Question

let n be a 3 digit number such that n=sum of the square of the digit of n the number of such n is​

Answer 1

Answer:

As the last digit of n^2 is 4 , then last digit of n is either 2 or 8 .

If we consider the last digit as 2 , then the digit before 4 can't be a odd number (as 5). It must be a even number ( it is a sum of the two same numbers, you can check this by taking any three digit ending with 2)

If we consider that last digit is 8 , then also the digit before 4 is a even number , as it is sun the of two same numbers + 6 .

So there is no value for n

Step-by-step explanation:

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Answer 2

Answer:

Assume that

$n = 100a + 10b + c$

is the desired integer if it exists.

$n = {a}^{2} + {b}^{2} + {c}^{2}$

Since

a., b, c <10, it implies that

${a}^{2} + {b}^{2} + {c}^{2} < 300$

Thus n<300

This implies that a<3. Since n is a three digit number, thus a>0

So a=1 or a=2

Even if b and c have maximum possible of 9, a2+b2+c2<200.

Thus a=1.

Thus

100+10b+c=b2+c2+1

99+(10b−b2)=c2−c

Since b and c are integers between 0 and 9 (inclusive), we know that 10b−b2≥1

Thus the LHS ≥100

Obviously RHS is less than c2, which means that RHS <81.

This leads to a contradiction.

Thus the desired integer n does not exist.

Step-by-step explanation:

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