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We are given the expression:
12! * 11! + 11! * 10!

Step 1: Factor common terms
Notice that 11! is common in both terms:
= 11! * (12! + 10!)

Step 2: Simplify inside the parentheses
We know:
12! = 12 × 11!
10! = 10! (as is)

So:
= 11! * (12 × 11! + 10!)
Now factor 10! from both terms inside the parentheses:

12 × 11! = 12 × 11 × 10!
So,
= 11! * [12 × 11 × 10! + 10!]
= 11! * 10! * (12 × 11 + 1)
= 11! * 10! * (132 + 1)
= 11! * 10! * 133

Step 3: Find the greatest prime factor of the expression
We have:
Expression = 11! * 10! * 133
So the prime factors come from:
- 11! (contains primes up to 11)
- 10! (contains primes up to 7)
- 133

Let’s factor 133:
133 = 7 × 19

So the prime factors in the expression include:
{2, 3, 5, 7, 11, 19}

⇒ Greatest prime factor = 19

Answer: (E) 19
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Bunuel
What is the greatest prime factor of 12! * 11! + 11! * 10! ?

(A) 7
(B) 11
(C) 13
(D) 17
(E) 19


­
Let’s first solve the given equation

12! * 11! + 11! * 10!

Taking 11! As common, we get

11! (12! + 10!)

again taking 10! as common, we get

11! * 10! (12*11 + 1)

11! * 10! * 133

11! * 10! * 19 * 7

The. Highest prime factor of the above equation is 19.

Option E
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12! * 11! + 11! * 10!
(11!)[(12!)+(10!)]
(11!)[(12*11*10!)+10!]
(11!*10!)(132+1)
(10!*11!)(133)
(11!*10!)(19*7)
Prime factors (2,3,5,7,11,19)
Greatest prime factor= 19

Answer: E
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12!*11!+11!*10!=12*11!+11*10!*10!=12*11*10!*11*10!+11*10!*10!=10!^2*11*(12*11+1)= 10!^2*11*133

133is divisible by 19 so greatest prime factor is 19.

Answer is E.
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I think in this question, I could not factor 133 right away so I tried 19 and 17 from the option itself.
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