mathschallenge.net

Problem

Using each of the digits 1, 2, 3, and 4, once and only once, with the basic rules of arithmetic (+, –, , ÷, and parentheses), express all of the integers from 1 to 25.

For example, 1 = 2 3 – (1 + 4)

Solution

Of course, there are may be other ways of arriving at each of these numbers:

1 = 2 3 – (1 + 4) 14 = 1 4 3 + 2 2 = 4 – 3 + 2 – 1 15 = 3 4 + 1 + 2 3 = 2 3 – (4 – 1) 16 = 2(1 + 3 + 4) 4 = 2 4 – (1 + 3) 17 = 3(2 + 4) – 1 5 = 2 4 – 1 3 18 = 3(2 + 4) 1 6 = 2 4 – 3 + 1 19 = 3(2 + 4) + 1 7 = 3(4 – 1) – 2 20 = 1 4 (2 + 3) 8 = 2 + 3 + 4 – 1 21 = 4(2 + 3) + 1 9 = 2 3 + (4 – 1) 22 = 2(3 4 – 1) 10 = 1 + 2 + 3 + 4 23 = 3 4 2 – 1 11 = 2 3 + (1 + 4) 24 = 1 2 3 4 12 = 3 4 (2 – 1) 25 = 2 3 4 + 1 13 = 3 4 + 1 + 2

Extensions

• If you are now permitted to use square roots, exponents, and factorials, can you produce all of the integers from 1 to 100?
• What is the first natural number that cannot be derived?
• Which is the first number that cannot be obtained if you are only permitted to use the basic rules of arithmetic (+, –, , and ÷)?
• What is the largest known prime you can produce?

Notes

Surprisingly it is possible to produce any finite integer using logarithms in a rather ingenious way.

We can see that,

2 = 2^1/2
2 = (2^1/2)^1/2 = 2^1/4
2 = ((2^1/2)^1/2)^1/2 = 2^1/8, and so on.

Therefore,

log₂(2) = 1/2 = (1/2)¹
log₂(2) = 1/4 = (1/2)²
log₂(2) = 1/8 = (1/2)³, et cetera.

Hence,

log_1/2(log₂(2)) = 1
log_1/2(log₂(2)) = 2
log_1/2(log₂(2)) = 3, ...

By using the integer part function, 1/[(3!)] = 1/[2.449...] = 1/2, we can obtain the required base 1/2, and using 4 to obtain the base 2, we can now produce any finite integer using the digits 1, 2, 3, and 4.