3.3: Divisibility rules revisited

Thinking out Loud

Can any integer \(n\) be written as a sum of distinct powers of \(2\)?

Example \(\PageIndex{1}\):

Express \(2019\) as a sum of distinct powers of \(2\)?

Note that, \(10 \equiv 1 ( mod \,\,3), 10 \equiv 1 ( mod\, 9),\) and \(10 \equiv (-1)( mod \,\,11),\).

Let \(x \in \mathbb{Z_+}\).

Then \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0,\)

which implies

\(x= 10( d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1)+d_0.\)

Thus we can express \(x\) as \(10 a+b\), where \(b=d_0\) is the ones digit of \(x\), and \(a= d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1\).

Divisibility by \(2=2^1:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(2 \mid x\) iff \(2 \mid d_0\). In other words, \(2\) divides an integer iff the ones digit of the integer is either \(0, 2, 4, 6,\) or \( 8\). That is. \(x \equiv 0 (mod \,2) \) iff \(d_0 \equiv 0 (mod \,2)\).

Proof:

Since \( 10 \equiv 0 (mod \,2) \) and \(x=10 a+b\), \(x \equiv 0 (mod \,2) \) iff \(d_0 \equiv 0 (mod \,2)\).\(\Box\)

Divisibility by \(5:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(5 \mid x\) iff \(5 \mid d_0\). In other words, \(5\) divides an integer iff the ones digit of the integer is either \(0,\) or \( 5\).

That is. \(x \equiv 0 (mod \,\,5) \) iff \(d_0 \equiv 0 (mod \,\,5)\).

Proof:

Since \( 10 \equiv 0 (mod \,5) \) and \(x=10 a+b\), \(x \equiv 0 (mod \,5) \) iff \(d_0 \equiv 0 (mod \,5)\).\(\Box\)

Divisibility by \(10:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(10 \mid x\) iff \(10 \mid d_0\). In other words, \(10\) divides an integer iff the ones digit of the integer is \(0,\).

That is. \(x \equiv 0 (mod \,10) \) iff \(d_0 \equiv 0 (mod \,10)\).

Proof:

Since \( 10 \equiv 0 (mod \,10) \) and \(x=10 a+b\), \(x \equiv 0 (mod \,10) \) iff \(d_0 \equiv 0 (mod \,10)\).\(\Box\)

Divisibility by \(4=2^2:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(4 \mid x\) iff \( 4 \mid d_1d_0\). That is. \(x \equiv 0 (mod \,4) \) iff \(d_1d_0 \equiv 0 (mod \,4)\).

Proof:

Let \(x\) be an integer. Then

\( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\), which implies

\(x= 100( d_n10^{n-2} +d_{n-1}10^{n-3}+ \cdots+ d_2)+ 10d_1+d_0=100( d_n10^{n-2} +d_{n-1}10^{n-3}+ \cdots+ d_2)+ d_1d_0\).

Since \( 100 \equiv 0 (mod \,4)\), \(x=100 a+ d_1d_0\), \(x \equiv 0 (mod \,4) \) iff \(d_1d_0 \equiv 0 (mod \,4)\).\(\Box\)

Divisibility by \(8=2^3:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(8 \mid x\) iff \( 8 \mid d_2d_1d_0\). That is. \(x \equiv 0 (mod \,8) \) iff \(d_2d_1d_0 \equiv 0 (mod \,8)\).

Proof:

Let \(x\) be an integer. Then

\( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\), which implies

\(x= 10^3( d_n10^{n-3} +d_{n-1}10^{n-4}+ \cdots+ d_3)+ 100d_2+10d_1+d_0=10^3( d_n10^{n-3} +d_{n-1}10^{n-4}+ \cdots+ d_3)+ d_2d_1d_0\).

Since \( 1000 \equiv 0 (mod \,8)\), \(x=1000 a+ d_2d_1d_0\), \(x \equiv 0 (mod \,8) \) iff \(d_2d_1d_0 \equiv 0 (mod \,8)\).\(\Box\)

A similar argument can be made for divisibility by \(2^n\), for any positive integer \(n\). The following results follows from the fact that,

\( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0 = d_n(10^n-1) +d_{n-1}(10^{n-1}-1)+ \cdots+ d_2 (10^2-1)+d_1(10^1-1)+ (d_n+d_{n-1}+ \cdots+ d_1+ d_0)\).

Divisibility by \(3:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(3 \mid x\) iff \(3\) divides sum of its digits.

Proof:

Let \(x\) be an integer. Suppose \(x=d_k10^k+d_{k-1}10^{k-1}+\cdots+d_110+d_0.\)

Since \(10\equiv1\pmod3, 10^j\equiv1\pmod3.\)

Therefore, \( N\equiv d_k+d_{k-1}+\cdots+d_1+d_0\pmod3. \)

Hence \(x\) is divisible by \(3\) exactly when the sum of its digits is
divisible by \(3\).

 

Divisibility by \(9=3^2:\)

Let \( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\in \mathbb{Z_+}\), then \(9 \mid x\) iff \(9\) divides sum of its digits.

Proof:

Let \(x\) be an integer. Then

\( x= d_n10^n +d_{n-1}10^{n-1}+ \cdots+ d_2 10^2+d_110^1+d_0\), which implies \(x= d_n(10^n-1) +d_{n-1}(10^{n-1}-1)+ \cdots+ d_2 (10^2-1)+d_1(10^1-1)+ (d_n+d_{n-1}+ \cdots+ d_1+ d_0)\).

Hence, \(x (mod \,9) \equiv d_n(10^n-1) +d_{n-1}(10^{n-1}-1)+ \cdots+ d_2 (10^2-1)+d_1(10^1-1)) (mod \,9)+ (d_n+d_{n-1}+ \cdots+ d_1+ d_0) (mod \,9).\)  Notice that \((10^n-1) \equiv 0 (mod \, 9), \) for all positive integer \(n.\)

Therefore, \(x (mod \,9) \equiv (d_n+d_{n-1}+ \cdots+ d_1+ d_0) (mod \,9).\)

 Thus, \(9 \mid x\) iff \(9\) divides sum of its digits.

Divisibility by \(7:\)

\(7 \mid x\) iff \(7\) divides the absolute difference between \(a-2b\), where \(x=10 a+b\), and, \(b=d_0\) is the ones digit of \(x\) and \(a= d_n10^{n-1} +d_{n-1}10^{n-2}+ \cdots+ d_2 10+d_1\).

Proof:

Divisibility by \(11:\)

\(11 \mid x\) iff \(11\) divides the absolute difference between alternate sum.

Proof:

Since \(10\equiv-1\pmod{11},\) we obtain \(10^2\equiv1\pmod{11}, 10^3\equiv-1\pmod{11},\)
and so forth.

Thus, a decimal number is congruent modulo \(11\) to the alternating sum of
its digits.