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

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

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New post 06 May 2020, 13:37
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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
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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 06 May 2020, 18:09
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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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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 06 May 2020, 18:33
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\(1152 = 2^7*3^2\)

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

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New post 06 May 2020, 18:35
1
Good one :thumbsup:

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.
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Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 06 May 2020, 18:57
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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 \(1+2^2+2^2*3^2 = 41\)
Answer is B
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Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 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
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Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 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
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Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 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 \(1+2^2+2^2*3^2 = 41\)
Answer is B



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
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Re: Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2  [#permalink]

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New post 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 \(1+2^2+2^2*3^2 = 41\)
Answer is B



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.
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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

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