Let n -1 - 2 - 3 - n for integer n-1 - If p-1-1 -2-2 -3-3 -10-10 - then p-2 when divided by 11 - leaves a remainder of

The correct option is B 1(1 × 1!)+(2 × 2!) + (3 × 3!)+…… (10 × 10!) =(2 1)1! + (3 1)2! + (4 1)3! + …… (11 1)10! =2! 1!+3! 2!+4! 3!+ …… 11! 10! =11! 1 Sum of ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Quantitative Aptitude

Remainder Theorem

Let n !=1 × 2...

Question

Let $n! = 1 \times 2 \times 3 \dots \dots n$ for integer n > 1. If $p = 1.1! + (2.2!) + (3.3!) + \dots \dots + (10.10!)$ , then p + 2 when divided by 11! leaves a remainder of

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 1
$(1 \times 1!) + (2 \times 2!) + (3 \times 3!) + \dots \dots (10 \times 10!)$
$= (2 - 1) 1! + (3 - 1) 2! + (4 - 1) 3! + \dots \dots (11 - 1) 10!$
$= 2! - 1! + 3! - 2! + 4! - 3! + \dots \dots 11! - 10!$
$= 11! - 1$
Sum of the given series, $p = 11! - 1$
So, $p + 2 = 11! - 1 + 2 = 11! + 1$ . Hence, the remainder is 1
So, $\frac{11! + 1}{11}$ , remainder = 1

Suggest Corrections

Similar questions

Q. Let

n! = 1 \times 2 \times 3 \times \dots \dots \times n

for integer n > 1. If

p = 1! + (2 \times 2!) + (3 \times 3!) + \dots . . + (10 \times 10!)

, then p + 2 when divided by 11! leaves a remainder of:

Q. Let

n! = 1 \times 2 \times 3 \times . . . . . . . . . . . \times n

for integer

n > 1

. If

p = 1! + (2 \times 2!) + (3 \times 3!) + . . . . . . . . . . . + (10 \times 10!),

then (p+2) when divided by 11! leaves a remainder of: (CAT 2005)

Q. Let

n! = 1 \times 2 \times 3 \times . . . . . . . . . . . \times n

for integer

n > 1

. If

p = 1! + (2 + 2!) + (3 + 3!) + . . . . . . . . . . . + (10 + 10!),

then

p + 2

when divided by 11! leaves a remainder of (CAT 2005)

Q. Let

p (n) = 4 \cdot 6^{n} + 5^{n + 1}

. When

P (n)

is divided by

20

, leaves the remainder

Q. If

n

leaves remainder 1 when divided by 3, then

n^{3}

also leaves 1 as remainder when divided by 3.

Join BYJU'S Learning Program

Let n!=1×2×3……n for integer n > 1. If p=1.1!+(2.2!)+(3.3!)+……+(10.10!), then p + 2 when divided by 11! leaves a remainder of

The correct option is D 1(1×1!)+(2×2!)+(3×3!)+……(10×10!) =(2−1)1!+(3−1)2!+(4−1)3!+……(11−1)10! =2!−1!+3!−2!+4!−3!+……11!−10! =11!−1 Sum of the given series, p=11!−1 So, p+2=11!−1+2=11!+1. Hence, the remainder is 1 So, 11!+111, remainder = 1

Let $n! = 1 \times 2 \times 3 \dots \dots n$ for integer n > 1. If $p = 1.1! + (2.2!) + (3.3!) + \dots \dots + (10.10!)$ , then p + 2 when divided by 11! leaves a remainder of