Question

Let a, b, c, be positive integers less than 10 such that (100a + 10b + c)2 = (a + b + c)^5", what is the value
of (a x b-c)?

Who can solve it​

Answer 1

Answer:

29 is the answer.....

Step-by-step explanation:

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

EXCELLENT QUESTIONS!

✨✨✨✨Answer:✨✨✨✨

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

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

so, we can say that,we have to find a three digits number, if we square the number,we will get (sum of digits)^ 5 .

with this we can conclude 2 Things;

1. 100² < ( a+b+c)^5 < (999²)

(highest three -digits number is 999)

2. (a+b+c). must be a square....

with 2 conclude we can say that, (a+b+c)

can be :- 1,4 ,9, 16 and 25.

Now , by 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

putting this value in given now:

(100a+10b+c) ² = (a+b+c )^5

(a+b+c ) = (9)^5. ...(here , abc= A three- digits number).

square root both sides.

→ a b c. = √ (9^5)

a b c. = √ { (3)²}^5

a b c. =√ (3)^ 10

a b c. = √ (3^5)²

now, we know that ( √ (a ) ² = a .

a b c. = 3^ 5

a b c. =3* 3* 3* 3* 3*

a b c. = 243.

hence, our three digits number is 243 , where a=2 , b=4 , c=3.

so,

→ ( a²+b²+c²)

→. (2²+4²+3²)

→. (4+16+9)

ANS →. 29

please make as breilient answer..

Thank you........

✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨

Answer 2

Answer:

Hey mate...

Your ans. will be 29.❤️