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⟹6×10d+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=2×105−813=15384

and n=15384×10+6=153846