Question

If n is a positive integer,let s(n) denote the integer obtained by removing the last digit of n and placing it in front.for example,s(731)=173.what is the smallest positive integer n ending in6 satisfying s(n)=4n?

Answer 1

Very good question , thanks !!

Can we assume n = 10k + 6 , because according to question number is ending in 6.

so, $S(n)=6.10^a+k$ , where a and k are positive integers.

according to question, 

S(n) = 4n = 4(10k + 6) = 40k + 24 

$\implies S(n)=40k+24=6.10^a+k$

$\implies39k+24=6.10^a$ 

$\implies3(13k+8)=3.2.10^a$

$\implies13k+8=2.10^a$

means, if 2.10^a is divided by 13 , getting reminder = 8. 

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Answer 2

Thank you for asking this question. Here is your answer:

First of all we will Let n = 10a + 6 and also that a has d digits

S(n) = 4n ⟹ 6 × 10^d + a = 4n = 4 (10a+6)

6×10^d−24 = 39a

2 × 10^d −8 = 13a

2×10^ d ≡ 8 (mod13)  

10d ≡ 4 (mod13) , (2,13) = 1

105 ≡ 4 (mod13)

5 will be the least number to satisfy this:

a = 2 × 105 − 813 = 15384

and n = 15384 × 10 + 6

= 153846

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