When divided by 5, x has a remainder of 2 and y has a

x = 5k+2 , y = 5t+1

x+y = 5 (k+t) + 3 thus remainder when divided by 10 depends on (k+t) if k+t are non -ve even integer r = 3 , if +ve odd remainder is 8

from choices 8 is the right choice

Given data can be represented as \(x = 5*p + 2\) and \(y = 5*q + 1\), where p and q are non negative integers.

Hence \(x+y = 5*k + 3\), k is some non negative integer.

Now observe that the possible remainders obtained when 5*k is divided by 10 are 0 and 5, depending on the parity of k. (0 when k is even, 5 when k is odd)

Hence it follows that the possible remainders when 5*k + 3 is divided by 10 are 3 and 8.
8 is in the given options.

Hence correct option is C

Can anybody explain how to use picking numbers strategy for such a problem?

x is of the form 5*p +2 and y is of the form 5*q + 1

Look at the options. All of them are greater than 5.

So choose x and y such that their sum is greater than 5 and less than 10 (because that will simply represent the remainder we are looking for.)

For instance x = 7 and y = 1 can be taken.

Similarly x = 2 and y = 6 can be taken.

Hope this answers your question.

Oh..k..
i was just considering, 7,6 or 2,1
so i was only getting 3 as the remainder for x + y,
thanks for the explanation

hi

i used this method to solve so hope it helps u even

"x divided by 5 gives remainder 2"
=>any multiple of 5 will have units digit 0 or 5 so x would have units digit 2 or 7

and since we are to calculate for remainder when divided by 10 our main concern is the units digit only

"y divided by 5 gives remainder 1"
=> y would have units digit 1 or 6


for units digit of x+y we can take all combinations of x and y
(2,1) units digit 3
(2,6) units digit 8
(7,1) units digit 8
(7,6) units digit 3

so x+y will have units digit either 3 or 8
so x+y wen divided by 10 wud give either 8 or 3 as remainder

given x-2 is divisible by 5 and y-1 is also divisible by 5,
there could be either a even or odd multiple of 5, so

consider all possible varieties we get
if both are even or if both are odd
x+y makes it even
so remainder is 3 => 2+1

but if one is add and other is even then we get

5 +2+1 = 8,

hence solved.

please ping me if u need any further explanation .....

least possible value of x=2
least possible value of y=1
(2+1)/10 gives a remainder of 3--not listed
next least value of y=6
(2+6)/10 gives a remainder of 8
C

We should recall that when dividing by 5, we only need to know the units digit of the divisor to determine the remainder. Thus, when x is divided by 5 and the remainder is 2, the units digit of x must be either 7 or 2. Also, when y is divided by 5 and the remainder is 1, the units digit of y must be either 6 or 1.

Likewise, when dividing by 10, we only need to know the units digit of the divisor to determine the remainder. Let’s now determine some possible sums of x and y.

x + y = 7 + 6 = 13, which has a remainder of 3 when divided by 10.

x + y = 7 + 1 = 8, which has a remainder of 8 when divided by 10.

x + y = 2 + 6 = 8, which has a remainder of 8 when divided by 10.

x + y = 2 + 1 = 3, which has a remainder of 3 when divided by 10.

Thus, the possible remainders when x + y is divided by 10 are 3 and 8. Since only 8 is given in the answer choices, the correct answer is C.

Answer: C

five least values of x=2 7 12 17 22 27
five least values of y= 1 6 11 16 21
note that all x+y values sum to a 3 or 8 units digit
C

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