If s and k are positive integers, what is the remainder when s is divi

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Competition Mode Question

If s and k are positive integers, what is the remainder when s is divided by 5?

(1)  6!*k = s 
(2)  s > 5!

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eakabuah · Answer

We are given that s and k are positive numbers. We are to determine the remainder when s is divided by 5.

Statement 1: 6!*k=s
sufficient. We can deduce from statement 1 that s is a multiple of 5. Since 6! is a multiple of 5, and k is a positive integer. Hence when s is divided by 5, the remainder is 0.

Statement 2: s>5!
Insufficient. Since s can be 5!+1 and result ina remainder of 1 when divided by 5 and s can be 5!+2 and result in a remainder of 2 when divided by 5.

The answer is A.

GMATinsight · Answer

If s and k are positive integers, what is the remainder when s is divided by 5?

(1) 6!∗k=s6!∗k=s
(2) s>5!

Question: When s is divided by 5 = ?

The unit digit of the number is sufficient to find remainder when number is divided by 5

Statement (1) 6!∗k=s

i.e. s is a multiple of 6! i.e. 720 which is divisible by 5
i.e. s when divided by 5 leaves remainder 0

SUFFICIENT

Statement (2) s>5!

s could be 121 which leaves remainder 1 when divided by 5 OR
s could be 122 which leaves remainder 2 when divided by 5 and so on hence

NOT SUFFICIENT

ANSWER: OPTION A

freedom128 · Answer

Known: s and k are positive integers

Q. what is the remainder when s is divided by 5?

1. s=6!*k =6*5*4*3*2*k  
Since 5 is a factor of integer s, the the remainder when s is divided by 5 is zero
SUFFICIENT

2. s>5! 
If s=5!+1, then the the remainder when s is divided by 5 is 1.
If s=5!+5, then the the remainder when s is divided by 5 is zero.
We don't know which one. 
NOT SUFFICIENT

FINAL ANSWER IS (A)

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ArunSharma12 · Answer

If s and k are positive integers, what is the remainder when s is divided by 5?

(1) 6!∗k=s
(2) s>5!

Given: s,k > 0

statement 1:
s = 6!*k = 5(x)
s is a multiple of 5. Remainder = 0
sufficient

statement 2:
s > 5!
s > 120
S can be  , remainder will change depending on the value of s.
not sufficient

Ans: A

exc4libur · Answer

(1) sufic

6!k/5, remainder is 0, since 6! is a multiple of 5.

(2) insufic

s=121 remainder 121/5 is 1
s=125 remainder 121/5 is 0

Ans (A)

unraveled · Answer

If s and k are positive integers, what is the remainder when s is divided by 5?

(1) 6!∗k=s
 \frac{s}{5} = \frac{{6!∗k}}{5} = \frac{6*5*4*3*2*1*k}{5} 
Remainder will always be '0' since 6! is always a multiple of 5.

SUFFICIENT.

(2) s>5!
If s = 5! + 1 = 121 then NO
If s = 6! = 720 YES

INSUFFICIENT.

Answer A.

randomavoidplease · Answer

E Since there can be multiple remainders I guess.

minustark · Answer

If s and k are positive integers, what is the remainder when s is divided by 5?

(1) 6!∗k=s
(2) s>5!

1) Since 6! contains 10 and is divisible by 5, s will always be divisible by 5. For instance, when k is 1, s = 120*6 = 720, divisible by 5. k= 4, s= 720*4 = 2880, also divisible by 5. Sufficient.

2) 5! is 120, which is divisible by 5. But we don't know the exact value of s. It can be 121 or 123, in which case the unit digit is the remainder. Again, for values like 130, 135, there is no remainder. Not sufficient.
E is the answer.

monikakumar · Answer

If s and k are positive integers, what is the remainder when s is divided by 5?

(1) 6!∗k=s6!∗k=s
(2) s>5!

1) 6! has 5 in it, so multiplied by anyother positive number will also be divisible by 5
2) 5!=120, s>120 implied it can 121 or whatever, so may or may not be divisible by 5

Ans A

adstudy · Answer

Prompt says, s and k are positive integers, that is s, k > 0 and are integers

We need to find remainder when s is divided by 5?

Statement 1  -  6! ∗ k = s

Here 6! has 5 and 2 in it. Hence will having a ZERO at the end of it.

If any integer, k, is multiplied also with it, we will still get a ZERO at the end.

This means, s = 6! * k = ........0

Hence when s is divided by 5, we will get remainder as 0.

Hence Statement 1 is sufficient

Statement 2  -  s > 5!

Here it just says s > 5!,

Now we know 5! = 120 (5 * 4 * 3 * 2 * 1),

so 's' can be anything greater than 120. It can be anything out of 121, 1022, 10023, 10025 ,...

Each number will have a different remainder with 5,

Hence Statement 2 is Insufficient

Hence answer is A, IMO

adstudy · Answer

Prompt says, s and k are positive integers, that is s, k > 0 and are integers

We need to find remainder when s is divided by 5?

Statement 1  -  6! ∗ k = s

Here 6! has 5 and 2 in it. Hence will having a ZERO at the end of it.

If any integer, k, is multiplied also with it, we will still get a ZERO at the end.

This means, s = 6! * k = ........0

Hence when s is divided by 5, we will get remainder as 0.

Hence Statement 1 is sufficient

Statement 2  -  s > 5!

Here it just says s > 5!,

Now we know 5! = 120 (5 * 4 * 3 * 2 * 1),

so 's' can be anything greater than 120. It can be anything out of 121, 1022, 10023, 10025 ,...

Each number will have a different remainder with 5,

Hence Statement 2 is Insufficient

Hence answer is A, IMO

firas92 · Answer

(1) 6!*k=s

So s=k*6*5*4! which means s is divisible by 5 and so reminder is 0

Sufficient

(2) s>5!

If s=5!+1, remainder is 1
If s=10!, remainder is 0

Not sufficient

Answer is (A)

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Kinshook · Answer

Asked: If s and k are positive integers, what is the remainder when s is divided by 5?

(1)  6!*k = s 
Since 5 is a factor in 6!
s is a multiple of 5
The remainder when s is divided by 5 = 0
SUFFICIENT

(2)  s > 5! 
If s = 5! + 1 ; s is not divisible by 5; remainder = 1
But if s = 5! + 5; s is divisible by 5; remainder = 0
s may or may not be a multiple of 5
NOT SUFFICIENT

IMO A

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If s and k are positive integers, what is the remainder when s is divi

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