Schools Mathematics Grand Challenge
Week five's Puzzles
Problem 9:
The ninth problem was:
The number 6 has the following property:
If you subtract two from it you get 4 which is a perfect square,
6 - 2 = 4 = (2 x 2) = 2^2
and if you add two to it you get 8 which is a perfect cube,
6 + 2 = 8 = (2 x 2 x 2) = 2^3.
What is the next whole number, strictly bigger than 6, with this
property? (meaning that when you subtract two from it, you get a perfect
square and when you add two to it, you get a perfect cube)
The solution was:
Let X be the number we seek. Then X + 2 is a perfect cube and X - 2
is a perfect square. One way to find the number we are looking for is
to find the next cube number (after 8) such that when we subtract four
from it we get a perfect square.
3^3 - 4 = 27 - 4 = 23 which is not a square
4^3 - 4 = 64 - 4 = 60 which is not a square
5^3 - 4 = 125 - 4 = 121 which is eleven squared.
So the answer is 125 - 2 = 123.
Problem 10:
The tenth problem was:
The factorial of a positive whole number n is written n! and is
calculated by multiplying together all of the whole numbers from n back
to one. For instance:
4! = 4x3x2x1 = 24
5! = 5x4x3x2x1 = 120
3! = 3x2x1 = 6
7! = 7x6x5x4x3x2x1 = 5040.
If you add up the factorials of all the whole numbers from 1
up to 1,000,000 (1 million) and divide the sum by 40, what will the
remainder be?
So in this question we are asking for the remainder when you divide
1! + 2! + 3! + 4! + 5! + ... + 1,000,000! (the sum of the factorials
of the first million positive whole numbers) by 40?
Remember that your answer must be a whole number between 0 and
39.
HINT: You do not need to work out the whole sum. Also be aware
that calculators do not always give the right answer when you work with
very large numbers!
The solution was:
The key point here is that if 40 divides evenly into n! for some positive
number n, it will also divide evenly into the factorials of all numbers
greater than n.
The first factorial into which 40 divides evenly is 5! = 5 x 4 x 3 x 2
x 1 = 120 = 40 x 3. Hence 40 divides evenly into
5! + 6! + 7! + ... + (1,000,000)!
and the remainder we seek is the same as the remainder when you divide 40 into
1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33.
So the answer is 33.
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