Respuesta :
1, 3, 5, 7, 9 are the possible digits for the tens digit. Notice they are all odd.
My reasoning:
When you multiply any 2 numbers together, the last digit of the result will be the last digit of the product of the original last digits (ex. 12*7 = 84: 7*2 = 14 : 4 is last digit)
The last TWO digits of the result will be the same as the last 2 digits of the product of just the last two digits. (ex, 1422*234 = 332,748: 22*48 = 748 : 48 are last two digits )
My reasoning:
When you multiply any 2 numbers together, the last digit of the result will be the last digit of the product of the original last digits (ex. 12*7 = 84: 7*2 = 14 : 4 is last digit)
The last TWO digits of the result will be the same as the last 2 digits of the product of just the last two digits. (ex, 1422*234 = 332,748: 22*48 = 748 : 48 are last two digits )
If is a perfect square ends in 6, the last digit squared must end in 6 as well.
The only single digit numbers squared that end in 6 are: 4*4=16 and 6*6=36
They are the only 2 so the number that is being squared must end in 4 or 6
Let the second last digit be X. X can be any digit.
So the last 2 digits of our number that is being squared are either X6 or X4
That is, the last 2 digits of the number are either (10X+6) or (10X+4),
this will result in X always being an odd number.
The possible values of the tens digit given that the units digit of the perfect square is 6 are; 9, 7, 5, 3, or 1
The reason for the above possible tens digit values are as follows:
The perfect square square formula is presented as follows;
a² + 2·a·b + b² = (a + b)²
Let a in the above formula represent the tens part of the number and let b represent the units part, we have;
When b ≠ 0
a·b = a × b is a multiple of 10, that has 0 as its last digit
a² is a multiple of 10, having a last digit of 0
Therefore, the term that determines the non-zero value of units digit is b².
The single digit numbers that when raised to the power of 2 gives a number with 6 as the units digits are 4 (4² = 16), and 6 (6² = 36)
Therefore, two number, tens = a and units = b, give a perfect square that has the unit digit of 6 only when b = 4 or 6
The possible value of the tens digit are;
0² + 2 × 0 × 4 + 10 = 10; The tens digit = 1
0² + 2 × 0 × 6 + 30 = 10; The tens digit = 3
10² + 2 × 10 × 4 + 10 = 190; The tens digit = 9
Therefore, the tens digit are given by adding 1 to multiples of 8 (2 × 4 = 8), or multiples of 12 (2 × 6 = 12), which are;
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144
Adding 1 to the above multiples of 8 and 12 gives the tens digit as either; 9, 7, 5, 3, or 1
The possible values of the tens digit are; 9, 7, 5, 3, or 1
Learn more about place values here:
https://brainly.com/question/1372231
