Question

There is a real m such that (m + 1)! + (m + 2)! = m!
440. Find the sum of digits of m.​

Answer 1

Answer:

Step-by-step explanation:

Given a number, find sum of its digits. Examples : Input : n = 687 Output : 21 Input : n = 12 Output : 3 ... Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M; Number formed by deleting digits such that sum of the digits becomes even and the number odd; Find smallest number with given number of digits and sum of digits

Answer 2

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$(m + 1)! + (m + 2)! = m! * 440$

$(m + 1) * m! + (m + 2)(m+ 1) * m! = m! * 440$

$(m + 1) + (m +2)(m + 1) = 440$

$m^2 + 4m +437 = 0$

$(m +23)(m - 19) =0$

$m= 19$

$So, sum=9+1=10$

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HOPE IT HELPS!!