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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

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


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