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Binary Squares
If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?
Learn about Number Bases
We are used to writing numbers in base ten, using 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Eg. 75 means 7 tens and five units. This article explains how numbers can be written in any number base.
Age 16 to 18
Challenge Level
Vassil of Lawnswood High School in Leeds sent in a good solution. He worked systematically through the possible values of c using the rearrangement c(c$^2$ -1) = b(10-b) + 99a to establish the range of possible values for a and then testing to find values of b which fitted.
Many thanks to Sue Liu who is in S6 at Madras College in St Andrews for the followingbeautifully neat solution.
To find all 3-digit numbers $abc$ (in base $10$) such that
$$ a + b^2 + c^3 = 100a + 10b + c $$ Rearranging gives $$c^3 - c - 99a = b (10 - b)$$ $$c(c+1)(c-1) - 99a = b (10 - b)$$
For any three consecutive integers one of them is divisible by
$3$. Since $3$ divides $99$ it follows that $3$ divides $b(10-b)$.
Since $3$ is a prime this limits the possible choices of $b$:
Either
| $b=0 10-b=10$ | $b(10-b) = 0$ | |
| $b=3 10-b=7$ | or $b=7 10-b=3$ | $b(10-b) = 21$ |
| $b=6 10-b=4$ | or $b=4 10-b=6$ | $b(10-b) = 24$ |
| $b=9 10-b=1$ | or $b=1 10-b=9$ | $b(10-b) = 9$ |
Hence the possible values of $b(10-b)$ are $0$, $9$, $21$ and $24$.
We now have to find a multiple of $99$ which when subtracted
from a product of $3$ consecutive natural numbers gives $0$, $9$,
$21$ or $24$.
Since $a$ is at least $1$ $c(c+1)(c-1)$ is at least $99$ so $c$ is
at least $5$.
| $c$ | $c(c+1)(c-1)$ | $a$ | $99a$ | $c(c+1)(c-1)-99a$ |
|---|---|---|---|---|
| $5$ | $120$ | $1$ | $99$ | $21$ |
| $6$ | $210$ | $2$ | $198$ |
$12$
|
| $7$ | $336$ | $3$ | $297$ | $39$ |
| $8$ | $504$ | $5$ | $495$ | $9$ |
| $9$ | $720$ | $7$ | $693$ | $27$ |
(Since $0$, $9$, $21$, $24 < 99$ only the multiple of $99$ which is closest to $(c+1)c(c-1)$ needs to be checked.)
From the table we can see that the following are the possibilities for $a$, $b$ and $c$:
-
$a=1$ $b=3$ or $b=7$ $c=5$
-
$a=5$ $b=1$ or $b=9$ $c=8$
giving the solutions $135$, $175$, $518$ and $598$.
NRICH is part of the family of activities in the Millennium Mathematics Project.