Positive integer x when divided by 5 leaves a remainder of 2 and posit

Question

Positive integer x when divided by 5 leaves a remainder of 2 and positive integer y when divided by 8 leaves a remainder of 3, if x + y = 136, what is the units digit of x*y?

A. 1
B. 2
C. 3
D. 7
E. 9

Divide 136 into two parts, one of which when divided by 5 leaves a remainder of 2 and the other when divided by 8 leaves a remainder of 3.

ScottTargetTestPrep · Accepted Answer

Solution:

Since, when x is divided by 5, there is a remainder of 2, the units digit of x must be either 2 or 7. Similarly, since, when y is divided by 8, there is a remainder of 3, the units digit of y must be odd.

Since x + y = 136 and the units digit of x is 2, then the units digit of y is 4. However, since we noted that the units of y must be odd, we see that the units digit of x can’t be 2. Therefore, the units digit of x must be 7, and therefore, the units digit of y must be 9. Since the product of the units digits of x and y is 7 * 9 = 63, the units digit of x * y is 3.

Answer: C

bond001 · Answer

Answer C
99+37=136
here is the solution 
x=5x+2
y=8y+3
5x+2+8y+3=136
5x+8y+5=136
5x+8y=131
now I have to find the solution that satisfies this equation
let us start with 
y=8 then 5x=123 no
y=16 then 5x=115(since it is ps we can safely tell that this is the answer gmat will not make answer with different last digit of couse there will be a different answer )
y=16+3=19 
x=115+2=117
therefore 9*7=63 
last digit =3

nisthagupta28 · Answer

X/5, r=2, then x can be 2,7,12,17,22,27,32,37...

(We can observe that the unit digit will be either 2 or 7)

Y/8, r=3, then y can be 3,11,19,27,35,43....

Since x+y= 136.

As x can be 2 or 7.. That means units digit of y needs to be either 4 (_2+_4) = 136. In the present case xy = 8 in the units digit

Scenario 2

If the unit digit of x is 7, then the unit digit for y = 9.

Xy= 9*7 = 63.

We have 2 options for unit digit either 8 or 3. Since 8 is not in the answer list. 
Answer is 3.

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MathRevolution · Answer

x = 5p + 2 and y = 8q + 3

=> x + y = 136

=> 5p + 2 + 8q + 3 = 136

=> 5p + 8q = 131

For p = 23 and q = 2

=> 5(23) + 8(2) = 131

Therefore, x = 5(23) + 2 = 117 and y = 8(2) + 3 = 19

=> x * y = 117 * 9 = 7 * 9 = unit digit 3

Answer C

BrushMyQuant · Answer

Positive integer x when divided by 5 leaves a remainder of 2.

Dividend = Divisor * Quotient + Remainder

=> x = 5*a + 2 (where a is the quotient)
=> x = 5a + 2 ...(1)

Positive integer y when divided by 8 leaves a remainder of 3.

=> y = 8*b + 3 (where b is the quotient)
=> y = 8b + 3 ...(2)

x + y = 136

=> 5a + 2 + 8b + 3 = 136
=> 5a + 8b = 131
=> a =  \frac{131 - 8b}{5}

Only those values of b which will make a an integer will give us the possible values for x and y

Possible values of b are b = 2, a = 23
=> x = 5*23 + 2 = 117
=> y = 8*2 + 3 = 19

=> Units' digit of 117 * 19 = 7 * 9 = 3

So,  Answer will be C 
Hope it helps!

Watch the following video to MASTER Remainders

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

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Positive integer x when divided by 5 leaves a remainder of 2 and posit

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Positive integer x when divided by 5 leaves a remainder of 2 and posit

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