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x and y are positive integers. When 16x is divided by y, the quotient
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06 Jul 2022, 03:13
06 Jul 2022, 03:13
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x and y are positive integers. When 16x is divided by y, the quotient is x, and the remainder is 4. What is the sum of all possible y-values?
A) 7
B) 12
C) 19
D) 26
E) 41
A) 7
B) 12
C) 19
D) 26
E) 41
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x and y are positive integers. When 16x is divided by y, the quotient
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06 Jul 2022, 07:20
06 Jul 2022, 07:20
1
Carcass wrote:
x and y are positive integers. When 16x is divided by y, the quotient is x, and the remainder is 4. What is the sum of all possible y-values?
A) 7
B) 12
C) 19
D) 26
E) 41
A) 7
B) 12
C) 19
D) 26
E) 41
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials (2022)
GRE Geometry Formulas for a Q170
GRE - Math Book
There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
------NOW ONTO THE QUESTION------------------------------
When 16x is divided by y, the quotient is x, and the remainder is 4.
Applying the above rule, we can write: 16x = (y)(x) + 4
Subtract xy from both sides: 16x - xy = 4
Factor: x(16 - y) = 4
Since x and (16 - y) are both positive integers, there are 3 possible solutions:
#1: x = 1 and (16 - y) = 4, in which case y = 12
#2: x = 2 and (16 - y) = 2, in which case y = 14
#3: x = 4 and (16 - y) = 1, in which case y = 15
What is the sum of all possible y-values?
SUM = 12 + 14 + 15
= 41
Answer: E
RELATED VIDEO FROM OUR COURSE
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Re: x and y are positive integers. When 16x is divided by y, the quotient
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03 Oct 2022, 02:59
03 Oct 2022, 02:59
1
When 16x is divided by y, the quotient is x, and the remainder is 4
When a positive integer m is divided by 17, the remainder is 3
Theory: Dividend = Divisor*Quotient + Remainder
16x -> Dividend
y -> Divisor
x -> Quotient
4 -> Remainders
=> 16x = y*x + 4 = xy + 4
=> y = \(\frac{16x - 4}{x}\) = \(\frac{16x}{x} - \frac{4}{x}\) = \(16 - \frac{4}{x}\)
Since y is an integer => \(\frac{4}{x}\) will be an integer
=> 4 should be divisible by x
=> x = 1, 2 or 4
x = 1
16x = xy + 4
=> 16*1 = 1*y + 4
=> y = 16 - 4 = 12
x = 2
16x = xy + 4
=> 16*2 = 2*y + 4
=> y = \(\frac{32 - 4}{2}\)= 14
x = 4
16x = xy + 4
=> 16*4 = 4*y + 4
=> y = \(\frac{64 - 4}{4}\)= 15
=> Sum of all values of y = 12 + 14 + 15 = 41
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Remainders
When a positive integer m is divided by 17, the remainder is 3
Theory: Dividend = Divisor*Quotient + Remainder
16x -> Dividend
y -> Divisor
x -> Quotient
4 -> Remainders
=> 16x = y*x + 4 = xy + 4
=> y = \(\frac{16x - 4}{x}\) = \(\frac{16x}{x} - \frac{4}{x}\) = \(16 - \frac{4}{x}\)
Since y is an integer => \(\frac{4}{x}\) will be an integer
=> 4 should be divisible by x
=> x = 1, 2 or 4
x = 1
16x = xy + 4
=> 16*1 = 1*y + 4
=> y = 16 - 4 = 12
x = 2
16x = xy + 4
=> 16*2 = 2*y + 4
=> y = \(\frac{32 - 4}{2}\)= 14
x = 4
16x = xy + 4
=> 16*4 = 4*y + 4
=> y = \(\frac{64 - 4}{4}\)= 15
=> Sum of all values of y = 12 + 14 + 15 = 41
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Remainders
