consider 3 digit number and square each number in that 3 digit numberand add each squar to get the same number

Question

example: 444=4^2+4^2+4^2=16+16+16= 48 not equal to 444
do it in similar way

Answer 1

no such numbers. (up to 999)

however if u try the same with cubes we get 153, 370, 371, 407

these are called Armstrong numbers

Answer 2

For any number of digits, this is only possible for 1 and 0. Why?

consider any number with more than three digits. Call these digits a1,a2,a3,a4, . . ., an and define (a1,a2, . . ., an) to be the number with these digits. Then the maximum of the sum of the squares of the digits occurs when a1, . . ., an all equal 9. At this point we get the sum of the squares equal to n•81. But n•81<10^(n-1) for all n>3, so the sum can not be an n digit number, and thus can not equal the original number.

OK, so let's look at a 3 digit number with digits a,b,c. So the number is (a,b,c)=a•10^2+b•10+c. The sum of the square of the digits can be maximum when a=b=c=9, and this maximum is 3•81=243. It is easy to see that the number ≤243 with the largest sum of the squares of it's digits is 199. The sum is then 2•81+1=163. We now have a new maximum. The largest sum of 3 digit numbers ≤263 is for 159, and this sum is 81+25+1=107. Again, a new maximum, but the largest sum of numbers ≤107 is for 107 and is 49+1=50. Thus there is no 3 digit number with this property.

Let's look at 2 digit numbers:

if the first digit is odd, and the second digit is even, then the number is even, but the sum is an odd number squared + an even number squared=an odd+even=odd. Therefore it can't work in this case. If the first number is an odd number and the second is an odd number, then the number is odd, but the sum is odd+odd=even, thus no number with the first digit odd can work.

Therefore the highest number possible is 89. If 8 is a digit then the other digit can not be greater than √(89-8^2)=√(89-64)=√25=5. Thus 85 is the highest number possible. The sum of 85 is 64+25=89, 84 is 64+16=80<81<82<83, thus 85,84,83,82,81,and 80 obviously don't work. And since no 70's work (because 7 is odd) there new highest number is 69.

If 6 is a digit, the other digit can not be greater than √(69-36)=√33<6. so 65 is the highest number. But the sum of 65 is 36+25=61, so 65,64, 63, 62, 61 don't work. and the sum of 60=36 so there are no numbers greater than 49 that work.

Repeating this process, we find that the highest number is now 45. The sum of 45 is 41, so the highest number is 40 (it's sum is 16, so highest number is 29).

Repeat again, and greatest number is 25. It's sum is 27, 24's sum is18<20. Thus no 2 digit numbers work.

If a one digit number works then the number is x and x^2=x, so x^2-x=x•(x-1)=0, so x=0 or x=1.

Answer 3

i think no possibility.