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

Question

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

JeffTargetTestPrep · Accepted Answer

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

puneetj88 · Answer

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

chetan2u · Answer

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

frndrobinster · Answer

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.

gracie · Answer

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

Psk13 · Answer

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

KarishmaB · Answer

When divided by 7, the remainder is 4 (Same as -3) When divided by 4, the remainder is 1 (Same as -3) Common remainder = -3 p = 28m - 3 All numbers of this format will satisfy these conditions. m can take 4 values: 1, 2, 3, 4 Answer (B) Here is a bit about negative remainders and their application: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/

BrushMyQuant · Answer

Given that When p is divided by 7, the remainder is 4. When p is divided by 4, the remainder is 1. And we need to find How many different values of p are less than 120

Theory: Dividend = Divisor*Quotient + Remainder

When p is divided by 7, the remainder is 4

p -> Dividend
7 -> Divisor
a -> Quotient (Assume)
4 -> Remainders
=> p = 7*a + 4 = 7a + 4

When p is divided by 4, the remainder is 1

p -> Dividend
4 -> Divisor
b -> Quotient (Assume)
1 -> Remainders
=> p = 4*b + 1 = 4b + 1

p = 7a + 4 = 4b + 1
=> b =  \frac{7a + 3}{4}

Only those values of "a" which will also give "b" as integer will give us the common values of p
a = 3, 7, 11, 15, 19,...
But a = 19 will give us a value of > 120 for p

=> 4 value of p which are less than 120 are possible

So,  Answer will be B 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

martinfzb · Answer

When time is short, we can list out the value to find some logic.

N 7*n+4 4*m+1
0 4 1
1 11 5
2 18 9
3  25  13
4 32 17
5 39 21
6 46   25 
7  53  29
8 61 33
9 60 37
10 74 41
11 81 45
12 88 49
13 95  53 
while n =3 and m=6, 7*3+4=25=4*6+1 
and while n=7 and m=13, 7*7+4=53=4*13+1
the logix is that 
n increase at a step of 7-3=4, 
m increase at a step of 13-6=7
120 /7 = 17 and 17/4=  4 step at most

bumpbot · Answer

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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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