Interview Question 2

Jun 29, 2022

Find the smallest natural number ending in \(6\), such that if the last \(6\) is removed and placed in the front, the new number is \(4\) times the original number.
Can you give a procedure to find the smallest number ending in digit \(d\), where \(d\) is between \(1\) and \(9\), such that if the last digit, \(d\), is removed and placed in front, you get a number that is \(k\) times as large? Is this possible for every value of \(d\) and \(k\)?

This question was used in CS interviews at Lady Margaret Hall in December 2019.

Update: This post was updated to use MathJax on July 25, 2022.

Tags: oxford, cs, maths, interview

Respond to this

2 responses

153846

def soln(d, k):
    old_num = d
    new_num = d
    i = 1
    while old_num * k != new_num:
        temp = old_num * k
        old_num = (temp % (10 ** i))*10 + d
        new_num = d * (10 ** i) + old_num // 10
        i += 1
    return old_num

Jingbo Zhao – July 25th, 2022

(i) 153846
(ii) Obviously many values of k and d are ridiculous, so first limit it to 1<=k<=9 and 1<=d<=9.
Then, incrementing n from 1, find the first value of n for which d(10^n -k)/(10k-1) (<= expression derived using algebra around the idea of breaking down a number (n-long) into digits by place value and fiddling around but a bit length to type) is an integer (brute force or by checking factors, etc) and also if numbers start going negative or bizarre then abort and call it impossible. The result of this division for smallest valid n gives you the starting (n-1) digits of the desired number, which you just need to append ‘d’ onto.

E.g. for the first part of question this happens on n=5, giving 6(10^5 - 4)/39 = 15384. Appending the 6 gives the desired answer of 153846 :)

Jay – October 7th, 2023