Question

Let x and y be two numbers, which leave remainders 4 and 5 respectively after dividing them by *7'. Then
what is the remainder when we divide 2x2 + 3y2 by 7?​

Answer 1

Answer:

2

Step-by-step explanation:

When we divide x by 7 then 4 is the reminder.

Let us assume that, x=7m+4 ........ (1)

Again, when we divide y by 7 then 5 is the reminder.

Let us again assume that, y= 7n+5 ........ (2) { Where m and n are two positive integers }

Now, 2x²+3y² =2(7m+4)²+3(7n+5)² {From equations (1) and (2)}

                       =2(49m²+56m+16)+3(49n²+70n+25)

                       =98m²+112m+32+147n²+210n+75

                       =7(14m²+16m+21n²+30n)+107

                       =7(14m²+16m+21n²+30n)+15×7+2

                       =7(14m²+16m+21n²+30n+15)+2

Hence, from the following expression, it is clear that if we divide the expression (2x²+3y²) by 7 then the reminder will be 2. (Answer)