4 = 40

Algebra Level 3

Let a,b,c and d be four positive integers ,

a < b < c < d
a + b + c + d = 40
You can get any positive integer up to 40, Using a,b,c and d (each integer you can use one time or zero) , and the operations { + , - } only !

For example : if d = 15 ,
● You can get 40 by : a + b + c + d
● Yoy can get 25 by : a + b + c
● You can get 10 by : a + b + c - d

So, what is : (2ad - bc)

38 20 18 23 32 27

1 solution

Henry U
Dec 8, 2018

The key is to write the integers up to 40 in base 3:

base 10	$1_{10}$	$2_{10}$	$3_{10}$	$4_{10}$	$5_{10}$	$6_{10}$	$\cdots$	$39_{10}$	$40_{10}$
base 3	$1_3$	$2_3$	$10_3$	$11_3$	$12_3$	$20_3$	$\cdots$	$1110_3$	$1111_3$

Then, every integer can be written using sums and differences of powers of 3, so $\boxed{1,3,9,27}$ by the following rules that are applied to each digit going from right to left

If there is a 0 in the base 3 representation, the corresponding power isn't used.
If there is a 1 in the base 3 representation, the corresponding power is used with a positive sign, so it is added.
If there is a 2 in the base 3 representation, the corresponding power of 3 is subtracted in the sum. Then, the base 3 representation is modified by adding this power of 3 – making the digit with a 2 to a 0 – and the other powers of 3 are determined by applying these rules to the modified number, starting with the next digit, not the rightmost digit again.

(A few examples are written in a comment to this solution)

So, $(a,b,c,d)=(1,3,9,27)$ are integers that can make all integers up to 40 and they satisfy the given conditions because $1<3<9<27$ and $1+3+9+27=40$ .

This means that the answer is $2 \cdot 1 \cdot 27 - 3 \cdot 9 = \boxed{27}$ .

A few examples on how to apply the rules

12:

$12_{10}=110_3$

From right to left:

$11{\color{#D61F06}0} \Rightarrow 0 \times 1$

$1{\color{#D61F06}1}0 \Rightarrow +1 \times 3$

${\color{#D61F06}1}10 \Rightarrow +1 \times 9$

$\boxed{12=+3+9}$

11:

$11_{10}=102_3$

From right to left:

$10{\color{#D61F06}2} \Rightarrow -1 \times 1$

Modify to $11+1=12 = 110_3$

$1{\color{#D61F06}1}0 \Rightarrow +1 \times 3$

${\color{#D61F06}1}10 \Rightarrow +1 \times 9$

$\boxed{12=-1+3+9}$

15:

$15_{10}=120_3$

From right to left:

$12{\color{#D61F06}0} \Rightarrow 0 \times 1$

$1{\color{#D61F06}2}0 \Rightarrow -1 \times 3$

Modify to $15+3=18 = 200_3$

${\color{#D61F06}2}00 \Rightarrow -1 \times 9$

Modify to $18+9=27= 1000_3$

${\color{#D61F06}1}000 \Rightarrow +1 \times 27$

$\boxed{15=-3-9+27}$

Henry U - 2 years, 6 months ago

4 = 40

1 solution

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