Number Puzzle Challenge
Earlier today I presented an elegant number puzzle. Here it is again, along with its solution.
There exists a number N starting with 4 such that moving the 4 to the end of the number produces a new number equal to one quarter of N.
In other words, N has the form 4[…], where […] represents a sequence of digits, and the equation N ÷ 4 = […]4 holds true.
The question is: what is the smallest possible value of N?
Approach to the Solution
My approach begins by testing numbers with increasing digits, starting from two-digit numbers and moving upwards until the condition is satisfied.
Two-digit numbers
Assume N = 4[?], where [?] is a single digit.
The only possible digit for [?] is 1 because a quarter of 4 is 1, and the quarter of 4[?] must be [?]4.
However, 14 is not a quarter of 41, so N must have more than two digits.
Three-digit numbers
Let N = 4[??]. By similar reasoning, the second digit must be 1, so N = 41[?].
We know that a quarter of 41[?] equals 1[?]4, which means 4 × 1[?]4 = 41[?].
We deduce the last digit of N must be 6 since 4 × 4 = 16.
However, a quarter of 416 is not 164, so N must have more than three digits.
Four-digit numbers
Let N = 4[???]. Following the same logic, N = 41[?]6.
We know a quarter of 41[?]6 equals 1[?]64, so 4 × 1[?]64 = 41[?]6.
The penultimate digit must be 5 because 4 × 64 = 256.
However, a quarter of 4156 is not 1564, so we continue.
Five-digit numbers
We have 4 × 1[?]564 = 41[?]56.
Since 4 × 564 = 2256, the antepenultimate digit must be 2.
But a quarter of 41256 is not 12564, so we proceed further.
Six-digit numbers
We know 4 × 1[?]2564 = 41[?]256.
Since 4 × 2564 = 10256, the digit [?] must be 0.
This satisfies the condition, so we have found our answer.
Conclusion
I hope you enjoyed this puzzle. I will return with another challenge in two weeks.
Source: Moscow Mathematical Olympiad 1983, via and Kevin Gately
I have been posting puzzles here on alternate Mondays since 2015. I am always looking for great puzzles. If you would like to suggest one, please email me.