What is the remainder when p is divided by 5?

Answer D
state 1:sufficient
p=LCM of 5,10 and q=50q
hence when divided by 5.remainder=0
state:2 sufficient
p=50q-4
hence when divided by 5 remainder=1

What is the remainder when p is divided by 5?
St 1: p leaves the same remainder when divided by 5, 10, and q.
St 2: p is 4 less than the LCM of 5, 10, and q.

ANs: E

Promp requires P=? and then reminder of p/5
1) this gives us three equations below

p=5Q1+R
P=10Q2+R
P=qQ3+R

Not sufficient

2) Two provides below equation
LCM of 5,10,q=10q
there fore 10q-4=P

Not sufficeint

together not sufficient aswell.
10Q-4=P

IMO B

What is the remainder when p is divided by 5?

St 1: p leaves the same remainder when divided by 5, 10, and q.
If q=2
n = 10K + 1 (K=integer)
Remainder (n/5)= 1

But if q =5
p=10K+X (X<5)
Remainder (n/5) = X (=0/1/2/3/4)

Not Sufficient

St 2: p is 4 less than the LCM of 5, 10, and q.
LCM/5 Leave remainder =0
(LCM-4)/5 Leave remainder -4 = (5-4) = 1

Sufficient.

Statement 1 :
Case 1 - if q is smallest number, then remainder can be 0, ...., q

Case 2 - if 5 is smallest number, then remainder can be 0,1,2,3 or 4

from this we can't find exact remainder

Statement 2:
Case 1 - q is less than 10, LCM will be >= 10 and multiple of 5. therefore p = LCM - 4 = 5k - 4 or 5m +1. therefore remainder is 1

Case 2 - If q is greater than 10, LCM will be >= q and multiple of 5, therefore p = LCM - 4 = 5k - 4 or 5m +1. therefore remainder is 1

from this we get exact remainder equal to 1

Thus OA is B

Find R(p/5) = ?

From St(1), we dont know value of 'q'

Case-1 : q = 2 or 4
R must be zero so that R(p/2) = R(p/4) = R(p/5) = 0 holds true.

Case-2 : q >= 5
To hold R(p/5) = R(p/q) true, R value must be 1, 2, 3 or 4

Lets say q = 20 & p = 41, R = 1

Considering both cases, we cannot find any specific value of R.

Hence, ST(1) is not sufficient.

From St(2), p = LCM (5, 10 , q) - 4
Hence, for any value of q, LCM is a multiple of 10.

Lets say p = 10(x+1) - 4 = 10x + 6
R(p/5) = R(10x/5) + R(6/5) = 1

Hence, ST(2) is sufficient.

S1: Remainder can be anything between 0 to 4. Not sufficient.

S2: LCM will always be a multiple of 10. So, 4 less than LCM implies that unit digit will always be 6. Thus, remainder will always be 1 when divided by 5. Sufficient.

Ans: B

GMATBusters

There is a discrepancy between the two statements in this question.

As per st.1, the remainder when divided by 10 can be between 0-4

But as per st.2, the remainder when divided by 10 would be 6

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Statement 1: p leaves the same remainder with 5,10 and q.
Numbers that leave the same remainder when divided by 5 and 10 are the numbers which have units places less than 5.
Ex: 1,2,3,4,11,12,13,14,.......91,92,93,94,...... infinite

But we don't have information about "q" so we can't exactly answer the question. Hence Statement 1 is NOT Sufficient

Statement 2:p is 4 more than the LCM of 5, 10, and q.
LCM of a set of numbers in which one of the numbers is 10 will always have Zero"0" in its unit's place
Ex: LCM of 5,10,13 = 130
LCM of 5,10,2= 10
LCM of 5,10,15=60

p is 4 more than LCM of 5, 10, and q. Hence p is some number with units digit as 4. Therefore the remainder when p divided by 5, is 4.

Implies Statement 2 is Sufficient

Answer is B

(1)

For statement 1, the remainder is the same if p is divided by 5, 10, or q. This statement does not give any information as to what is the value of p. Clearly insufficient.

(2) Interesting!

In (2), p = LCM of 5, 10, and q + 4

5 -> 5^1
10 -> 5^1 x 2^1

LCM is equivalent to the unique prime factors raised to the maximum exponent of each unique prime factor in 5, 10, and q

Regardless of the value of q, we know 2 and 5 will be among the unique factors of their LCM.

LCM = 2^1 x 5^2 x q OR 2^(1+x) x 5^(1+y) x some unique factors of q raised to their respective exponents if q is divisible by 2 and 5.

p = 2^(1+x) x 5^(1+y) x a (a is what is left of q when we remove 2^x and 5^y as its prime factors) + 4

The first term of p is clearly divisible by 5. Therefore the remainder will always be equal to 4.

SUFFICIENT

The answer is B

hi Bunuel, could you please help to solve this? thank you.

Statement 1:
Let q=3, implying that p yields the same remainder when divided by 5, 10, or 3.
Case 1: p=31 --> dividing p by 5, 10, or 3 yields a remainder of 1
Case 2: p=32 --> dividing p by 5, 10, or 3 yields a remainder of 2
Since each case yields a different remainder when p is divided by 5, INSUFFICIENT.

Statement 2:
Since p must be equal to 4 more than a multiple of 5, dividing p by 5 will always yield a remainder of 4.
SUFFICIENT.

In statement 2, isn't everyone assuming that q is positive?

If q = -1, then
p = -10-4 = -14
p/5 leaves a remainder of -4

If q = 1
p = 10-4 = 6
p/5 leaves a remainder of 1

Any thoughts?

GMAT problems involving remainders are constrained to NONNEGATIVE INTEGERS.
From the math review in the OG:
If x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y = xq + r and 0 ≤ r < x.

Got it, thank you!

My pleasure.
Also:
The LCM of two or more integers is considered the smallest POSITIVE multiple of the integers.
No value could ever be considered the least common NEGATIVE multiple.
While -10 can be divided cleanly by 5, 10 and -1, so can every value in the following list:
-20, -30, -40, -50...
As the list above illustrates, negative multiples of 5, 10 and -1 can be infinitely small, with the result that there is no LEAST common negative multiple.
Thus, p=-14 is not a valid case in Statement 2.

Statement 1: Let r be the remainder.
\(P = 5k + r\)
\(P = 10k + r\)
\(P = qk + r\)

When P is divided by 5, 10 or q, then the remainder is \(\frac{r}{5}\)

\(\frac{p}{5} = \frac{5k}{5} + \frac{r}{5}\)

\(\frac{p}{5} = \frac{10k}{5} + \frac{r}{5}\)

\(\frac{p}{5} = \frac{qk}{5} + \frac{r}{5}\)

where \(k ≥ 1\)

This only tells us that q is a multiple of 5. Nothing about \(r\).

Insufficient



Statement 2: What can be the LCM of 5, 10, q?
What is the LCM of 5 and 10? 10
What is the LCM of 35 and 10? 70
What is the LCM of 23 and 10? 230
What is the LCM of 17 and 10? 170

Whenever a 10 is there, the LCM will always be in the form \(10x\)
where \(x ≥ 1\)

So, LCM of 5, 10, and q is 10x

We are given that P is 4 more than the LCM of 5, 10, and q.

So, \(P = 10x - 4\)
This equation is very similar to \(P = 10k + r\)

When we divide P by 5, we get

\(\frac{p}{5} = \frac{10k}{5} - \frac{4}{5}\)

There is a remainder of 4 in \(\frac{4}{5}\)

Sufficient


Therefore, B is the answer.

The remainder when dividing by 5 corresponds to the Units Digit of the Dividend.

(Units Digit of Dividend) / 5 = remainder when the Dividend is divided by 5


S1: p divided by 5, 10, and Q all leave the same remainder.

We can find different values of P in the ranges:

(11 - 14) and (20 - 24) etc.


Case 1:

Let P = 21 and Q = 20

P / 10 —— Rem of 1

P / 5 ——- Rem of 1

P / Q = 21 /20 —— Rem of 1

Case 2:
Let P = 22 and Q = 20

P / 10 ——— Rem of 2

P / 5 ———- Rem of 2

P / Q = 22 / 20 ——- Rem of 2

S1 is not sufficient to determine the remainder when P is divided by 5


Statement 2:

The LCM of a number means that each term of the set is evenly divisible into the LCM.

Whatever the LCM is of the 3 Numbers, it must be evenly divisible by 10.

Therefore, the LCM will have at least one trailing zero - the units digit will be 0

If P is -4 less than this value, then the units digit must be 6

Since we have a unique value of P’s units digit, we have a unique value of the remainder when P is divided by 5

S2 is sufficient

B



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