basic question

Algebra Level 1

What is the value of $n$ for which the expression $n!+10$ becomes a perfect square?

9 6 20 3

3 solutions

Steven Jim
Dec 19, 2017

Note that a perfect square, when divided by 4, has a remainder of 0 or 1. Thus, $n<4$ .

If $n=1$ , then $n!+10=11$ , which is not a perfect square.

If $n=2$ , then $n!+10=12$ , which is not a perfect square.

If $n=3$ , then $n!+10=16$ , which is a perfect square.

Thus $n=3$ .

Munem Shahriar
Dec 19, 2017

If $n= 3$ $3! +10 = 16 \longrightarrow \text{Perfect square}$

If $n = 6$

$6! + 10 = 730 \longrightarrow \text{Not a perfect square}$

If $n = 9$

$9! +10 = 362890 \longrightarrow \text{Not a perfect square}$

If $n =20$

$\sqrt{20! +10} \approx 1559776268.64 \longrightarrow \text{Not a perfect square}$

Hence $n =\boxed{3}$

How can $20!+10<4$ ?

Besides, how can you be sure if values of $n>4$ will satisfy or not?

Steven Jim - 3 years, 5 months ago

I fixed it. Thanks.

Actually,I just checked the given options.

Munem Shahriar - 3 years, 5 months ago

Oh, I get it. Still, what if this isn't the "multiple answer" type? I'd like to see how you solve it.

Steven Jim - 3 years, 5 months ago

@Steven Jim – You can use the fact that a perfect square ends with 1,4,6,9,25 or 00.

This shows that the $n$ is between $1 \leq n \leq 10$ .

Because If $n \geq 10,$ $n! + 10$ will always end with 10.

So it seems that $n =3$ is the only solution.

Munem Shahriar - 3 years, 5 months ago

@Munem Shahriar – Good one :)

Steven Jim - 3 years, 5 months ago

Sayak Dutta
Dec 19, 2017

if we put 1,2,3,......as the values of n ,then 1!+10=11,2!+10=12,3!+10=16. but again 4!+10=34 which is not a perfect square.Also for n=1,2 the expression is not a perfect square. so the correct answer is 3.

How can you be sure if values of $n>4$ will satisfy or not?

Steven Jim - 3 years, 5 months ago

basic question

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