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# When p is divided by 7, the remainder is 4. When p is divided by 4, th

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Joined: 02 Sep 2009
Posts: 50619
When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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05 Nov 2017, 02:06
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55% (hard)

Question Stats:

68% (02:16) correct 32% (02:29) wrong based on 121 sessions

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When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?

(A) 2

(B) 4

(C) 5

(D) 11

(E) 18

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Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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05 Nov 2017, 02:52
ONLY AN ODD NUMBER WILL LEAVE REMAINDER 1 WHEN DIVIDED BY 4, SO WE ACTUALLY WANT ODD NUMBERS WHICH WILL LEAVE REMAINDER 4 WHEN DIVIDED BY 7 AND REMAINDER 1 WHEN DIVIDED BY 4.
25,53,81,109

ANS:B
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Joined: 02 Aug 2009
Posts: 7035
When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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05 Nov 2017, 03:22
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Bunuel wrote:
When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?

(A) 2

(B) 4

(C) 5

(D) 11

(E) 18

Hi..

an algebraic method..

p is divided by 7, the remainder is 4 MEANS $$p = 7x+4$$
p is divided by 4, the remainder is 1 MEANS $$p = 4y+1$$

so $$7x+4=4y+1.......7x+4-1=4y...........7x+3=4y$$
so 7x+3 should be a MULTIPLE of 4 as RHS is 4y..
first value of x as 3 fits in 7x+3=7*3+3=24..
next value of x will be 3+4 and then 3+2*4 and so on.. formula = 3+4n where n is 0,1,2,3.....

the equation as per restriction of <120
$$7x+4<120.......7(3+4n)<120.......21+28n<120.........28n<99..........n<\frac{99}{28}$$.
$$n<3.abc$$
so n can take values from 0 to 3 : 0,1,2,3...... 4 values

or else, after finding first value as 25 keep adding 28 to it till you get >120
B
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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05 Nov 2017, 03:29
2
Let the number be 7a+4 = 4b+1 ( When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1)
Equating we get, 4b-7a = 3
Solving for integral value of a, b ; we get min value of (a, b) = (3, 6)
hence number is 7*3+4 = 25
the smallest number is 25
for other numbers add (LCM of 7 & 4)28 to 25 to get other numbers. hence others nos are 25, 53,81,109,137 and so on...
Numbers smaller than 120 will be 25, 53,81,109.
hence there will be 4 such numbers.
VP
Joined: 07 Dec 2014
Posts: 1114
When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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05 Nov 2017, 13:28
Bunuel wrote:
When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?

(A) 2

(B) 4

(C) 5

(D) 11

(E) 18

let quotients=x and y
p-4=7x
p-1=4y
7x=4y-3
after testing 0,1, and 2 for x,
least possible values for x and y are 3 and 6 respectively
least value of p=25
25+(n-1)(7*4)<120
28n<123
n<4.4
n=4
B
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Posts: 2830
Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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08 Nov 2017, 16:32
Bunuel wrote:
When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?

(A) 2

(B) 4

(C) 5

(D) 11

(E) 18

We have the following two equations:

p = 7Q + 4 where Q is a whole number.

So, p can be values such as 4, 11, 18, 25, …, 7(16) + 4 = 116.

and

p = 4S + 1 where S is a whole number.

So, p can be values such as 1, 5, 13, 14, 21, 25, …, 4(29) + 1 = 117.

We see that 25 is the first matching value of p. Since the LCM of 7 and 4 is 28, the next will be 25 + 28 = 53, and the next will be 53 + 28 = 81, and the final matching value will be 81 + 28 = 109, so there are a total of 4 values of p less than 120 that satisfy the two criteria.

Alternate Solution:

We have

p = 7Q + 4

for some Q and

p = 4S + 1

for some S.

Then, p + 3 = 7Q + 7 = 4S + 4 is divisible by both 7 and 4. Let’s list the numbers less than 120 that are divisible by 7 and 4, i.e., by 28: 0, 28, 56, 84, 112. However, these are values for p + 3; thus, the corresponding values for p are -3, 25, 53, 81, and 109. p can take any value in this list except for -3; therefore, there are four values in total.

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Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th  [#permalink]

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18 Nov 2017, 04:53
JeffTargetTestPrep wrote:
Bunuel wrote:
When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. How many different values of p are less than 120?

(A) 2

(B) 4

(C) 5

(D) 11

(E) 18

We have the following two equations:

p = 7Q + 4 where Q is a whole number.

So, p can be values such as 4, 11, 18, 25, …, 7(16) + 4 = 116.

and

p = 4S + 1 where S is a whole number.

So, p can be values such as 1, 5, 13, 14, 21, 25, …, 4(29) + 1 = 117.

We see that 25 is the first matching value of p. Since the LCM of 7 and 4 is 28, the next will be 25 + 28 = 53, and the next will be 53 + 28 = 81, and the final matching value will be 81 + 28 = 109, so there are a total of 4 values of p less than 120 that satisfy the two criteria.

Alternate Solution:

We have

p = 7Q + 4

for some Q and

p = 4S + 1

for some S.

Then, p + 3 = 7Q + 7 = 4S + 4 is divisible by both 7 and 4. Let’s list the numbers less than 120 that are divisible by 7 and 4, i.e., by 28: 0, 28, 56, 84, 112. However, these are values for p + 3; thus, the corresponding values for p are -3, 25, 53, 81, and 109. p can take any value in this list except for -3; therefore, there are four values in total.

hi
your first method is quite interesting
i want to ask is this some kind of rule/ some basic knowledge (adding the lcm to the first common number to get other possibilities)
it can be applicable to every similar problem
and
i didnt get the hang of your second method
can you please elaborate it a bit
thanku
your
Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th &nbs [#permalink] 18 Nov 2017, 04:53
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