Investigate the relationship between the digit sum of a number and the digit sum of its cube. | Step-by-Step Solution

Solving This Fascinating Number Theory Problem! 🔢

What We're Solving: We need to determine if there exists a positive integer n where the sum of its digits equals 100, AND the sum of the digits of n³ equals 100³ = 1,000,000.

The Approach: This is a beautiful problem that combines digit sums with modular arithmetic! We'll use a powerful tool called the "digital root" property and modular arithmetic to investigate whether such a number can exist. The key insight is that digit sums have special properties when we work modulo 9.

Step-by-Step Solution:

Step 1: Understanding the Digital Root Property Here's a crucial fact: For any positive integer, the sum of its digits has the same remainder when divided by 9 as the original number itself!

For example:

• 23 ≡ 5 (mod 9), and its digit sum is 2+3 = 5 ≡ 5 (mod 9)

Step 2: Apply This Property to Our Problem If n has digit sum 100, then:

• n ≡ 100 (mod 9)
• Since 100 = 11×9 + 1, we have 100 ≡ 1 (mod 9)
• Therefore: n ≡ 1 (mod 9)

Step 3: What About n³? If n ≡ 1 (mod 9), then:

• n³ ≡ 1³ ≡ 1 (mod 9)

By our digital root property, this means the sum of digits of n³ must also be congruent to 1 modulo 9.

Step 4: Check Our Required Condition We need the sum of digits of n³ to equal 1,000,000. Let's check: 1,000,000 ≡ ? (mod 9)

1,000,000 = 111,111 × 9 + 1 So 1,000,000 ≡ 1 (mod 9) ✓

Step 5: Consider the Magnitudes If n has digit sum 100, the smallest such n would be something like 199999999999999... (nineteen 9's after the 1), and the largest would be much bigger.

But here's the key insight: even for very large values of n with digit sum 100, when we cube them, the resulting n³ would need to have a digit sum of exactly 1,000,000. This is an enormous constraint!

Step 6: The Magnitude Problem Consider that if n ≈ 10^k for some k, then n³ ≈ 10^(3k). The maximum possible digit sum for a number around 10^(3k) would be roughly 9×(3k+1). For this to equal 1,000,000, we'd need k to be enormous, but then n would be so large that having a digit sum of only 100 becomes practically impossible.

The Answer: While the modular arithmetic is consistent (both conditions give remainder 1 when divided by 9), the magnitude constraints make this impossible. The number n would need to be simultaneously small enough to have digit sum 100, yet when cubed, produce a number whose digits sum to the enormous value of 1,000,000. This creates an impossible situation.

Therefore, NO such positive integer n exists.

Memory Tip: Remember the "Rule of 9s" - any number and the sum of its digits have the same remainder when divided by 9! This is your first check for digit sum problems. But don't forget to also consider whether the magnitudes make sense - sometimes the arithmetic works but the scale doesn't! 🎯

Great job working through this challenging problem - you've just used some graduate-level number theory concepts!