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Question

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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Solution

100r=0(r2+4r+4)(r+1)!=100r=0(r+2)(r+2)!=100r=0(r+31)(r+2)!=100r=0(r+3)!100r=0(r+2)!=(3!+4!++103!)(2!+3!++102!)=(103)!2a+b=105
Hence sum of digits of (a+b) is 6

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