No because it is impossible.
Let mod(x.3) denote the remainder when x is divided by 3. Let n be any integer. Then mod(n,3) = 0,1 or 2.
When mod(n,3) = 0, mod(n2,3) = 0
When mod(n,3) = 1, mod(n2,3) = 1
When mod(n,3) = 2, mod(n2,3) = 4 and, equivalently mod(n2,3) = 1.
So, there are no integers whose squar leaves a remainder of 2 when divided by 3.
506
11.0503
23.25
To find the number, multiply the divisor and quotient, then add the remainder. 9 (divisor) times 6 is 54. 54 plus 7 is 61. The number is 61.
273
If it leaves no remainder when divided by 2 then it is an even number.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
506
35
find the smallest number by which 9408 must be multiplied to get a perfect square/ also find the square of the number
11.0503
127.6
12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.
To perform division with a remainder, divide the dividend (the number being divided) by the divisor (the number you are dividing by) to find the quotient (the whole number result). Multiply the quotient by the divisor, and then subtract this product from the original dividend to find the remainder. The final result can be expressed as: Dividend = (Divisor × Quotient) + Remainder. The remainder must always be less than the divisor.
30/21 = 1 with a remainder of 9
23.25