Determine-the-smallest-positive-integer-x-whose-last-digit-is-6-and-if-we-erase-this-6-and-put-it-in-left-most-of-the-number-so-obtained-the-number-becomes-4x-

Question Number 18884 by Tinkutara last updated on 31/Jul/17

Determine the smallest positive integer

x, whose last digit is 6 and if we erase

this 6 and put it in left most of the

number so obtained, the number

becomes 4 x .

Answered by mrW1 last updated on 01/Aug/17

{let}^{'} s say the number has n + 1 digits

let x = [abcd \dots pqr 6] = 10 u + 6

let y = [6 abcd \dots pqr] = 6 \times 10^{n} + u

y = 4 x

\Rightarrow 6 \times 10^{n} + u = 4 (10 u + 6)

\Rightarrow 6 (10^{n} - 4) = 39 u

\Rightarrow 2 (2^{n} \times 5^{n} - 2^{2}) = 13 u

\Rightarrow 2^{3} (2^{n - 2} \times 5^{n} - 1) = 13 u

\Rightarrow 2^{3} (25 \times 10^{n - 2} - 1) = 13 u

\Rightarrow u = \frac{2^{3} (25 \times 10^{n - 2} - 1)}{13}

\Rightarrow 25 \times 10^{n - 2} - 1 = 13 k

the smallest n = 5 :

25 \times 10^{5 - 2} - 1 = 25000 - 1 = 24999 = 13 \times 1923

\Rightarrow \min . u = 2^{3} \times 1923 = 15384

\Rightarrow \min . x = 153846

the further numbers are :

x = 153846153846

x = 153846153846153846

x = 153846153846153846153846

\dots \dots

Commented by Tinkutara last updated on 01/Aug/17

Please also explain why 25 \times 10^{n - 2} - 1

= 13 k because it should be \frac{13 k}{8} .

Commented by mrW1 last updated on 01/Aug/17

yes, u should = \frac{2^{3} (25 \times 10^{n - 2} - 1)}{13} . I had just a typo .

for u = \frac{2^{3} (25 \times 10^{n - 2} - 1)}{13} and 13 is prime

25 \times 10^{n - 2} - 1 must be a multiple of 13,

otherwise u woulde be no integer .

i . e . 25 \times 10^{n - 2} - 1 = 13 k with k = integer .

then u = 2^{3} k .

so our task is to solve the equation

25 \times 10^{n - 2} - 1 = 13 k with n, k \in N .

Commented by mrW1 last updated on 01/Aug/17

certainly you can solve this question

directly like this :

abcdef 6

\times 4

- - - - - - - - -

= 6 abcdef

\Rightarrow f = 4

\Rightarrow e = 8

\Rightarrow d = 3

\Rightarrow c = 5

\Rightarrow b = 1

\Rightarrow a = 6! {that}^{'} s right . \Rightarrow our first number is 153846

just go on with these steps if you want to get the

next suitable number .

Commented by Tinkutara last updated on 01/Aug/17

Thank you very much mrW 1 Sir!

Commented by mrW1 last updated on 01/Aug/17

I tried to use an equation to solve the

question, because the question can

be generalised :

Find such numbers when we erase their last

digit and put it before the first digit,

the new number will be m times as

the original number .

m can be 2 til 9 .

Commented by Tinkutara last updated on 01/Aug/17

Which equation ? Can you give its

derivation ?