A question from Number Theory book of Thomas Koshy?

Question

I have a question here from the book I have mentioned above

The question is - There are 4 nos between 100 and 1000, which are equal to the sum of the cubes of its digits. Three of them are 153, 371 and 407. Find the fourth no.

The answer is 370.

I still actually didn't get as to how to proceed to solve the problem. Can anyone explain the solution to me.

I found a pattern here - In each of the nos, the transition of the nos from one to the next, results in the change of only one digit

153
} 7
371
} 1
370
} 4
407

But I need a more plausible explanation as to how I can go about solving this and not use mere brute force!

Answer 1

Such numbers are called Narkisis numbers.
153 -- 1^3 + 5^3 + 3^3
370 -- 3^3 + 7^3 + 0 ^3
371 -- 3^3) + 7^3 + 1^3
407 -- 4^3 + 0^3 + 7^3

Some other examples
1634 = 1^4 + 6^4 + 3^4 + 4^4
4150 = 4^4 +1^4 +5^4 + 0^4

1741725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7

Answer 2

Suppose the digits are a,b,c in decreasing order. The conditions tell us that a^3 + b^3 + c^3 = 100a + 10b + c. Let us rewrite the equation in this form:

(b^3 - 10b) + (c^3 - c) = 100a - a^3.

Let us build a table of values of x^3 - 10x, x^3 - x, and 100x - x^3 for x = 0, 1, ... 9.

x______|_0_|_1_|_2_|_3_|_4_|_5_|_6_|_7_|_8_|_9_|
x^3____|_0_|_1_|_8_|_27|_64|125|216|343|512|729|
x^3 - 10x|_0_|_-9|_-12|_-3|_24|_75|156|273|432|639|
x^3 - x__|_0_|_0_|_6_|_24|_60|120|210|336|504|720|
100x-x^3|_0_|_99|192|273|336|375|384|357|288|171|

Now let me show you by an example how to use the table. Let us consider the case when a = 3. By the above displayed equation, we must have (b^3 - b) + (c^3 - c) = 273. We look for entries in the third and fourth row of the table whose sum is 273. We find 273 in the third row for x = 7, and both 0 and 1 have entries 0 in the fourth row. Thus, b = 7 and c = 0 or 1. This gives us the solutions 370 and 371.

You should consider the other cases for other values of a, and you will find exactly the numbers he gave you.

Answer 3

This is the Exact answer to ur Ques;

153 -- 1(3) + 5(3) + 3(3)

370 -- 3(3) + 7(3) + 0 (3)

371 -- 3(3) + 7(3) + 1(3)

470 -- 4(3) + 0(3) + 7(3)

Answer 4

you have to use 3 digit number bigger than 100 so it means:
- you must at least one digit equal or greater than 5
0^3=0, 1^3=1, 2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=216, 7^3=343, 8^3=512, 9^3=729
so for one digit you have possible numbers: 5-6-7-8-9
so for:
9 -> you need two other number whose sum equal or less than 271
------> you can use only 6 and below
-------->if you use 6 you have left 55 than just 0,1,2,3 left for use
-------->if you use 5 you have left 146 than just 0,1,2,3,4,5 left for use
-------->if you use 4 you have left 207 than just 0,1,2,3,4,5 left
-------->if you use 3 you have left 244 than just 0,1,2,3,4,5,6 left
-------->if you use 2 you have left 263 than just 0,1,2,3,4,5,6 left
-------->if you use 1 you have left 270 than just 0,1,2,3,4,5,6 left
-------->if you use 0 you have left 271 than just 0,1,2,3,4,5,6 left

İf you go by this method for the below you have a number theorem three,

8-> you need two other number whose equal or sum less than 488
------> you can use only 7 or below
7-> you need two other number whose equal or sum less than 657
6-> you need two other number whose equal or sum less than 784
5->you need two other number whose equal or sum less than 675

And if you do them you will see the repeating point (its 5) which is the intersection of these combinations, and you will see the intersection of all possible ways is some group of numbers, then the combinations of these numbers will give you the solution.

Answer 5

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