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

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

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06 May 2020, 13:37
10
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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
Joined: 02 Aug 2009
Posts: 8753
Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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06 May 2020, 18:09
6
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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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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06 May 2020, 18:33
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
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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06 May 2020, 18:35
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
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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06 May 2020, 18:57
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
Joined: 28 Mar 2018
Posts: 44
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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07 May 2020, 01:40
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
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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12 May 2020, 02:40
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
Joined: 17 Jul 2018
Posts: 1
Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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13 May 2020, 06:53
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
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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15 Jun 2020, 06:25
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