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]