined by J. Fabrykowski, B. Wolk and R. Padmanabhan ( University of Manitoba ), with 125 being the smallest of them.

These numbers got their name from the perfect numbers , for which the sum of their nontrivial divisors of, however, is and for which the sum of the squares of their digits does not matter. ${\displaystyle n}$ ${\displaystyle n-1}$

Examples

• ${\displaystyle 125}$is a Canada-perfect number. It has only two nontrivial factors, namely and . Thus: ${\displaystyle 5}$ ${\displaystyle 25}$

${\displaystyle 5+25=1^{2}+2^{2}+5^{2}=30}$

• ${\displaystyle 581}$is a Canada-perfect number. It also has only two nontrivial factors, namely and . Thus: ${\displaystyle 7}$ ${\displaystyle 83}$

${\displaystyle 7+83=5^{2}+8^{2}+1^{2}=90}$

• ${\displaystyle 8549}$is a Canada-perfect number. It also has only two nontrivial factors, namely and . Thus: ${\displaystyle 83}$ ${\displaystyle 103}$

${\displaystyle 83+103=8^{2}+5^{2}+4^{2}+9^{2}=186}$

• ${\displaystyle 16999}$is a Canada-perfect number. It also has only two nontrivial factors, namely and . Thus: ${\displaystyle 89}$ ${\displaystyle 191}$

${\displaystyle 89+191=1^{2}+6^{2}+9^{2}+9^{2}+9^{2}=280}$

properties

• There are no Canada-perfect numbers greater than . ${\displaystyle 236196}$

Proof:

• Proposition 1: The sum of the squares of the digits of a natural number is at most 81 times its number of digits . ${\displaystyle n\in \mathbb {N} }$

Proof:

Be a -digit number. The sum of the squares of the digits of this number is maximally large if the number consists exclusively of ern. Thus is the maximum sum of the squares of the digits . So: ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle k}$ ${\displaystyle n}$ ${\displaystyle 9}$ ${\displaystyle k\cdot 9^{2}}$

Sum of the squares of the digits in a number Number of digits in that number ${\displaystyle \leq k\cdot 9^{2}=81k=81\cdot }$ ${\displaystyle \Box }$

• Proposition 2: The sum of the nontrivial divisors of a composite number is at least equal to its square root . ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle {\sqrt {n}}}$

Idea of ​​proof:

The more prime divisors a number has, the higher the sum of its nontrivial divisors. For example, the sum of the nontrivial divisors of a number with only two prime divisors is at least twice its square root, so it is (the proof would be an extreme value problem ). But if is a number , it has a single non-trivial divisor, namely . For all other composite numbers with more prime factors, the sum of their nontrivial factors is greater than . ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle n=p\cdot q}$ ${\displaystyle p,q\in \mathbb {P} }$ ${\displaystyle p+q\geq 2{\sqrt {n}}}$ ${\displaystyle n=p^{2}}$ ${\displaystyle n}$ ${\displaystyle p={\sqrt {n}}}$ ${\displaystyle n}$ ${\displaystyle {\sqrt {n}}}$ ${\displaystyle \Box }$

A number is a Canada-perfect number if the sum of its nontrivial divisors is equal to the sum of the squares of its digits. Let this value be the same . Consider a few examples: ${\displaystyle n\in \mathbb {N} }$ ${\displaystyle s}$

• If the number has five digits (i.e. ), then the sum of the squares of its digits is at most because of subordinate clause 1 . The sum of their nontrivial factors is at least because of Proposition 2 . So it is, and solutions to this problem are possible. ${\displaystyle n}$ ${\displaystyle 10^{4}<n<10^{5}}$ ${\displaystyle s\leq 5\cdot 9^{2}=5\cdot 81=405}$ ${\displaystyle s\geq {\sqrt {n}}>{\sqrt {10^{4}}}\approx 31{,}6}$ ${\displaystyle 31{,}6<s\leq 405}$
• If the number has six digits (i.e. ), the sum of the squares of its digits is at most due to subordinate clause 1 . The sum of their nontrivial factors is at least because of Proposition 2 . So it is, and solutions to this problem are possible. ${\displaystyle n}$ ${\displaystyle 10^{5}<n<10^{6}}$ ${\displaystyle s\leq 6\cdot 9^{2}=6\cdot 81=486}$ ${\displaystyle s\geq {\sqrt {n}}>{\sqrt {10^{5}}}\approx 316{,}2}$ ${\displaystyle 316{,}2<s\leq 486}$
• If the number is , then the number has 6 digits and thus the sum of the squares of its digits is at most because of the auxiliary clause 1 . The sum of their nontrivial factors is at least because of Proposition 2 . So it is with what no longer a solution is possible. ${\displaystyle n>236196}$ ${\displaystyle s\leq 6\cdot 9^{2}=6\cdot 81=486}$ ${\displaystyle s>{\sqrt {236196}}=486}$ ${\displaystyle 486<s\leq 486}$
• If the number had seven digits (i.e. ), then the sum of the squares of its digits would be at most because of proposition 1 . The sum of their nontrivial divisors would be at least because of Proposition 2 . So it would have to be what is no longer possible. With even higher ones there would be no suitable interval for more. ${\displaystyle n}$ ${\displaystyle 10^{6}<n<10^{7}}$ ${\displaystyle s\leq 7\cdot 9^{2}=6\cdot 81=567}$ ${\displaystyle s\geq {\sqrt {n}}>{\sqrt {10^{6}}}=1000}$ ${\displaystyle 1000<s\leq 567}$ ${\displaystyle n}$ ${\displaystyle s}$

So it must be so that the sum of the nontrivial divisors is equal to the sum of the squares of their digits. ${\displaystyle n<236196}$ ${\displaystyle \Box }$

• There are exactly four Canada-perfect numbers: 125, 581, 8549, 16999

Proof:

Because you know because of the above property that there are no Canada-perfect numbers that are greater than , you only have to examine all cases and thus only try out a finite number of possibilities. A not particularly fast computer that tests all variants is sufficient. You get exactly these four solutions 125, 581, 8549 and 16999. ${\displaystyle 236196}$ ${\displaystyle n\leq 236196}$ ${\displaystyle \Box }$

Individual evidence

1. ↑ Follow A070308 in OEIS

Web links

• Jean-Marie De Koninck , Armel Mercier : 1001 Problems in Classical Number Theory. American Mathematical Society , 2007, pp. 76-77 , accessed November 25, 2018 .