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 in 6 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

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

$\implies39k+24=6.10^a$

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

$\implies13k+8=2.10^a$......(iii)

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

now ,we can write it as $2.10^a=8(mod\:13)$

$\implies10^a=4(mod\:13)$.....(i)

[ note :- for understanding , a basic logic is that if we assume 190 is a random number and now 190 is divided by 13 , then remainder = 8

now half of 190 = 95 is divided by 13 then remainder = 4 , so $10^a=4(mod\:13)$ ]

now we see,

if 1 is divided by 13 , getting remainder = 1

so, we can write 1 Ξ 1 (mod 13)

10 is divided by 13 , getting reminder = -3

so, 10 Ξ -3

similarly, 100 Ξ 9 Ξ -4

1000 Ξ 12 Ξ -1

so, we get, 100 × 1000 Ξ (-4) × (-1) Ξ 4

hence, 100000 = 4(mod 13)

e.g., $10^5=4(mod\:13)$......(ii)

compare eqs. (i) and (ii),

so, a = 5

now, $k=\frac{2.10^5-8}{13}=15384$ [ from equation (iii)]

hence, n = 10k + 6

n = 10 × 15384 + 6

n = 153840 + 6 = 153846

hence, S(153846) = 615384

Answer 2

Let n=10a+6, and 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)

10^d≡4(mod13),(2,13)=1

10^5≡4(mod13)

and 5 is the minimum number to fulfill this.

a=2×10^5−813=15384

and n=15384×10+6

= 153846