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Intern  B
Joined: 28 Mar 2018
Posts: 44
Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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10 00:00

Difficulty:   65% (hard)

Question Stats: 47% (01:59) correct 53% (02:31) wrong based on 58 sessions

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Find the remainder when $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ is divided by 1152, if N is 1153?

A. 10
B. 41
C. 540
D. 576
E. 1140

Very similar to this this question, with a small trick
Math Expert V
Joined: 02 Aug 2009
Posts: 8753
Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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2
tam87 wrote:
Find the remainder when $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ is divided by 1152, if N is 1153?

A. 10
B. 41
C. 540
D. 576
E. 1140

Very similar to this this question, with a small trick

All terms in $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ except 1! will be EVEN, so O+E+E+.......+E=O

So when we divide an Odd number by an even number 1152, we will get an odd number as the remainder.

Only 41 is the odd number, so B.

But have you checked if 41 will come as remainder or just added one random ODD choice... . kidding. Good question but could have added some more odd choices.
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DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1994
Location: India
Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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1
3
$$1152 = 2^7*3^2$$

Hence $$(6!)^2 = [2^4*3^2*5]^2 = 2^8*3^4*5^2$$ and higher terms are divisible by 1152.

Rem $$\frac{(5!)^2}{1152}$$

$$= Rem \frac{(2^3*3*5)^2}{1152}$$

$$= Rem \frac{(2^6*3^2*5^2)}{2^7*3^2}$$

$$= 2^6*3^2 Rem\frac{5^2}{2 }$$

$$= 2^6*3^2 = 576 = 1152/2 = (4!)^2$$

Hence $$(4!)^2 + (5!)^2$$ is divisible by 1152.

the remainder when $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ is divided by 1152, if N is 1153 =$$(1!)^2 + (2!)^2 + (3!)^2 = 41$$

tam87 wrote:
Find the remainder when $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ is divided by 1152, if N is 1153?

A. 10
B. 41
C. 540
D. 576
E. 1140

Very similar to this this question, with a small trick
DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1994
Location: India
Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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1
Good one chetan2u wrote:
tam87 wrote:
Find the remainder when $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ is divided by 1152, if N is 1153?

A. 10
B. 41
C. 540
D. 576
E. 1140

Very similar to this this question, with a small trick

All terms in $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ except 1! will be EVEN, so O+E+E+.......+E=O

So when we divide an Odd number by an even number 1152, we will get an odd number as the remainder.

Only 41 is the odd number, so B.

But have you checked if 41 will come as remainder or just added one random ODD choice.. . kidding. Good question but could have added some more odd choices.
Manager  G
Joined: 24 Sep 2014
Posts: 73
Concentration: General Management, Technology
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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1
Find the remainder when $$(1!)^2+(2!)^2+(3!)^2+...+(N!)^2$$ is divided by 1152, if N is 1153?
$$1152 = 2^7*3^2$$
Now, $$\frac{(1!)^2+(2!)^2+(3!)^2+...+(N!)^2}{(2^7x3^2)}$$
$$=\frac{1+2^2+2^2*3^2+2^6*3^2+2^6*3^2*5^2+.......+(1153)^2}{(2^7*3^2)}$$
adding $$(4!)^2+(5!)^2 = 2^6*3^2+2^6*3^2*5^2 = 2^7*3^2*13$$ (which is divisible by 1152), and all other terms from $$(6!)^2 to (1153!)^2$$ are divisible by 1152
$$= \frac{1+2^2+2^2*3^2 + (multiple of 1152)}{(1152)}$$
so, the remainder is $$1+2^2+2^2*3^2 = 41$$
Intern  B
Joined: 28 Mar 2018
Posts: 44
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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1
chetan2u wrote:

All terms in $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ except 1! will be EVEN, so O+E+E+.......+E=O

So when we divide an Odd number by an even number 1152, we will get an odd number as the remainder.

Only 41 is the odd number, so B.

But have you checked if 41 will come as remainder or just added one random ODD choice... . kidding. Good question but could have added some more odd choices.

The answer 41 is genuine, as confirmed by others...but this method is probably the best one - using the options here with multiple odd choices, there is no other way other than solving it fully, I guess
Intern  B
Joined: 18 Jun 2018
Posts: 27
Location: India
Schools: ISB
WE: Corporate Finance (Retail Banking)
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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chetan2u wrote:
tam87 wrote:
Find the remainder when $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ is divided by 1152, if N is 1153?

A. 10
B. 41
C. 540
D. 576
E. 1140

Very similar to this this question, with a small trick

All terms in $$(1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2$$ except 1! will be EVEN, so O+E+E+.......+E=O

So when we divide an Odd number by an even number 1152, we will get an odd number as the remainder.

Only 41 is the odd number, so B.

But have you checked if 41 will come as remainder or just added one random ODD choice... . kidding. Good question but could have added some more odd choices.

Amazing technique and presence of mind
Intern  B
Joined: 17 Jul 2018
Posts: 1
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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kris19 wrote:
Find the remainder when $$(1!)^2+(2!)^2+(3!)^2+...+(N!)^2$$ is divided by 1152, if N is 1153?
$$1152 = 2^7*3^2$$
Now, $$\frac{(1!)^2+(2!)^2+(3!)^2+...+(N!)^2}{(2^7x3^2)}$$
$$=\frac{1+2^2+2^2*3^2+2^6*3^2+2^6*3^2*5^2+.......+(1153)^2}{(2^7*3^2)}$$
adding $$(4!)^2+(5!)^2 = 2^6*3^2+2^6*3^2*5^2 = 2^7*3^2*13$$ (which is divisible by 1152), and all other terms from $$(6!)^2 to (1153!)^2$$ are divisible by 1152
$$= \frac{1+2^2+2^2*3^2 + (multiple of 1152)}{(1152)}$$
so, the remainder is $$1+2^2+2^2*3^2 = 41$$

Hi Kris19, could you please explain as to how did (4!)^2 + (5!)^2 become divisible by 1152, could you please elaborate on the additions of (4!)^2 + (5!)^2
Manager  G
Joined: 24 Sep 2014
Posts: 73
Concentration: General Management, Technology
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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varunkhosla747 wrote:
kris19 wrote:
Find the remainder when $$(1!)^2+(2!)^2+(3!)^2+...+(N!)^2$$ is divided by 1152, if N is 1153?
$$1152 = 2^7*3^2$$
Now, $$\frac{(1!)^2+(2!)^2+(3!)^2+...+(N!)^2}{(2^7x3^2)}$$
$$=\frac{1+2^2+2^2*3^2+2^6*3^2+2^6*3^2*5^2+.......+(1153)^2}{(2^7*3^2)}$$
adding $$(4!)^2+(5!)^2 = 2^6*3^2+2^6*3^2*5^2 = 2^7*3^2*13$$ (which is divisible by 1152), and all other terms from $$(6!)^2 to (1153!)^2$$ are divisible by 1152
$$= \frac{1+2^2+2^2*3^2 + (multiple of 1152)}{(1152)}$$
so, the remainder is $$1+2^2+2^2*3^2 = 41$$

Hi Kris19, could you please explain as to how did (4!)^2 + (5!)^2 become divisible by 1152, could you please elaborate on the additions of (4!)^2 + (5!)^2

I expanded:
$$4! = 4*3*2*1 = 2^3*3^1$$
$$(4!)^2 = (2^3*3^1)^2$$
$$(4!)^2 = 2^6*3^2$$

$$5! = 5*4*3*2*1 = 2^3*3^1*5^1$$
$$(5!)^2 = (2^3*3^1*5^1)^2$$
$$(5!)^2 = 2^6*3^2*5^2$$

Now, $$(4!)^2 + (5!)^2 = 2^6*3^2 + 2^6*3^2*5^2$$
$$= 2^6*3^2*(1 + 5^2)$$
$$= 2^6*3^2*(1+25)$$
$$= 2^6*3^2*26$$
$$= 2^6*3^2*(2*13)$$
$$= 2^7*3^2*13$$, which is divisible by 1152, I hope this helps you. Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2   [#permalink] 15 Jun 2020, 06:25

# Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  