Question

Hey

plz help me out!
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Answer 1

Given a three digit number $XYZ$ such that,

$\longrightarrow XYZ=X!+Y!+Z!$

This implies each of the three $X,\ Y$ and $Z$ is a one digit number.

We see that,

$\longrightarrow n!>1000\quad\forall\,n>6$

like $7!=5040,\ 8!=40320,\ 9!=362880$ and so on.

This implies $0\leq X,\ Y,\ Z\leq 6$ since $XYZ$ is a three digit number.

If anyone among the three is equal to 6, then,

$\longrightarrow XYZ=X!+Y!+Z!\,\geq\,6!=720$

which implies any among the three $X,\ Y$ and $Z,$ other than that one which is equated to 6, must be greater than or equal to 7 (since hundreds digit of 720 is 7).

This contradicts the earlier implication $0\leq X,\ Y,\ Z\leq 6.$

Therefore,

$\longrightarrow0\leq X,\ Y,\ Z\leq 5.$

If we take $X=Y=Z=5,$ we see that,

$\longrightarrow XYZ=555\neq360=5!+5!+5!=X!+Y!+Z!$

If we take $X=Y=Z=4,$ we see that $X!+Y!+Z!=72$ won't be a third digit number. This implies atleast one among $X,\ Y$ and $Z$ should be equal to 5 else $X!+Y!+Z!$ won't be a three digit number.

What if two are equal to 5?

$\longrightarrow5!+5!+Z!=240+Z!$

This condition is not possible else there should exist 255 satisfying the condition but not (there's no possible $n\in\mathbb{N}$ such that $n!=15$).

This implies exactly one among $X,\ Y$ and $Z$ should be equal to 5.

Now we're sure $XYZ$ does not exceed 200 since $2\times4!<100,$ so the hundreds digit of $XYZ,$ i.e., $X,$ should be 1.

Now $XYZ\geq5!+1!=121$ where $X=1$ and one among $Y$ and $Z$ is equal to 5.

Let us assume $XYZ=154.$

$\longrightarrow 154-121=33$

Let us assume $XYZ=151.$

$\longrightarrow 151-121=30$

There's no possible $n\in\mathbb{N}$ such that $30\leq n!\leq33.$

Let us assume $XYZ=145.$

$\longrightarrow 145-121=24=4!$

But in 145, $Y=4.$ So we get that,

$\longrightarrow 145=1!+4!+5!$

Therefore,

• $X=1$

• $Y=4$

• $Z=5$

And so,

$\longrightarrow X+Y+Z=1+4+5$

$\longrightarrow\underline{\underline{X+Y+Z=10}}$

Hence 10 is the answer.