1 hour ago
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Earlier today I set this 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.
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.
Six digits
N = 41[?]256
We know that 4 x 1[?]2564 = 41[?]256.
Since 4 x 2564 = 10256, we know that the [?] must be 0.
This works! We have our answer:
N = 410256 = 4 x 102564
I hope you enjoyed this puzzle. I’ll be back in two weeks.
Source: Moscow Mathematical Olympiad 1983, via @mathematicsproblems and Kevin Gately
I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
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