The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
In 15 years' time my age will be the square of my age 15 years ago.
Can you work out my age, and when I had other special birthdays?
Libby Jared helped to set up NRICH and this is one of her favourite
problems. It's a problem suitable for a wide age range and best
Published January 1998,February 2011.
Let us look again at the problem posed at the end of the last article. We asked there whether or not the number $123456$ ends up at $0$ or in a cycle of four numbers? Using a calculator, we see that $123456$ is not a multiple of $101$ so that (by what we have just shown) it cannot end up at $0$.
What about the number $12345678987654321$? This number is too big to put on a calculator so we need to find another approach for it would clearly be a very long task indeed to keep on applying the rule to this number! How do we handle this number?
We want to decide whether or not the number $12345678987654321$ is a multiple of $101$. Of course if a whole number $X$ is a multiple of $101$ then the number $12345678987654321 - X $ is also a multiple of $101$ and conversely, and using this we see that it is enough to subtract multiples of $101$
from $12345678987654321$ and then check whether or not the answer is a multiple of $101$. Better still, if P is any whole number then:
$$10000P = 9999P + P = (99 \times101)P + P$$ so that $10000 P$ is a multiple of $101$ plus $P$.
Thus, $$12345678987654321 = (1234567898765 x 10000) + 4321 = 1234567898765 + 101Q + 4321$$ for some whole number $Q$. Applying this again, we get $$1234567898765 = (123456789 x 10000) + 8765 = 123456789 + 101S + 8765$$ for some whole number $S$, and again, $$123456789 = (12345 x 10000) + 6789 = 12345 + 101T + 6789$$ for some whole number $T$.
Putting all these together, we find that the two numbers $12345678987654321$ and $( 4321 + 8765 + 6789 + 12345)$ differ by a multiple of $101$. It is enough, therefore, to check whether $(4321+8765+6789+12345)$ is, or is not, a multiple of $101$, and we have now reduced the problem to one that we can do on a calculator.
You can now answer the question : does $12345678987654321$ eventually reach $0$ or not ?
A final question : does $8765432123456789$ eventually reach $0$ or not?