Adhiraj's sum

All perfect square doesn't end with $3$ . (You can prove this by listing out all the squares of integers from $0$ to $9$ ).

$1! = 1$ $1!+2! = 3$ $1! +2! +3! = 9$ $1! +2! + 3! + 4! = 33$ $1! +2! + 3! + 4! +5! = 153$

Note that since $x!$ , where $x \geq 5$ , is a multiple of $10$ , it will have unit digit $0$ . Hence, all sums of factorial including $x!$ $(x\geq 5)$ will have a unit digit of $3$ . Hence, all of these factorial sums (with unit digit $3$ cannot be a perfect square). Thus, the largest $n$ whereby $1! +2! + \ldots + n!$ is a perfect square is $\boxed3$

Let $S_n = 1! + 2! + 3! + \ldots + n!,$ and consider $S_n$ modulo $5.$ In mod $5,$ we have $1! \equiv 1,$ $2! \equiv 2,$ $3! \equiv 1,$ $4! \equiv 4,$ and $n! \equiv 0$ for all other $n \ge 5.$ Thus, if $n \ge 5,$ then

$S_n \equiv 1 + 2 + 1 + 4 + 0 + 0 + \ldots + 0 \equiv 3 \pmod{5}.$

Since no square is congruent to $3 \pmod 5,$ we must have $n \le 4.$ Testing $n = 4, 3, 2, 1,$ we find that $n = \boxed{3}$ is the largest working value. $\square$

The question demands the summation of factorials of first n natural numbers. Note that for all $n \geq 5$ , its factorial have a zero at units place. Therefore, finding the summation for $n=4$ may yield us some pattern or method to approach the problem as for all $n \geq 5$ the units place will have same digit as that for $n=4$ .

The n=4 value gives a value of 33 which means that 3 will be the units place for every greater value of n. Therefore no $n \geq 4$ will be a perfect square ( no perfect square has 3 as its unit place) Therefore the answer is less than 4 which on inspection comes out to be 3.

It is easy to see that $1! + 2! + 3! = 9$ , which is a perfect square as $9 = 3^{2}$ .

Now we prove that $1! + 2! + 3! + ... + n!$ , with $n \ge 3$ , is not a perfect square.

$1! + 2! + 3! + 4! = 9 + 24 = 33$ , which is obviously not a perfect square as no perfect squares end in $3$ .

Now, as $5!$ = 120, any $n!$ with $n \ge 5$ will end in $0$ .

Hence, $1! + 2! + 3! + ... + n!$ will always end in $3$ when $n \ge 5$ and is not a perfect square.

Ans: $\boxed {n = 3}$ .

Observe that 5! ends with a zero. 1!+2!+3!+4!=33. No further addition can change the unit's digit. No perfect square has unit-digit 3. So, 1 & 3 are the only possible choices. Larger is 3, hence 3 is answer !

Note $3! + 2! + 1! = 9$ clearly statisfies. We will prove no greater $n$ satisfies. We see $4! + 3! + 2! + 1! = 3$ doesn't satisfy. We also see $33 \equiv 3 \pmod 5$ . Additionally, for $n \geq 5$ , $n!$ is a multiple of 5, and hence $n! \equiv 0 \pmod 5$ . This gives us $1! + 2! + 3! + \cdots + n! \equiv 3 \pmod 5$ , but 3 isn't a quadratic residue modulo 5, and hence that is never a square.

This is a pretty cool problem. If you're looking for a bigger challenge, try finding the largest $n$ such that $1! + 2! + \cdots + n! = a^b$ for some positive integers $a$ and $b$ and $b > 1$ . It is actually not that difficult, and solving this problem is the first step to that other one.

Clearly n= 1 holds. Now, note that a perfect square can only end in 0, 1, 4, 5, 6 and 9.

We observe that from 5!, n! (n > = 5) will have last digit 0. So the last digit of entire sum is determined by 1! + 2! + 3! + 4! which comes out to be 3. But this definitely isn't a perfect square.

But 1! + 2! + 3! = 9 which satisfies the condition.

Hence answer is 3.

lets assume initially that n>=5. then the required sum of factorials of the noumbers will be having 3 as the unit digit (as the unit digit of the factorial of a no. greater than 5 is 0) that can"t be a perfect square.so n is obviously less than 5. and perfect squre comes when n=3.

Since m = 0 , 1, 2, 3 or 4 mod 5 by squaring each we can see that m^2 = 0 or 1 or 4 mod 5.

If r >= 5 then r! is divisible by 5 so r! = 0 mod 5.

Then 1! + 2! + 3! + 4! + ...+n! = 1! + 2! + 3! + 4! mod 5 = 1 + 2 + 6 + 24 = 3 mod 5, which is impossible so n <= 3 and 1 + 2! + 3! = 9 a square, so n = 3.

Let n greater than or equal to 4, the sum's last digit will be 3 and a number with last digit as 3 cannot be a square. Now we have to check for n less than or equal to 3. At 3 the sum is equal to 9 which is a square. Hence, n= 3. Note: last digit will be constant after n greater than or equal to 4 as the the numbers after 4! have 0 as their last digit.

All sums for n >3 end up with 3 as the last digit and thus cannot be perfect squares

Note that for $n \geq 4$ , the expression ends with $3$ , and there are no perfect squares ending with $3$ . So, evaluating the first $3$ , we have $n = 1$ or $n = 3$ . As we're looking for the greater, then $n = \boxed {3}$ .

1!=1,2!=2,3!=6,4!=24,5!=120,6!=720,7!=5040........... 1+2+6+24=33.......

from now if you add 33 to 120 i.e., 5!.....33+120 =153

if you add 153 to 720...............................153+720 =873

if you add 873 to 5040.............................873+5040=5913

if you observe all the results have 3 at its units place......a perfect square does not have a 3 in its units place..... so there is no possibility of answer if we go forward....so come back

1+2+6+24=33

1+2+6=9 and this is the largest perfect square in series....

Finding the Factorial and expanding we have 1!=1 2!=2 3!=6 4!=4 5!=120 and so on Therefore after 4! adding the factorial produces unit digit 3 in the number.A number cannot be a perfect square if it contains 2,3,7 and 8 in units digit.Hence the required answer is 1!+2!+3!=9

for n> 3 , unit digit of sum of factorials is always 3 , like 1! +2! +3! +4! = 33 , so only possible perfect square is 1! +2! +3! = 9. A perfect square is possible only when the unit digit is either 1,4,5,6 ,9 or 0.

k^2 = 1, 4, 5,6,9,0 mod 10 for all k integer. but for each N bigger than 3 it's obvious that S= 1+2!+3! +...+n! = 3 mod 10 cases n=1,2,3: n=1: S=1 n=2: S=7 n=3: 1+2!+3! =9 the largest value of n such S is a perfect square is n=3.

for n=1 then 1! = 1 is a perfect square for n=2 then 1! + 2! = 3 is not a perfect square for n=3 then 1! + 2! + 3! = 9 is a perfect square for n=4 then 1! + 2! + 3! + 4! = 33 is not a perfect square we see that for n>= 5 then $1! + 2! + ... + n! \equiv 1! + 2! + 3! + 4! \equiv 3(mod\ 5)$ since for k>=5 then 5|k! but we know that every perfect squares is never congruent 3 mod 5. So the largest value of n is 3

I think that only 1! + 2! + 3! that have perfect square number. So, 1! + 2! + 3! = 1 + 2 + 6 = 9 (3^2). Then, the answer is 3.

Modular Arithmetic The key to this problem was finding the right number to take mod of. I tried $4$ and $8$ but they didn't work. When you have finished reading solution go back and check why. So: All squares are congruent to either $0, 1$ or $-1mod5$ 1!=1 mod 5 2!=2 3!=6=1 4!=24=-1 5!=0 6!=0 . . . The rest will also be 0 mod 5 since there will always be the factor of 5 there.

1!=1 and 1!+2!+3!=9 work upon checking. Now proving that 9 is the highest.

9=-1 mod 5

9+4!=-1-1=-2mod5 We just showed that n!=0mod 5 for n>4 So if we choose to go past 4! the sum will still be -2mod5

But no squares exist that are congruent to -2 mod5.

And so 9 is the highest square and this occurs when n = $\boxed{3}$

Sorry no latex using my phone and it's just too long

One can easily observe that $5!+6!+7!....+n!$ is divisible by 5,for $n \geq 5$ . $1!+2!+3!+4!=33 \equiv 3 (mod5).1!+2!+3!+.....+n! \equiv3(mod5).$ But every integer is $\equiv0,1,4(mod5)$ .Which shows that $n< 5$ .So n=3 works.

Note that perfect squares are always $\equiv 1,2,9,6,5,6,9,4,1,0 \pmod {10}$

(I bashed this since it's not TOO bad).

Also, note that after $5!$ , the units digit of any number factorial is zero, thus the units digit of the SUM of the first $n$ factorials (where $n \ge 5$ of course), will always be $1+2+6 + 4 = 3$ . (these numbers came from the units digit of the first four factorial numbers).

Checking the list, we see that 3 is NOT one of the residues for perfect squares mod 10. This means that we only need to check up to 4!, and we easily see the greatest value of $n$ is $\boxed{3}$ , which generates $1! + 2! +3! \implies 1+2+6 = 9$ , which is a perfect square.

1! + 2! +3! = 9 1! + 2! +3! +4! =33 and r! ends with 0 for r>4 so the sum 1! + 2! +3! +4! +5! +......... will end with 3 which cannot be a square

1+2+3=6+3=9 Nine is a perfect square because 3*3=9 A perfect square is a number that can be expressed as the product of two equal integers.

5! last digt =0 so n=4 or 3 1!+2!+3! =9 and 1!+2!+3!+4! = 33 so n = 3

If we examine the sum for some small n's, we will soon see that every value of n beginning with $n=4$ will result in a sum ending in 3 because every factorial higher than 5 will end in a zero because it contains both a factor of 2 and a factor of 5. There are no perfect squares that end in the number 3, and therefore no sum that ends in 3 can be a perfect square. This means that n must be equal to 3.