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Discuss the problem of representing a number in the power of another number. We are given two numbers, x, and y. We need to tell whether y can be represented in the power of x where each power of x can be used once, for example

Input: x = 4, y = 11 Output: true Explanation: 4^2 - 4^1 - 4^0 = 11 Hence y can be represented in the power of x. Input: x = 2, y = 19 Output: true Explanation: 2^4 + 2^1 + 2^0 =19 Hence y can be represented in the power of x. Input: x = 3, y = 14 Output: false Explanation: 14 can be represented as 3^2 + 3^1 + 3^0 + 3^0 but we cannot use one term of power of x twice.

From examining the example on how 19 is represented in the terms of the power of 2 we can form an equation −

c0(x^0) + c1(x^1) + c2(x^2) + c3(x^3) + … = y ….(1),

Where c0, c1, c2 can be -1, 0, +1 to represent whether (-1)term to be subtracted, (+1)term to be added, (0)term not to be included −

c1(x^1) + c2(x^2) + c3(x^3) + … = y - c0,

Taking x as common,

c1(x^0) + c2(x^1) + c3(x^2) + … = (y - c0)/x ….(2),

From eq (1) and (2) we can represent the number in the way again and for a solution to exist (y - Ci) should be divisible by x and Ci can only contain -1, 0 and +1.

So finally we need to check till y>0 whether [(y-1) % x == 0] or [(y) % x == 0] or [(y+1) % x == 0] or whether a solution does not exist.

#include <bits/stdc++.h> using namespace std; int main(){ int x = 2, y = 19; // checking y divisibility till y>0 while (y>0) { // If y-1 is divisible by x. if ((y - 1) % x == 0) y = (y - 1) / x; // If y is divisible by x. else if (y % x == 0) y = y / x; // If y+1 is divisible by x. else if ((y + 1) % x == 0) y = (y + 1) / x; // If no condition satisfies means // y cannot be represented in terms of power of x. else break; } if(y==0) cout<<"y can be represented in terms of the power of x."; else cout<<"y cannot be represented in terms of the power of x."; return 0; }

y can be represented in terms of the power of x.

In this tutorial, we discussed how to check if the representation of a number is possible in terms of the power of another number. We discussed a simple approach to solve this problem by checking current, preceding, and succeeding number divisibility with y.

We also discussed the C++ program for this problem which we can do with programming languages like C, Java, Python, etc. We hope you find this tutorial helpful.

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