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chetan2u
What is the remainder when 25!+5! is divided by 100?
(1) 0
(B) 5
(C) 25-5
(D) 25
(E) 25+5


New question!!!..

25! is divisible by 100 (since it consists of 25x4).
5!=120, which leaves a remainder of 20, when divided by 100.
So remainder 20.
Answer (C).
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25! will be divisible by 100 hence the remainder will be determined by 5!. 5! = 120 hence when divided by 100, remainder will be 20. IMO C.
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When 25! Is divided by 100, remainder would be 0. So all depends on 5! Which is 120. When 120 is divided by 100, remainder is 20. THUS C.

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How do you guys determine that 25! is always divisible by a hundred without doing the drawn out calculation?

Someone enlighten me please :)
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25!+5! / 100 = remainder?
Forget 25! as whenever a number crosses 10!, at least two zeros are added to its tail which makes it divisible by 100 (10! = 10*...*5*...2*1 = 100*...).
So, the actual question is, what will be the remainder if 5! = 120 is divided by 100.
Our answer is 20 i.e. 25-5 (C)
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got logic correct but i did 5! / 100, leaving remainder 1, because 120/100 = 6/5 (is this incorrect, if the answer choice has 1?) can some one please clarify?
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got logic correct but i did 5! / 100, leaving remainder 1, because 120/100 = 6/5 (is this incorrect, if the answer choice has 1?) can some one please clarify?

You left out R/20. The equation becomes 5!=100q+R.when you divide through by 20, you have 6= 5q + R/20.The remainder when 6 is divided by 5 = 1 as (as you rightly indicated). Thus R/20 = 1 Therefore, R=20.

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Asked: What is the remainder when 25!+5! is divided by 100?


Power of 10 in 25! = maximum multiple of 5 = 5+1 = 6
5! = 120
Since 25! will have 6 zeros at the end , it will have remainder 0 when divided by 100.
The remainder when 5!=120 is divided by 100 = 20

IMO C

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25! is exactly divisible by 100 . Since 100 = \(5^2 * 2^2\) and there are more than two 5's and two 2's in 25!.

5! = 120 when divided by 100 will give a remainder 20.

So 25! + 5! will give a remainder 20 when divided by 100.

Option C is the answer.

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Simple logic about the question is

(25!+5!)/100

We can see the remainder separately like, 25!/100 + 5!/100 (Adding the remainder of both)
now
100 = 5^2*2^2
1) 25! consists of 5^6 and 2^22, hence it's definitely divisible by 100 so Remainder by this = 0
2) 5! = 120, when divided by 100 it leaves the remainder = 20

so final Remainder = 20+0 = 20

now, 20/100, Remainder = 20

Answer is C
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What is the remainder when 25!+5! is divided by 100?

Maximum power of 100 in 25! = (5 + 1)/2 = 3
The remainder when 25! is divided by 100 = 0

Maximum power of 5 in 5! = 1
The remainder when 5!=120 is divided by 100 = 20 = 25 - 5

IMO C
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧



Remainder when 25! + 5! by 100 = Remainder of 25! by 100 + Remainder of 5! by 100

Remainder of 25! by 100
25! = 25*24*...*1
=> 25! is a multiple of 100
=> Remainder of 25! by 100 = 0

Remainder of 5! by 100 = Remainder of 5*4*3*2*! by 100 = Remainder of 120 by 100 = 20

=> Remainder of 25! + 5! by 100 = 0 + 20 = 20

So, Answer will be C
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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25! Will always be divisible by 100.

So we have 5! Remaining.. 5!=120.

120÷100=20

Answer: C
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