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Given a number (ab)3
, 1 a 9, 0 b 9. Find without use of calculator the number
ab.
Solution by Mikolaj Hajduk:
The aforementioned math problem is a symbolic description of the following game: an illusion-
ist asks a randomly chosen spectator to pick (secretly) one of the natural numbers smaller
than 100 and calculate its cube and give the result speaking it aloud. The illusionist imme-
diately gives an answer, i.e. the original number chosen by the spectator. This trick doesn’t
require the magician to know all cubes of the numbers between 1 and 100. It’s enough to remember
only cubes of the first ten numbers from 1 to 10 because the method of determination of the original
number is based on this information as it will be shown in the further parts of this paper.
The cube of any two-digit number and the cube of the last digit of the original number are congruent
modulo 10, more precisely
(ab)3
≡ b3
(mod 10)
indeed, we have
(ab)3
= (10a + b)3
= (10a)3
+ 3(10a)2
b + 3(10a)b2
+ b3
=
= 1000a3
+ 300a2
b + 30ab2
+ b3
=
= 10(100a3
+ 30a2
b + 3ab2
) + b3
So the last digit of the cube of the original number gives us a way to determine the second digit of the
original number. Let’s take a look at the following array:
c 2015/04/28 22:18:03, Mikolaj Hajduk 1 / 3 next
a 1 2 3 4 5 6 7 8 9 10
a3
1 8 27 64 125 216 343 512 729 1000
a3
mod 10 1 8 7 4 5 6 3 2 9 0
The first digit of the original number chosen by the spectator may be determined as follows: we discard
the last three digits of the cube and take the integer part of the cube root from the number that has
left, in other words we get that
a =
3 (ab)3
1000
and we will show below how we can explain this result. At first, from the definition of the floor function
we get the following useful feature:
A
x ∈ R
A
m ∈ Z
(m x =⇒ m x ) (∗)
now, bearing in mind the feature (∗), we can prove correctness of the formula for the first digit of the
guessed number:
10a 10a + b = ab < 10a + 10 = 10(a + 1) ⇐⇒
1000a3
(ab)3
< 1000(a + 1)3
⇐⇒
a3 (ab)3
1000
< (a + 1)3 =⇒
(∗)
c 2015/04/28 22:18:03, Mikolaj Hajduk 2 / 3 next
a3 (ab)3
1000
< (a + 1)3
⇐⇒
a
3 (ab)3
1000
< a + 1 =⇒
(∗)
a
3 (ab)3
1000
< a + 1
Because the number
3 (ab)3
1000
is integer, greater than or equal to a and less than a + 1, then it has
to be equal to a (the is no integer n such that a < n < a + 1). Finally we get the postulated formula
for the first digit of the guessed number.
Example of use
Let (ab)3
= 250047. The second number of the guessed number ab is 3 as it corresponds in our array
to 7 = (ab)3
mod 10. The first digit of the original number is equal to
3 (ab)3
1000
=
3
√
250 = 6
so the original number is equal to 63. Let’s check it out. Indeed 633
= 250047.
c 2015/04/28 22:18:03, Mikolaj Hajduk 3 / 3

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Cube root

  • 1. Given a number (ab)3 , 1 a 9, 0 b 9. Find without use of calculator the number ab. Solution by Mikolaj Hajduk: The aforementioned math problem is a symbolic description of the following game: an illusion- ist asks a randomly chosen spectator to pick (secretly) one of the natural numbers smaller than 100 and calculate its cube and give the result speaking it aloud. The illusionist imme- diately gives an answer, i.e. the original number chosen by the spectator. This trick doesn’t require the magician to know all cubes of the numbers between 1 and 100. It’s enough to remember only cubes of the first ten numbers from 1 to 10 because the method of determination of the original number is based on this information as it will be shown in the further parts of this paper. The cube of any two-digit number and the cube of the last digit of the original number are congruent modulo 10, more precisely (ab)3 ≡ b3 (mod 10) indeed, we have (ab)3 = (10a + b)3 = (10a)3 + 3(10a)2 b + 3(10a)b2 + b3 = = 1000a3 + 300a2 b + 30ab2 + b3 = = 10(100a3 + 30a2 b + 3ab2 ) + b3 So the last digit of the cube of the original number gives us a way to determine the second digit of the original number. Let’s take a look at the following array: c 2015/04/28 22:18:03, Mikolaj Hajduk 1 / 3 next
  • 2. a 1 2 3 4 5 6 7 8 9 10 a3 1 8 27 64 125 216 343 512 729 1000 a3 mod 10 1 8 7 4 5 6 3 2 9 0 The first digit of the original number chosen by the spectator may be determined as follows: we discard the last three digits of the cube and take the integer part of the cube root from the number that has left, in other words we get that a = 3 (ab)3 1000 and we will show below how we can explain this result. At first, from the definition of the floor function we get the following useful feature: A x ∈ R A m ∈ Z (m x =⇒ m x ) (∗) now, bearing in mind the feature (∗), we can prove correctness of the formula for the first digit of the guessed number: 10a 10a + b = ab < 10a + 10 = 10(a + 1) ⇐⇒ 1000a3 (ab)3 < 1000(a + 1)3 ⇐⇒ a3 (ab)3 1000 < (a + 1)3 =⇒ (∗) c 2015/04/28 22:18:03, Mikolaj Hajduk 2 / 3 next
  • 3. a3 (ab)3 1000 < (a + 1)3 ⇐⇒ a 3 (ab)3 1000 < a + 1 =⇒ (∗) a 3 (ab)3 1000 < a + 1 Because the number 3 (ab)3 1000 is integer, greater than or equal to a and less than a + 1, then it has to be equal to a (the is no integer n such that a < n < a + 1). Finally we get the postulated formula for the first digit of the guessed number. Example of use Let (ab)3 = 250047. The second number of the guessed number ab is 3 as it corresponds in our array to 7 = (ab)3 mod 10. The first digit of the original number is equal to 3 (ab)3 1000 = 3 √ 250 = 6 so the original number is equal to 63. Let’s check it out. Indeed 633 = 250047. c 2015/04/28 22:18:03, Mikolaj Hajduk 3 / 3