If the value of - r -0100 r 2-4 r -4 r -1 - is a - b- where 0 - b -10- then the sum of digits of a - b- is

∑^100_r=0(r^2+4r+4)(r+1)! = ∑^100_r=0(r+2)(r+2)! =∑^100_r=0(r+3 1)(r+2)! =∑^100_r=0(r+3)! ∑^100_r=0(r+2)! =(3!+4!+⋯ +103!) (2!+3!+⋯ +102!) =(103)! 2 ∴ nb ...

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Standard XI

Mathematics

Permutation: n Different Things Taken All at a Time When All Are Not Different

If the value ...

Question

If the value of $100 \sum r = 0 (r^{2} + 4 r + 4) (r + 1)!$ is $(a)! - b$ , where $0 \leq b < 10$ , then the sum of digits of $a + b$ , is

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Solution

$100 \sum r = 0 (r^{2} + 4 r + 4) (r + 1)! = 100 \sum r = 0 (r + 2) (r + 2)! = 100 \sum r = 0 (r + 3 - 1) (r + 2)! = 100 \sum r = 0 (r + 3)! - 100 \sum r = 0 (r + 2)! = (3! + 4! + \dots + 103!) - (2! + 3! + \dots + 102!) = (103)! - 2 ∴ a + b = 105$
Hence sum of digits of $(a + b)$ is $6$

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If the value of 100∑r=0(r2+4r+4)(r+1)! is (a)!−b, where 0≤b<10, then the sum of digits of a+b, is

100∑r=0(r2+4r+4)(r+1)!=100∑r=0(r+2)(r+2)!=100∑r=0(r+3−1)(r+2)!=100∑r=0(r+3)!−100∑r=0(r+2)!=(3!+4!+⋯+103!)−(2!+3!+⋯+102!)=(103)!−2∴a+b=105 Hence sum of digits of (a+b) is 6

If the value of $100 \sum r = 0 (r^{2} + 4 r + 4) (r + 1)!$ is $(a)! - b$ , where $0 \leq b < 10$ , then the sum of digits of $a + b$ , is

$100 \sum r = 0 (r^{2} + 4 r + 4) (r + 1)! = 100 \sum r = 0 (r + 2) (r + 2)! = 100 \sum r = 0 (r + 3 - 1) (r + 2)! = 100 \sum r = 0 (r + 3)! - 100 \sum r = 0 (r + 2)! = (3! + 4! + \dots + 103!) - (2! + 3! + \dots + 102!) = (103)! - 2 ∴ a + b = 105$
Hence sum of digits of $(a + b)$ is $6$