Flexible Numbers

Let our three digit number be $abc$ . Then first of all it must be divisible by 8 so we have $100a + 10 b + c \equiv 0 \mod 8 \implies 4a + 2b + c \equiv 0 \mod 8$ . (1)

Then it must be that $8abc$ is divisible by 8, but $8abc \equiv 8 \times 1000 + abc \equiv abc \mod 8$ so no new properties are obtained.

Then it must be that $a8bc \equiv 0 \mod 8$ so $1000 \times a + 10b + c \equiv 0 \mod 8 \implies 2b + c \equiv 0 \mod 8$ . (2) .

Then it must be that $ab8c \equiv 0 \mod 8$ so $1000a + 100b + c \equiv 0 \mod 8 \implies 4b + c \equiv 0 \mod 8$ . (3)

Finally it must be that $abc8 \equiv 0 \mod 8$ so $1000a + 100b + 10c \equiv 0 \mod 8 \implies 4b + 2c \equiv 0 \mod 8$ (4)

Subtracting (2) from (1) we get $4a \equiv 0 \mod 8 \implies a \equiv 0 \mod 2$ .

Subtracting (2) from (3) we get $2b \equiv 0 \mod 8 \implies b \equiv 0 \mod 4$ .

Subtracting (3) from (4) we get $c \equiv 0 \mod 8$ .

Therefore any choice of: $a \in \{2,4,6,8 \}, b \in \{0,4,8 \}, c \in \{0,8 \}$ yields a flexible number. And there are $4 \times 3 \times 2 = 24$ ways of picking.

First, let the desired three-digit number be $\overline{abc}$ .

And, let the single-digit number be p.

The possible outcomes are $\overline{pabc}$ , $\overline{apbc}$ , $\overline{abpc}$ , $\overline{abcp}$ .

Strategically, let us consider $\overline{abcp}$ first

$\overline{abcp}$ = $10\overline{abc}+p$

$\overline{abc}$ is divisible by 8

$\Rightarrow$ $\overline{abc}\equiv0\pmod{8}$

$\Rightarrow$ $\overline{abc0}\equiv0\pmod{8}$

$\Rightarrow$ $\overline{abcp}\equiv p\pmod{8}$

Since $\overline{abcp}$ is divisible by 8

$\Rightarrow$ $\overline{abcp}\equiv 0\pmod{8}$

Therefore, $\Rightarrow$ $p\equiv 0\pmod{8}$

$p=0$ or $p=8$

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$\overline{abc}$ is divisible by 8

$\Rightarrow$ $\overline{abc}$ is divisible by 4

$\Rightarrow$ $\overline{bc}$ is divisible by 4

In $\overline{apbc}$ ,

$\overline{apbc}$ is divisible by 8

$\overline{pbc}$ is divisible by 8

$\Rightarrow$ $\overline{pbc}\equiv 0\pmod{8}$

$\Rightarrow$ $p\equiv 0\pmod{8}$

$\Rightarrow$ $100p\equiv 0\pmod{8}$

$\Rightarrow$ $100p+\overline{bc}\equiv \overline{bc}\pmod{8}$

$\Rightarrow$ $\overline{pbc}\equiv \overline{bc}\pmod{8}$

$\Rightarrow$ $\overline{bc}\equiv 0\pmod{8}$

Therefore, $\overline{bc}$ is divisible by 8.

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In $\overline{abpc}$

$\overline{abpc}$ is divisible by 8

$\overline{bpc}$ is divisible by 8

$\overline{bc}\equiv 0\pmod{8}$

$\overline{b0c}=100b+c=(10b+c)+90b=\overline{bc}+90b$

$\overline{bc}\equiv 0\pmod{8}$

$\overline{b0c}\equiv 90b\pmod{8}$ ------(1)

$p\equiv 0\pmod{8}$

$10p\equiv 0\pmod{8}$ -----------------------(2)

(1)+(2): $\overline{bpc}\equiv 90b\pmod{8}$

$\overline{bpc}\equiv 0\pmod{8}$

So, $90b\equiv0\pmod{8}$

$90b$ is divisible by 8

Let $90b=8x$ , such that x is positive integer.

$\Rightarrow$ $45b=4x$

$\Rightarrow$ $x=\cfrac{45b}{4}$

$x$ is a positive integer.

Hence, $b$ is divisible by 4

$b=0$ or $b=4$ or $b=8$

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We have shown $\overline{bc}$ is divisible by 8

$\overline{bc}=10b+c$

$b\equiv 0\pmod{4}$

$\Rightarrow$ $2b\equiv 0\pmod{8}$

$\Rightarrow$ $10b\equiv 0\pmod{8}$

$\Rightarrow$ $10b+c\equiv c\pmod{8}$

$\overline{bc}$ is divisible by 8

$\Rightarrow$ $10b+c\equiv 0\pmod{8}$

Therefore, $c\equiv 0\pmod{8}$

$c$ is divisible by $8$

$c=0$ or $c=8$

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Combining the results we have,

$p=0$ or $p=8$

$b=0$ or $b=4$ or $b=8$

$c=0$ or $c=8$

$\overline{abc}$ is divisible by 8

$\overline{bc}$ is divisible by 8

$\overline{abc}=100a+\overline{bc}$

$100a$ is divisible by 8

$100\equiv4\pmod{8}$

$200\equiv8\pmod{8}$

$200\equiv0\pmod{8}$

$100a=200,400,600,800$

$a=2,4,6,8$

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Finally, we have

$a=2$ or $a=4$ or $a=6$ or $a=8$

$b=0$ or $b=4$ or $b=8$

$c=0$ or $c=8$

So, the number of possible values of $\overline{abc}$ = $4\times 3\times 2$ = $24$

The total number of positive three-digit flexible numbers= $24$