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Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th
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05 Nov 2017, 03:52
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
When p is divided by 7, the remainder is 4. When p is divided by 4, th
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05 Nov 2017, 04:22
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1
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
_________________
Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th
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05 Nov 2017, 04:29
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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.
When p is divided by 7, the remainder is 4. When p is divided by 4, th
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05 Nov 2017, 14:28
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
Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th
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08 Nov 2017, 17:32
1
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.
Re: When p is divided by 7, the remainder is 4. When p is divided by 4, th
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18 Nov 2017, 05:53
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.
Answer: B
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
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69% (02:18) correct 
