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tam87
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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 06 May 2020, 14:37
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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85% (hard)

Question Stats:

47% (02:13) correct 53% (02:39) wrong based on 184 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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B
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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 06 May 2020, 19:09
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tam87
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
Show more

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 06 May 2020, 19:33
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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
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
Show more
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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 06 May 2020, 19:35
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Good one :thumbsup:

chetan2u
tam87
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.
Show more
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kris19
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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 06 May 2020, 19:57
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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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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 07 May 2020, 02:40
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chetan2u


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.
Show more

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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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 12 May 2020, 03:40
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chetan2u
tam87
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.
Show more

Amazing technique and presence of mind
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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 13 May 2020, 07:53
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kris19
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
Show more


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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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 15 Jun 2020, 07:25
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varunkhosla747
kris19
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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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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Find the remainder when (1!)^2 + (2!)^2 + (3!)^2 + ... + (N!)^2 18 Aug 2024, 05:14
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