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New Scientist Enigma 518 – Day of Reckoning

by Chris Maslanka

From Issue 1670, 24th June 1989

Nugel, Chancellor of the Exchequer of Planet-X, was to bring before the Prime Minister the results of his latest calculations of the planetary trade deficit. The calculation had proved awkward (but not as awkward, he thought ruefully, as trying to account for them was going to be) and he had asked the Minister for More Mathematics (who had attended the same school many moons ago) to process the figures in secret on his Megathon computer. The result arrived while the Chancellor was already waiting to be admitted to The Presence an he read the missive at once:

“A boring task, my dear Nugel, but an aesthetic result: your trade deficit in Zloots is represented by the smallest integer which is simultaneously half a perfect square, a third of a perfect cube, a fifth of a perfect fifth power and a seventh of a perfect seventh power.”

Seventh heaven was where Nugel wasn’t when close inspection of the letter failed to reveal the number. He felt let down by his ex-prefect. What on Planet-X was the old coot on about. With a sigh and resourceful to the last, he took out his battered Clackulator and set to work. In the background he could hear the voice of the only minister (Prime or Composite) to have survived 19 terms of office.

Can you find the trade deficit in Zloots (give your answer as a product of powers of prime numbers) before Nugel becomes the ex-Chancellor of the Exchequer?

One Comment Leave one →
  1. Brian Gladman permalink

    Consider the power of 2 in the number. Since the number is a third of a cube, a fifth of a fifth power and a seventh of a seventh power, the power of two in the number must be a multiple of 3, 5 and 7, so the factor of 2 in the number is of the form \(2^{3\cdot 5\cdot 7\cdot a}\) for some integer \(a\).

    Similar arguments for 3, 5 and 7 lead to the form of number as:\[\begin{align}N&=2^{3\cdot 5\cdot 7\cdot a} \cdot 3^{2\cdot 5\cdot 7\cdot b} \cdot 5^{2\cdot 3\cdot 7\cdot c} \cdot 7^{2\cdot 3\cdot 5\cdot d}\\&= 2^{105a} \cdot 3^{70b} \cdot 5^{42c} \cdot 7^{30d}\end{align}\]

    for integer \(a\), \(b\), \(c\) and \(d\). Now:

    1) \(2N\) is a square which means that:\[105a+1=0\mod2\implies a=1\mod2\]

    2) \(3N\) is a cube which means that:\[70b+1=0\mod3\implies b=2\mod3\]

    3) \(5N\) is a fifth power which means that:\[42c+1=0\mod5\implies c=2\mod5\]

    4)  \(7N\) is a seventh power which means that:\[30d+1=0\mod7\implies c=3\mod7\]Since we need the smallest value of \(N\), we hence obtain:\[N=2^{105}\cdot3^{140}\cdot5^{84}\cdot7^{90}\]

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