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We can factor each of these numbers.

1! + 2! = 1! ( 1 + 2 ) 

2! + 3! = 2! ( 1+ 3 )

3! + 4! = 3! ( 1 + 4 )

4! + 5! = 4! ( 1 + 5 )

5! + 6! = 5! ( 1 + 6 )

6! + 7! = 6! ( 1 + 7 )

7! + 8! = 7! ( 1 + 8 )

8! + 9! = 8! ( 1 + 9 )

Once again:

1! + 2! = 1! ( 3 ) 

2! + 3! = 2! ( 4 )

3! + 4! = 3! ( 5 )

4! + 5! = 4! ( 6 )

5! + 6! = 5! ( 7 )

6! + 7! = 6! ( 8 )

7! + 8! = 7! ( 9 )

8! + 9! = 8! ( 10 )

We can split the number into two parts, a! * b

The LCM of 1!, 2!, 3!, ... , 8! is 8!.

The LCM of 3, 4, 5, ... , 10 is 2520

The LCM is 2520 * 8! 

a = 2520

b = 8

a + b = 2528

I hope this helped,

Gavin

The lcm of[1!+2!, 2!+3!, 3!+4!, 4!+5!, 5!+6!, 6!+7!, 7!+8!, 8!+9! ] =10! =3,628,800.

3,628,800 =10 x 9! =So, a + b = 10 + 9 =19. OR:

Find the least common multiple:
lcm(3, 8, 30, 144, 840, 5760, 45360, 403200)

Find the prime factorization of each integer:
The prime factorization of 3 is:
3 = 3^1

The prime factorization of 8 is:
8 = 2^3

The prime factorization of 30 is:
30 = 2×3×5

The prime factorization of 144 is:
144 = 2^4×3^2

The prime factorization of 840 is:
840 = 2^3×3×5×7

The prime factorization of 5760 is:
5760 = 2^7×3^2×5

The prime factorization of 45360 is:
45360 = 2^4×3^4×5×7

The prime factorization of 403200 is:
403200 = 2^8×3^2×5^2×7
Find the largest power of each prime factor.
The largest power of 2 that appears in the prime factorizations is 2^8.
The largest power of 3 that appears in the prime factorizations is 3^4.
The largest power of 5 that appears in the prime factorizations is 5^2.
The largest power of 7 that appears in the prime factorizations is 7^1.
Therefore lcm(3, 8, 30, 144, 840, 5760, 45360, 403200) = 2^8×3^4×5^2×7^1 = 3628800:

 lcm(3, 8, 30, 144, 840, 5760, 45360, 403200) = 3628800

The LCM of {3, 8, 30, 144, 840, 5,760, 45,360, 403,200} =3,628,800 = 10!. As above, 10! can be written as: 1 x 10! =a + b! =1 + 10 =11, which is probably the more accurate answer, since the question wants "b" to be "as large as possible".

[Courtesy of Mathematica 11 Home Edition]