Number Puzzle Solved: Find N Where Moving 4 to End Quarters It

Earlier today I set an elegant number puzzle. Here it is again with a solution.

Nose to tail

There is a number N beginning with 4 such that moving the 4 to the end of it creates a new number that is a quarter of N. In other words, N is of the form 4[...], where [...] is a sequence of digits, and N ÷ 4 = [...]4. What is the lowest possible value of N?

Solution

My strategy is to start by trying to find a two-digit N, and gradually increase the number of digits until we’re done.

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Two digits

Say N = 4[?], where [?] is a digit. The only possible value for [?] is 1 because we know that a quarter of 4[?] is [?]4 and a quarter of 4 is 1. But 14 is not a quarter of 41, so we conclude N has more than two digits.

Three digits

Let N = 4[??]. For the same reason as before, the second digit of N must be 1. So N = 41[?]. We know that a quarter of 41[?] = 1[?]4, which is the same as saying that 4 x 1[?]4 = 41[?]. We deduce that the final digit of N must be 6, since 4 x 4 is 16. However, a quarter of 416 is not 164, so N has more than three digits.

Four digits

Let N = 4[???]. For the same reasons as above, N = 41[?]6. We know that a quarter of 41[?]6 = 1[?]64, which is the same as saying that 4 x 1[?]64 = 41[?]6. We deduce that the penultimate digit of N must be 5, since 4 x 64 = 256. However, a quarter of 4156 is not 1564, so onwards we go.

Five digits

N = 41[?]56. We know that 4 x 1[?]564 = 41[?]56. Since 4 x 564 = 2256, the antepenultimate digit of N must be 2. But a quarter of 41256 is not 12564, so we need to carry on.