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[The Problem of the Week] Powers

This week they are offered to apprentices the following problem:

A Margarita Felipe and like to compete with their calculators. Like many other powers, 7 ⁵⁹ not fit on the screen, but Philip says that the result ends in 1, while Margaret thinks that ends in 43. Which of us is right?

The solution here at this same screen, but a little lower.

[Here is a picture of one of the first mechanical calculator, the Pascaline , which was invented by Pascal, hence the name, in 1645. The Pascaline could only add and subtract numbers up to six figures, and Pascal was built thinking of helping his father, an accountant of the French Treasury, and promoting his work with commercial arithmetic.]

Solution:

problems in the higher powers that give us and ask us what number you end up trying to solve all the former powers and finding the sequence of terminations.

For the powers of 7 is easy to find the sequence:

7 ⁰ = 1

^{1 7 = 7}

^{2 7 = 49}

^{7 3 = 343}

^{7 4 = 2401}

^{7 5 = 16,807}

^{7 6 = 117,649}

^{7 7 = 823,543}

^{7 8 = 5,764,801}

etc.

As you can see, the number of units follows the sequence 1, 7, 9, 3, 1, 7, 9, 3, ... powers so that the end all multiples of 4 to 1.

power we are looking for is 59, the multiple of 4 Previous Next is the 56, 7 56 ends in 1, 7 57 ends in 7, 7 58 ends in 9 and 7 59 ends at 3. Felipe therefore not reason.
But more than in 3, Margarita says that ends in 43 exactly. Is that true? If we look at the succession of powers, see also the figure of the tens is following a particular pattern: 01, 07, 49, 43, 01, 07, 49, 43, then we can ensure that Margaret does have reason .

In fact, using a calculator that can show all numbers, for example Web2.0Calc , we get the full power we are looking for:

7 = 72,574,551,534,231,909 59.

331,741,171,093,173,785,967,490,646,405,143
Note: This problem has been excerpted from the book Text 4 º ESO SM publisher.