Question

Pls answer this question quickly ........
let a,b,c be positive integers less than 10 such that (100a+10b+c)^2=(a+b+c)^5 what is (a^2+b^2+c^2)?

Answer 1

||✪✪ QUESTION ✪✪||

Let a,b,c be positive integers less than 10 such that (100a+10b+c)^2=(a+b+c)^5 what is (a^2+b^2+c^2) ? [ Excellent Question. ]

|| ✰✰ ANSWER ✰✰ ||

Given that, (100a+10b+c)^2=(a+b+c)^5 ...

Now, we know that, any three - Digit Number can be written in the form of (100a + 10b + c).

So, we can say that, we have To Find a Three - Digit Number , if we Square the Number , we will get, (sum of digits )^5.

with This we can conclude 2 Things :-

⓵ (100)² ≤ (a+b+c)^5 ≤ (999)² [ As Highest Three digit Number is 999].

❷ (a + b + c) must be a Square ..

with 2nd conclusion we can say that, (a + b + c) can be :- 1 , 4 , 9 , 16 and 25 .

Now, By Inequality we can say that, (a + b + c) can be 1 , 4 or 9.

But, with 1st Conclusion { (100)² ≤ (a + b+c)², we can conclude That, (1)^5 and (4)^5 are very Small..

So, We conclude that (a + b + c) = 9.

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Putting This value in Given now :-

➪ (100a+10b+c)²=(a+b+c)^5

➪ (abc)² = (9)^5 { Here, abc = A three - digit Number .}

Square - Root both sides ,

➪ abc = √(9^5)

➪ abc = √((3)²)^5

➪ abc = √(3)^10

➪ abc = √(3^5)²

Now, we know that , √(a)² = a .

➪ abc = 3^5

➪ abc = 3 * 3 * 3 * 3 * 3

➪ abc = 243 .

Hence, our Three Digit Number is 243, where a = 2, b = 4 and c = 3.

___________________________

So,

☞ (a² + b² + c²)

☞ (2² + 4² + 3²)

☞ (4 + 16 + 9)

☙ ☛ 29 (Ans). ☘

Answer 2

Step-by-step explanation:

$\huge\underline\blue{\sf</p><p>Answer}$

Let N = ( 100a+10b+c)² = (a+b+c)^5

For Prime 'P' and positive integer 'n'

___________________________

Let ep(n) denote the highest power of 'P' dividing 'n'

Since 'N' is both a square and a fifth power ep(n) is divisible by both 2 & 5 and hence by 10 for each prime 'p' dividing 'N' .

___________________________

Thus,

N=

${m}^{10}$

and So 100a+10b+c = m^5

While a+b+c= m²

Now,

2^5 < 111 ≤ 100a + 10b+ c ≤ 999 < 3^5

$\large\implies{\sf }$ 100a + 10b+ c = 3^5 = 243

From above we verify that,

(243)² = (3^5)² = (3²)^5 = (2+4+3)^5

So, Therefore a=2, b= 4, c=3

Now putting the values of a , b and c in equation (a²+b²+c²)

$\large\implies{\sf }$(2²+4²+3³) = (4+16+9) = 29 ( Answer)