"A number becomes $57$ times smaller when the first digit is deleted." Explaining the solution.

Question

Puzzle: Find a number that becomes 57 times smaller when the first digit is deleted.

Answer: 7125

Explanation:

ab...z = 57 * b...z

X = b...z


aX = 57X

X + a0...0
    \__  _/
       \/
        k

above is X + a with k zeros

aX = a . 10^k + X

I don't understand from the a0...0 part.

If I relate it to something I do know then I can't see how it makes any sense.

eg

x=3
a=7

so:

ax = 21

imagine we don't know that a = 7 but we do know:

ax = 21

using the logic above we say:

ax = x + a0...0

ax = 3 + a0...0

21 = 3 + 70...0

but 3 + 7 with any number of zeros does not get you 21???

and:

aX = a . 10^k + X

aX = 21

a * 10^k + X

7 * 10^k + 3

but 7 multiplied by any number + 3 does not give us 21???

Answer 1

Let : $$ 1<a<9$$ $$ 0 \le n $$ $$ 0 < b < 10^n $$ Then the equation to be resolved is: $$ 10^n a + b = 57 b $$ $$ 2^n 5^n a = 2^3 \cdot 7 \cdot b $$ $$ 2^{n-3} \cdot 5^n \cdot a = 7 \cdot b $$ $$ 5^3 \cdot 10^{n-3} \cdot a = 7 \cdot b $$ If $ a = 7 $ : $$ 5^3 \cdot 10^{n-3} \cdot 7 = 7 \cdot b $$ $$ 5^3 \cdot 10^{n-3} = b $$ $$ 125 \cdot 10^{n-3} = b $$ $$ a=7, b \in \{125, 1250, 12500, ...\} $$ If $ a \neq 7 $(and $a \neq 0$ precondition): No solutions, the left part of equation is not divisible by 7 and the right always divisible by 7.

Answer 2

$$10^na+\underbrace{b}_{\text{ has } n(>1)\text{ digits}}=57b\iff56b=10^n a$$

Now $\dfrac{10^na}7=8b$ which is an integer

As $(7,10^n)=1,7|a\implies a=7$ as $0<a<10$

$\implies8b=10^n\implies b=\dfrac{10^n}{2^3}=10^{n-3}\cdot5^3=5^n2^{n-3}$

As $b$ is an integer, $n\ge3$

Answer 3

Here, $aX$ doesn't mean $a$ times $X$; it means "the decimal number with first digit $a$ followed by the digits of $X$". So in your example, $aX$ would be $73$, not $21$.

It's true that there is nothing in the notation itself to tell you this; but it's the only interpretation that makes sense in context. (And to add to the confusion, when they write $57X$, they do mean $57$ times $X$!)

Answer 4

Let the number be $\overline{a_1a_2...a_n}$. The condition is: $$\overline{a_1a_2...a_n}=57\overline{a_2...a_n} \Rightarrow \\ a_1\cdot 10^{n-1}+\overline{a_2...a_n}=57\overline{a_2...a_n} \Rightarrow \\ a_1\cdot 10^{n-1}=56\overline{a_2...a_n} \Rightarrow \\ \overline{a_2...a_n}=\frac{a_1\cdot 10^{n-1}}{56}=\frac{a_1\cdot 2^{n-1}\cdot 5^{n-1}}{7\cdot 2^3} \Rightarrow \\ a_1=7; 2^{n-4}\in \mathbb N \Rightarrow n\ge 4.$$