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The question asks if (15! + 13) divisible by any of the four choices. I can rewrite this as (a+b)/c = a/c + b/c. Here a = 15!, b=13 and c = one of the four answer choices.

We know 15! will contains all integers from 15 through 1----> so 15! is divisible by all choices.

Now 13 is divisible by only 13 and 1 because it is a prime number.

Therefore, 13 divides both 15! and obviously 13------> HENCE the answer.

hey the 15! = (15*14*13*12*11*10!) so it is clear that 13 is a divisor of 15!, and adding 13 to it makes the quotient increase by 1 thus 13 is a divisor of (15!+13) .

The question asks if (15! + 13) divisible by any of the four choices. I can rewrite this as (a+b)/c = a/c + b/c. Here a = 15!, b=13 and c = one of the four answer choices.

We know 15! will contains all integers from 15 through 1----> so 15! is divisible by all choices.

Now 13 is divisible by only 13 and 1 because it is a prime number.

Therefore, 13 divides both 15! and obviously 13------> HENCE the answer.

While you can use math to explain it simple logic will also suffice.

Talking 15! automatically makes the end result divisible by all the components (so 15! is divisible by 13). So merely adding 13 to this end results guarantees to make it a multiple of 13.

I found it to be really easy (sub 15 sec question) when looked at in that light.

For instance, \(12*13 - 13 = 13*(12 - 1)\) is divisible by 13 (no zeros!). \(12*13*10 - 13 = 13*(12*10 - 1)\) is divisible by 13. \(12*13*100 - 13 = 13*(12*100 - 1)\) is divisible by 13.

\((15! - 13)\) is divisible by 13 because, although obvious, it is a multiple of 13. You can know that after performing the factorization \(13*(\frac{15!}{13} - 1)\)

Is \(254000 + 13\) divisible by 13? _________________

This question is brilliant as it tests the two concepts of remainder theorm -

1. (Multiplication Rule): This rule is partially applicable here, i would still write it for people who dont know it yet. In an expression (a*b*c)/d, if any of the individual members of the numerator is divisible by d, then the whole numerator is divisible by d. In other words, the remainder of the expression in such a case is zero.

2. (Addition Rule): This one is applicable here. Again in an expression (a+b+c)/d, the remainder will be zero if a, b and c, each, when individually divided by d, gives a remainder zero OR when these individual remainder add up to d itself. Lets take an example to understand better. We all know that 27 is divisible by 9. If i were to write 27 as 9+9+9, then each of the 9's leaves a remainder zero. However if i were to write 27 as 21+2+4, then 21, 2 and 4, when divided by 9, will leave a remainder of 3, 2 and 4 respectively. These remainders add up to 9.

Now lets look at the question: 13! + 13 is always going to be divisible by 13. Why? Apply the fundamentals given above. 13! is nothing but 13*12*11*10 ..... *3*2*1. This is divisible by 13 (Use Multiplication rule on 13! part).

13, the second half, is only divisible by 13 & 1 (It's a prime number). All answer choices EXCEPT 13, leave some remainder. ONLY 13 divides both parts of the expression, leaving an overall remainder zero.

If \(n=15!+13\), which of the following is a divisor of \(n\)?

A. 15 B. 14 C. 13 D. 7 E. 2

\(15!=1*2*3*...12*13*14*15\), so \(n=1*2*3*...12*13*14*15+13\) --> factor out 13: \(n=13*(1*2*3*...12*14*15+1)\), hence 13 is definitely a divisor of \(n\).

Factorials can be rewritten as follows: n! = n*(n-1)*(n-2)*(n-3)*...*(n-k)!

Where k can be any number from 1 to n. We can rewrite (15!+13) as follows: 15!+13 = (15*14*13*12!)+13 = 13*(15*14*12!+1)

Therefore, (15!+13) in its final form has a divisor of 13 in it but not 15, 14, 7, or 2.

I'm having a difficult time understanding this explanation. Can anyone please help me understand. Thank you in advance.

(15!= 15x14x13x12x11x.......x2x1) is a multiple of 13 . When you add 13 to a multiple of 13, it is still a multiple of 13 and divisible by 13. So answer is C.