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
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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???
Best Answer
$$10^na+\underbrace{b}_{\text{ has } n(>1)\text{ digits}}=57b\iff56b=10^n a$$
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, $n\ge3$